## Saturday, December 22, 2012

### Axiomatic definition of the center of mass

This  was prompted by a nice Mathoverflow question.    $\newcommand{\bR}{\mathbb{R}}$  $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bp}{{\boldsymbol{p}}}$ $\newcommand{\Div}{\mathrm{Div}}$ $\newcommand{\supp}{\mathrm{supp}}$ $\newcommand{\bm}{\boldsymbol{m}}$ $\newcommand{\eC}{\mathscr{C}}$ $\newcommand{\bc}{\boldsymbol{c}}$ $\newcommand{\bq}{{\boldsymbol{q}}}$

We define an effective divisor   on $\bR^N$ to be a   function with finite support $\mu:\bR^N\to\bZ_{\geq 0}$. Its mass, denoted by $\bm(\mu)$, is the  nonnegative integer

$$\bm(\mu)=\sum_{\bp\in\bR^N} \mu(\bp).$$

We denote by $\Div_+(\bR^N)$ the set of effective divisors.  Note that $\Div_+(\bR^N)$ has a natural structure of Abelian semigroup.

For any $\bp\in\bR^N$ we denote by $\delta_\bp$ the Dirac divisor of mass $1$ and supported at  $\bp$.   The  Dirac divisors generate  the  semigroup $\Div_+(\bR^N)$.     We have a natural  topology on  $\Div_+(\bR^N)$ where $\mu_n\to \mu$ if and only if

$$\bm(\mu_n)\to \bm(\mu),\;\; {\rm dist}\,\bigr(\;\supp(\mu_n),\; \supp(\mu)\;\bigr)\to 0,$$

where ${\rm dist}$ denotes the Haudorff distance.

A center of mass   is a map

$$\eC:\Div_+(\bR^N)\to\Div_+(\bR^N)$$

satisfying the following conditions.

1. (Localization) For any divisor $\mu$ the support of $\eC(\mu)$ consists of  a single point $\bc(\mu)$.

2.  (Conservation of mass)

$$\bm(\mu)=\bm\bigl(\;\eC(\mu)\;\bigr),\;\;\forall\mu \in\Div_+(\bR^N),$$

so that

$$\eC(\mu)=\bm(\mu)\delta_{\bc(\mu)},\;\;\forall\mu \in\Div_+(\bR^N).$$

3. (Normalization)

$$\bc(m\delta_\bp)=\bp,\;\;\bc(\delta_\bp+\delta_\bq)=\frac{1}{2}(\bp+\bq),\;\;\forall \bp,\bq\in\bR^N,\;\;m\in\bZ_{>0}.$$

$$\eC(\mu_1+\mu_2)= \eC\bigl(\,\eC(\mu_1)+\eC(\mu_2)\,\bigr),\;\;\forall \mu_1,\mu_2\in \Div_+(\bR^N).$$

For example, the   correspondence

$$\Div_+ \ni \mu\mapsto \eC_0(\mu)=\bm(\mu)\delta_{\bc_0(\mu)}\in\Div_+,\;\;\bc_0(\mu):=\frac{1}{\bm(\mu)}\sum_\bp \mu(\bp)\bp$$

is a center-of-mass  map.  I want to show that this is the only center of mass map.

Proposition  If $\eC:\Div_+(\bR^N)\to \Div_+(\bR^N)$ is a  center-of-mass map, then $\eC=\eC_0$.

Proof.      We carry the proof in several steps.

Step 1 (Rescaling).   We can write the additivity property as

$$\bc(\mu_1+\mu_2) =\bc\bigl(\; \bm(\mu_1)\delta_{\bc(\mu_1)} +\bm(\mu_2)\delta_{\bc(\mu_2)}\;\bigr).$$

In particular, this implies that the rescaling property

$$\bc( k\mu)=\bc(\mu),\;\;\forall\mu \in\Div_+,\;\; k\in\bZ_{>0}. \tag{R}\label{R}$$

This follows by induction $k$. For $k=1$ it is obviously true.  In general

$$\bc( k\mu)=\bc\bigl(\;(k-1)\bm(\mu)\delta_{\bc(\;(k-1)\mu)}+\bm(\mu)\delta_{\bc(\mu)}\;\bigr) =\bc\bigl(\; k\bm(\mu)\delta_{\bc(\mu)}\;\bigr)={\bc(\mu)}$$

Step 2. (Equidistribution)   For any $n>0$ and any collinear points $\bp_1,\dotsc,\bp_n$ such that

$$|\bp_1-\bp_2|=\cdots=|\bp_{n-1}-\bp_n|$$

we have

$$\eC\Bigl(\sum_{k=1}^n\delta_{\bp_k}\;\Bigr)=\eC_0\Bigl(\sum_{k=1}^n\delta_{\bp_k}\;\Bigr) \tag{E}\label{E}.$$

Equivalently, this means that

$$\bc\Bigl(\sum_{k=1}^n\delta_{\bp_k}\;\Bigr)=\bc_0\bigl(\sum_{k=1}^n\delta_{\bp_k}\;\bigr)={\frac{1}{n}(\bp_1+\cdots+\bp_n)}.$$

We  will prove  (\ref{E}) arguing by induction on $n$. For $n=1,2$ this follows from the normalization property. Assume that (\ref{E}) is true for any $n< m$. We want to prove it is true for $n=m$.

We distinguish two cases.

(a)  $m$ is even, $m= 2m_0$. We set

$$\mu_1=\sum_{j=1}^{m_0} \delta_{\bp_j},\;\;\mu_2=\sum_{j=m_0+1}^{2m_0}\delta_{\bp_j}.$$

Then

$$\bc(\mu_1+\mu_2)= \bc\bigl(\; m_0\delta_{\bc(\mu1)}+m_0\delta{\bc(\mu_2)}\;\bigr) =\bc\bigl( \delta_{\bc(\mu_1)}+\delta_{\bc(\mu_2)}\;\bigr). \tag{1}\label{2}$$

By induction

$$\bc(\mu_1)=\bc_0(\mu_1),\;\;\bc(\mu_2)=\bc_0(\mu_2).$$

The normalization  condition now implies that

$$\bc\bigl( \delta_{\bc(\mu_1)}+\delta_{\bc(\mu_2)}\;\bigr)=\bc_0\bigl( \delta_{\bc_0(\mu_1)}+\delta_{\bc_0(\mu_2)}\;\bigr).$$

Now run the  arguments in (\ref{2}) in reverse, with $\bc$ replaced by $\bc_0$.

(b) $m$ is odd, $m=2m_0+1$.  Define

$$\mu_1=\delta_{\bp_{m_0+1}},\;\;\mu_2'=\sum_{j<m_0+1}\delta_{\bp_j},\;\;\mu_2''=\sum_{j>m_0+1}\delta_{\bp_j},\;\;\mu_2=\mu_2'+\mu_2''.$$

(Observe that $\bp_{m_0+1}$ is the mid-point in the string of equidistant collinear points $\bp_1,\dotsc,\bp_{2m_0+1}$. ) We have

$$\eC(\mu_2'+\mu_2'')=\eC\bigl( \; \eC(\mu_2')+\eC(\mu_2'')\;\bigr).$$

By induction

$$\eC(\mu_2')+\eC(\mu_2'')= \eC_0(\mu_2')+\eC_0(\mu_2'') =m_0\delta_{\bc_0(\mu_2')}+m_0\delta_{\bc_0(\mu_2'')}=m_0\bigl(\;\delta_{\bc_0(\mu_2')}+\delta_{\bc_0(\mu_2'')}\;\bigr).$$

Observing that

$$\frac{1}{2}\bigl(\bc_0(\mu_2')+\bc_0(\mu_2'')\;\bigr)=\bp_{m_0+1}$$

we deduce

$$\eC(\mu_2)= \eC(\mu_2'+\mu_2'')=m_0\eC\bigl( \delta_{\bc_0(\mu_2')}+\delta_{\bc_0(\mu_2'')}\;\bigr)=m_0\eC_0\bigl( \delta_{\bc_0(\mu_2')}+\delta_{\bc_0(\mu_2'')}\;\bigr)=2m_0\delta_{\bp_{m_0+1}}=2m_0\mu_1.$$

Finally  we deduce

$$\eC(\mu)=\eC\bigl(\;\eC(\mu_1)+\eC(\mu_2)\;\bigr)=\eC\bigl(\;\eC(\mu_1)+2m_0\eC(\mu_1)\;\bigr)= (2m_0+1)\delta_{\bp_{m_0+1}}=\eC_0(\mu).$$

Step 3. (Replacement) We will show that for any distinct points $\bq_1,\bq_2$ and any positive integers $m_1,m_2$ we  can find  $(m_1+m_2)$ equidistant points  $\bp_1,\dotsc,\bp_{m_1+m_2}$ on the line determined by $\bq_1$ and $\bq_2$  such that

$$m_1\delta_{\bq_1}=\eC_0\Bigl(\sum_{j=1}^{m_1} \delta_{\bp_j}\Bigr)=\eC\Bigl(\sum_{j=1}^{m_1} \delta_{\bp_j}\Bigr),\;\;\;m_2\delta_{\bq_2}=\eC_0\Bigl(\sum_{j=m_1+1}^{m_1+m_2} \delta_{\bp_j}\Bigr)=\eC\Bigl(\sum_{j=m_1+1}^{m_1+m_2} \delta_{\bp_j}\Bigr).$$

This is elementary. Without  restricting the generality we can assume  that $\bq_1$ and $\bq_2$ lie on an axis (or geodesic) $\bR$ of $\bR^N$, $\bq_0=0$ and $\bq_2=q>0$.       Clearly we can find  real numbers $x_0, r$, $r>0$, such that
$$\frac{1}{m_1}\sum_{j=1}^{m_1}(x_0+j r)=0,\;\;\frac{1}{m_2}\sum_{j=m_1+1}^{m_1+m_2}(x_0+jr)=q.$$

Indeed, the above two equalities can be rewritten as

$$x_0+\frac{m_1+1}{2} r=0,$$

$$q=x_0 +m_1 r+\frac{m_2+1}{2}=x_0+\frac{m_1+1}{2} r+\frac{m_1+m_2}{2} r.$$

Now place the points  $\bp_j$ at the locations $x_0+jr$.

Step 4. (Conclusion)   We argue on  by induction on mass that

$$\eC(\mu)=\eC_0(\mu),\;\;\forall \mu\in \Div_+\tag{2}\label{3}$$

Clearly, the normalization condition  shows that (\ref{3})  is true if $\supp\mu$ consists of a single point, or if  $\bm(\mu)\leq 2$.

In general if $\bm(\mu)>2$ we write $\mu=\mu_1+\mu_2$ where $m_1=\bm(\mu_1),m_2=\bm(\mu_2)<\bm(\mu)$.

By induction we have

$$\eC(\mu)= \eC\bigl( \eC(\mu_1)+\eC(\mu_2)\bigr)=\eC(\;\eC_0(\mu_1)+\eC_0(\mu_2)\;\bigr).$$

If $\bc_0(\mu_1)=\bc_0(\mu_2)$  the divisors $\eC_0(\mu_1)$ $\eC_0(\mu_2)$ are supported at the same point and we are done. Suppose that  $\bq_1=\bc_0(\mu_1)\neq\bc_0(\mu_2)=\bq_2$. By Step 3, we can  find     equidistant points $\bp_1,\dotsc,\bp_{m_1+m_2}$ such that

$$m_1\delta_{\bq_1}=\eC\Bigl(\sum_{j=1}^{m_1} \delta_{\bp_j}\Bigr)= \eC_0\Bigl(\sum_{j=1}^{m_1} \delta_{\bp_j}\Bigr)$$
$$m_2\delta_{\bq_2}=\eC\Bigl(\sum_{j=m_1+1}^{m_1+m_2} \delta_{\bp_j}\Bigr)=\eC_0\Bigl(\sum_{j=m_1+1}^{m_1+m_2} \delta_{\bp_j}\Bigr).$$

We deduce that

$$\eC(\mu)=\eC\Bigl(\sum_{k=1}^{m+1+m_2}\delta_{\bp_k}\Bigr),\;\; \eC_0(\mu)=\eC_0\Bigl(\sum_{k=1}^{m+1+m_2}\delta_{\bp_k}\Bigr).$$

The conclusion now follows from  (\ref{E}).  q.e.d

Remark.    The above proof    does not really use the  linear structure. If we uses only the fact that any two points in $\bR^N$ determine a unique geodesic.  The  Normalization condition can be replaced by the equivalent one

$$\bc(\delta_\bp+\delta_\bq)= \mbox{the midpoint of the geodesic segment [\bp,\bq]}.$$

If we replace $\bR^N$ with a hyperbolic space the same arguments show that  there exists at most  one center of mass map.

## Tuesday, December 18, 2012

### The 11/8-conjecture was resuscitated back to life. Yep, there was a flaw

Apparently Tom Mrowka found a flaw in  Bauer's "proof" of the  11/8-th conjecture.

## Monday, December 17, 2012

### A really nifty linear algebra trick

I've been stuck on this statistics problem for quite a while, I could taste where it was going but I could never put into words my   intuition. Today I discovered how to neatly  bypass the obstacle.  Discovered is not the appropriate word, because someone else  figured it out  long before me.  It's a really, really elementary linear algebra trick  that I have never encountered in my travels. Very likely, more experienced statisticians than myself    would  smile at my ignorance.

The earliest occurrence of this trick I could trace is    in a Russian paper  by  R. N. Belyaev published in  Teoryia Veroyasnosti i eio  Primenenyia, 1966.    $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bx}{\boldsymbol{x}}$ $\newcommand{\by}{\boldsymbol{y}}$

Suppose that  $S$ is an invertible $n\times n$ matrix and $\bx,\by\in\bR^n$ are vectors which I regard as column vectors, i.e., column matrices.  Denote by $(-, -)$ the  natural inner product in  $\bR^n$

$$(\bx,\by)= \bx^\dagger\cdot \by,$$

where ${}^\dagger$ denotes the transpose of a matrix.

Let $r\in\bR$. The name of the game is to compute the scalar

$$r- (\bx, S^{-1} \by)= r-\bx^\dagger\cdot S\cdot \by.$$

Such computations are often required in the neck of the woods where I've been spend the best  part of the last three years namely,  geometric probability. So here is the trick. I'll name it after Belyaev because I am sure he was not the first to observe it. (He even refers to an old book by H. Cramer on statistics.) $\newcommand{\one}{\boldsymbol{1}}$

Belyaev's Trick.

$$r-\bx^\dagger\cdot S\cdot \by =\frac{\det\left[\begin{array}{cc} S &\by\\ \bx^\dagger & r \end{array}\right]}{\det S}=\frac{\det\left[\begin{array}{cc} r &\bx^\dagger\\ \by & S \end{array}\right]}{\det S}.$$

Here is the disappointingly simple proof. Note that

$$\left[ \begin{array}{cc}\one_n & S^{-1}\by\\ 0 & r-\bx^\dagger S^{-1} \by \end{array} \right]= \left[ \begin{array}{cc} \one_n & 0\\-\bx^\dagger & 1 \end{array} \right]\cdot \left[ \begin{array}{cc} S^{-1} & 0\\ 0 & 1 \end{array}\right] \cdot \left[ \begin{array}{cc} S &\by\\ \bx^\dagger & r \end{array}\right].$$

Now take the determinants of  both sides to obtain the first equality.  The second equality  follows  from the first by permuting the rows and columns of the matrix at numerator. $\DeclareMathOperator{\Cov}{\boldsymbol{Cov}}$ $\newcommand{\bsE}{\boldsymbol{E}}$

Here is how it works in practice.   Suppose that   $(X_0, X_1, \dotsc, X_n)\in \bR^{n+1}$ is a centered random  Gaussian, with covariance matrix

$$\Cov(X_0, X_1, \dotsc, X_n)= \Bigl( \;\bsE\bigl( X_i\cdot X_j\,\bigr)\;\Bigr)_{0\leq i,j\leq n}.$$

Assume  that the Gaussian vector $(X_1,\dotsc, X_n)$ is nondegenerate, i.e., the  symmetric matrix  $S=\Cov(X_1,\dotsc, X_n)$ is invertible.

We can then define in an unambiguous way the conditional random variable $\DeclareMathOperator{\var}{\boldsymbol{var}}$

$$(X_0|\; X_1=\cdots =X_n=0).$$

This is a  centered Gaussian random variable with variance given by the   the regression formula

$$\var(X_0|\; X_1=\cdots =X_n=0)= \var(X_0) - \bx^\dagger\cdot S\cdot \bx,$$

where $\bx^\dagger$ is the row vector

$$\bx^\dagger =\left(\; \bsE(X_0X_1),\cdots ,\bsE(X_0 X_n)\;\right).$$

If we now use  Belyaev's trick we deduce

$$\var(X_0|\; X_1=\cdots =X_n=0)=\frac{\det\Cov(X_0, X_1, \dotsc, X_n)}{\det\Cov(X_1,\dotsc, X_n)}.$$

In this form it is used in the related paper of Jack Cuzick (Annals of Probability, 3(1975), 849-858.)

### On a "Car-Talk" problem

While driving back home I heard an interesting math question from all places, the Car Talk show on NPR.  This   made me think of a generalization of the trick they used and in particular, formulate the following problem.  $\newcommand{\bR}{\mathbb{R}}$

Problem  Determine all, reasonably well  behaved compact domains  $D\subset \bR^2$  with the following  property: any line through the origin divides  $D$ into two regions of equal areas.  We will refer to this as property $\boldsymbol{C}$ (for cut).

I know that "reasonably well behaved" is a rather fuzzy   requirement.  At this moment I don't want to think of Cantor like weirdos.  So let's assume that $D$ is semialgebraic.

We say that a domain $D$ satisfies property $\boldsymbol{S}$ (for symmetry) if it is invariant with respect to the  involution

$$\bR^2\ni (x,y)\mapsto (-x,-y) \in \bR^2.$$

It is not hard to see that

$$\boldsymbol{S}\Rightarrow \boldsymbol{C}.$$

Is the converse true?

I describe below one situation when this happens.

A special case.    I'll assume that $D$ is semialgebraic,  star-shaped with respect to the origin and satisfies $\boldsymbol{C}$.  We can  then describe $D$ in polar coordinates by an inequality of the form

$$(r,\theta)\in D \Longleftrightarrow 0\leq f(\theta),\;\;\theta\in[0,2\pi],$$

where $f:[0,2\pi]\to (0,\infty)$ is a semialgebraic function  such that $R(0)=R(2\pi)$. We can extend $f$ by $2\pi$-periodicity to a function  $f:\bR\to [0,\infty)$ whose restriction to any finite interval is semialgebraic.

For any $\phi\in[0,2\pi]$ denote by $\ell_\phi:\bR^2\to \bR$ the linear functiondefined by

$$\ell_\phi (x,y)= x\cos\phi +y\sin\phi.$$

Denote by  $A(\phi)$ the area of the region $D\cap \bigl\{ \ell\phi\geq 0\bigr\}$.    Since $D$ satisfies $\boldsymbol{C}$ we deduce

$$A(\phi)=\frac{1}{2} {\rm area}\;(D),$$

so that

$$A'(\phi)=0,\;\;\forall \phi.$$

Observe that

$$A(\phi+\Delta \phi)-A(\phi) =\int_{\phi+\frac{\pi}{2}}^{\phi+\frac{\pi}{2}+\Delta\phi} \left(\int_0^{f(t)} r dr\right) dt -\int_{\phi+\frac{3\pi}{2}}^{\phi+\frac{3\pi}{2}+\Delta\phi} \left(\int_0^{f(t)} r dr\right) dt.$$

For simplicity we set $\theta=\theta(\phi)=\theta+\frac{\pi}{2}$. We can then rewrite the  above equality  as

$$A(\phi+\Delta \phi)-A(\phi)=\int_\theta^{\theta+\Delta\theta}\left(\int_0^{f(t)} r dr\right) dt -\int_{\theta+\pi}^{\theta+\pi+\Delta\theta} \left(\int_0^{f(t)} r dr\right) dt.$$

Hence

$$0=A'(\phi)= \frac{}{2}\Bigl( f(\theta)^2-f(\theta+\pi)^2\Bigr).$$

Hence $f(\theta)= f(\theta+\pi)$, $\forall \theta$. This shows that   $D$ satisfies the symmetry condition $\boldsymbol{S}$.

$$\ast\ast\ast$$

Here is a simple instance when $\boldsymbol{C}$   does not imply $\boldsymbol{S}$.

Suppose that $D$ is semialgebraic and has the annular description

$$f(\theta)\leq r\leq F(\theta). \tag{1}\label{1}$$.

Using the same notations as above we deduce that

$$0=A'(\phi)= \frac{1}{2} \Bigl(F^2(\theta)-f^2(\theta)\Bigr)- \frac{1}{2} \Bigl(F^2(\theta+\pi)-f^2(\theta+\pi)\Bigr).$$

Thus, the domain (\ref{1}) satisfies $\boldsymbol{C}$ iff the function $G(\theta)=F^2(\theta)-f^2(\theta)$ is $\pi$-periodic.  Note that

$$F(\theta)= \sqrt{f^2(\theta)+ G(\theta)}.$$

If we choose

$$f(\theta)= e^{\sin \theta},\;\; G(\theta)=e^{\cos 2\theta},$$

then we obtain the domain bounded by the two closed curves in the   figure below. This domain obviously violates the  symmetry condition $\boldsymbol{S}$.

## Wednesday, December 12, 2012

### 12.12.12-Once in a century

I had to do this. It the last time this century one can do this, and I could not pass this opportunity to immortalize it.

### Geometry conference in the memory of Jianguo Cao

It's been a bit over a year and a half now since my dear friend Jianguo Cao unexpectedly passed away. I miss him for many reasons. He   was  my gentle, wise and always wellcoming    Riemann geometry guru.   Our department is organizing a conference in his memory  (March 13-17, 2013).  The least I could do  is to spread the word. Unfortunately, I cannot attend. In any case here is a picture of Jianguo from 2004.  He is the leftmost person in the row, I am the  only bearded  guy.

### On conditional expectations.

$\newcommand{\bR}{\mathbb{R}}$  I  am still struggling with the idea of conditioning.   Maybe this public  confession will help clear things out.   $\newcommand{\bsP}{\boldsymbol{P}}$ $\newcommand{\eA}{\mathscr{A}}$ $\newcommand{\si}{\sigma}$

Suppose that $(\Omega, \eA, \bsP)$ is a probability space, where $\eA$ is a $\si$-algebra of subsets of $\Omega$ and  $\bsP:\eA\to [0,1]$ is a probability measure. We assume that $\eA$ is complete with respect to $\bsP$, i.e.,  subsets of $\bsP$-negligible subsets are measurable. $\newcommand{\bsU}{{\boldsymbol{U}}}$ $\newcommand{\bsV}{{\boldsymbol{V}}}$ $\newcommand{\eB}{\mathscr{B}}$

Assume that $\bsU$ and $\bsV$ are two finite dimensional real vector spaces equipped with the $\si$-algebras of Borel subsets, $\eB_{\bsU}$ and respectively $\eB_{\bsV}$. Consider two random  variables  $X:\Omega\to \bsU$ and $Y:\Omega\to \bsV$ with probability   measures

$$p_X=X_*\bsP,\;\;p_Y=Y_*\bsP.$$

Denote by $p_{X,Y}$ the joint probability measure $\newcommand{\bsE}{\boldsymbol{E}}$

$$p_{X,Y}=(X\oplus Y)_*\bsP.$$

The expectation $\bsE(X|Y)$ is a new $\bsV$-valued  random variable $\omega\mapsto \bsE(X|Y)_\omega$,  but on a different probability space $(\Omega, \eA_Y, \bsP_Y)$ where $\eA_Y=Y^{-1}(\eB_\bsV)$, and $\bsP_Y$ is the restriction of $\bsP$ to $\eA_Y$. The events in $\eA_Y$ all have the form $\{Y\in B\}$, $B\in\eB_{\bsU}$.

This  $\eA_Y$-measurable random variable is defined uniquely by the  equality

$$\int_{Y\in B} E(X|Y)_\omega \bsP_Y(d\omega) = \int_{Y\in B}X(\omega) \bsP(d\omega),\;\;\forall B\in\eB_\bsV.$$

Warning: The truly subtle thing in the above   equality is  the integral in the left-hand-side which is performed with respect to the restricted measure $\bsP_Y$.

If we denote by $I_B$ the indicator function of  $B\in\eB_\bsV$, then we can rewrite the above  equality as

$$\int_\Omega \bsE(X|Y)_\omega I_B(Y(\omega)) \bsP_Y(d\omega)=\int_\Omega X(\omega) I_B(Y(\omega) )\bsP(d\omega).$$

In particular  we deduce that for any  step function $f: \bsV \to \bR$ we have

$$\int_\Omega \bsE(X|Y)_\omega f(Y(\omega)) \bsP_Y(d\omega) =\int_\Omega X(\omega) f(Y(\omega) )\bsP(d\omega).$$

The random variable  $\bsE(X|Y)$  defines  a   $\bsU$-valued  random variable $\bsV\ni y\mapsto \bsE(X|y)\in\bsU$ on the probability space $(\bsV,\eB_\bsV, p_Y)$   where

$$\int_B \bsE(X| y) p_Y(dy)=\int_{(x,y)\in B\times\bsV} x p_{X,Y}(dxdy).$$

Example 1.   Suppose that $A, B\subset \Omega$,  $X=I_A$, $Y=I_B$.   Then  $\eA_Y$ is the $\si$-algebra generated by $B$. The random variable $\bsE(I_A|I_B)$  has a constant value $x_B$ on $B$ and a constant value $x_{\neg B}$ on $\neg B :=\Omega\setminus B$. They are determined by the equality

$$x_B \bsP(B)= \int_B I_A(\omega)\bsP(d\omega) =\bsP(A\cap B)$$

so that

$$x_B=\frac{\bsP(A\cap B)}{\bsP(B)}=\bsP(A|B).$$

Similarly

$$x_{\neg B}= \bsP(A|\neg B).$$

$$\ast\ast\ast$$

Example 2.   Suppose $\bsU=\bsV=\bR$ and that $X$ and  $Y$ are discrete random variables with ranges $R_X$ and $R_Y$.   The random variable  $\bsE(X|Y)$ has a constant value $\bsE(X|y)$ on the set $\{Y=y\}$, $y\in R_Y$. It is determined from the equality

$$\bsE(X|Y=y)p_Y(y)=\int_{Y=y} \bsE(X|Y)_\omega \bsP_Y(\omega) =\int_{Y=y} X(\omega) d\bsP(\omega).$$

Then $\bsE(X|Y)$ can be viewed as a random variable $(R_Y, p_Y)\to \bR$,  $y\mapsto \bsE(X|Y)_y=\bsE(X|Y=y)$, where

$$\bsE(X|Y=y) =\frac{1}{p_Y(y)}\int_{Y=y} X(\omega) d\bsP(\omega).$$

For this reason  one should think of $\bsE(X|Y)$ as a function of $Y$.    From this point of view, a more appropriate notation would be $\bsE_X(Y)$.

The joint probability  distribution  $p_{X,Y}$ can be viewed as a function

$$p_{X,Y}: R_X\times R_Y\to \bR_{\geq 0},\;\;\sum_{(x,y)\in R_X\times R_Y} p_{X,Y}(x,y)= 1.$$

Then

$$\bsE(X|Y=y)= \sum_{x\in R_X} x\frac{P_{X,Y}(x,y)}{p_Y(y)}.$$

We introduce   new $R_X$-valued random variable  $(X|Y=y)$ with probability distribution $p_{X|Y=y}(x)=\frac{p_{X,Y}(x,y)}{p_Y(y)}$.

Then $\bsE(X|Y=y)$ is  the expectation of the  random variable $(X|Y=y)$.

$$\ast\ast\ast$$

Example 3. Suppose that  $\bsU, \bsV$ are equipped with  Euclidean metrics,  $X,Y$ are centered Gaussian random vectors with  covariance forms $A$ and respectively $B$.   Assume  that the covariance pairing between $X$ and $Y$ is $C$ so that the covariance  form of $(X, Y)$ is

$$S=\left[ \begin{array}{cc} A & C\\ C^\dagger & B \end{array} \right].$$

We have $\newcommand{\bsW}{\boldsymbol{W}}$

$$P_{X,Y}(dw) =\underbrace{\frac{1}{\sqrt{\det 2\pi S}} e^{-\frac{1}{2}(S^{-1}w,w)} }_{=:\gamma_S(w)}dw,\;\;w=x+y\in \bsW:=\bsU\oplus \bsV$$

$$P_Y(dy) = \gamma_B(y) dy),\;\;\gamma_B(Y)=\frac{1}{\sqrt{\det 2\pi B}} e^{-\frac{1}{2}(B^{-1}w,w)} .$$

For any  bounded measurable function $f:\bsU\to \bR$ we have

$$\int_{\Omega} f(X(\omega)) \bsP(d\omega)=\int_\bsV \bsE(f(X)|Y) \bsP_Y(d\omega)=\int_\bsV \bsE(f(X)| Y=y) dp_Y(y).$$

We deduce

$$\int_{\bsU\oplus \bsV} f(x) \gamma_S(x,y) dx dy= \int_\bsV \bsE(f(X)| Y=y) \gamma_B(y) dy.$$

Now observe that

$$\int_{\bsV}\left(\int_\bsU f(x) \gamma_S(x,y) dx\right) dy = \int_\bsV \bsE(f(X)| Y=y) \gamma_B(y) dy.$$

This implies that

$$\bsE(f(X)| Y=y) =\frac{1}{\gamma_B(y)} \left(\int_\bsU f(x) \gamma_S(x,y) dx\right).$$

We obtain a probability measure  $p_{X|Y=y}$ on the affine plane $\bsU\times \{y\}$ given by

$$p_{X|Y=y}(dx)= \frac{\gamma_S(x,y)}{\gamma_B(y)} dx.$$

This is a Gaussian    measure on $\bsU$. Its   statistics are described by  the regression formula.  More precisely, its mean is

$$m_{X|Y=y}= Cy,$$

and its covariance form is

$$S_{X|Y=y}= A- CB^{-1}C^\dagger.$$

$$\ast\ast\ast$$

In general, if we think of $p_{X,Y}$ as a density on $\bsU\oplus \bsV$,  of $p_Y$ as a density on $\bsV$ and we denote by $\pi_\bsV$ the natural projection $\bsU\oplus\bsV\to\bsV$, then the conditional probability distribution  $p_{X|Y=y}$ is a a probability density on $\pi^{-1}_\bsV(y)$. More precisely  it is the density $p_{X,Y}/\pi^*_Vp_Y$ defined as in  Section 9.1.1 of my lectures,  especially Proposition 9.1.8 page 350 of the lectures.

## Monday, December 3, 2012

### Degeneration of Gaussian measures

$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\ve}{{\varepsilon}}$ $\newcommand{\bsV}{\boldsymbol{V}}$  Suppose that $\bsV$ is an $N$-dimensional real Euclidean space   equipped with an orthogonal  direct sum $\newcommand{\bsU}{\boldsymbol{U}}$ $\newcommand{\bsW}{\boldsymbol{W}}$

$$\bsV =\bsU\oplus \bsW. \tag{1}\label{1}$$

Suppose that $S_n: \bsU\to\bsU$ and $C_n:\bsW\to \bsW$ are symmetric positive definite  operators such that

$$S_n\to 0,\;\;C_n\to C,\;\;\mbox{as}\;\;n\to \infty$$

where $C$ is a symmetric positive definite operator on $\bsW$.     We set

$$A_n=S_n\oplus C_n :\bsV\to \bsV$$

and we think of $A_n$ as the covariance   matrix  of a  Gaussian measure on $\bsV$ $\newcommand{\bv}{\boldsymbol{v}}$

$$\gamma_{A_n}(|d\bv|)=\frac{1}{\sqrt{\det 2\pi A_n}} e^{-\frac{1}{2}(A_n^{-1} \bv,\bv)} |d\bv|.$$

Suppose that $f:\bsV\to \bR$ is a  locally Lipschitz function,   positively homogeneous of degree $k\geq 1$.

I am interested in the  behavior as $n\to \infty$ of the expectation

$$E_n(f):=\int_{\bsV} f(\bv)\gamma_{A_n}(|d\bv|).$$

$\newcommand{\bu}{\boldsymbol{u}}$  $\newcommand{\bw}{\boldsymbol{w}}$ We respect to the decomposition (\ref{1}) a vector  $\bv\in \bv_0$  can be written as an orthogonal sum $\bv=\bu+\bw$.

Define

$$\bar{f}_n:\bsW\to [0,\infty),\;\; \bar{f}_n(\bw)= \int_{\bsU} f(\bu+\bw) \gamma_{S_n}(|d\bu|),$$

where $d\gamma_{S_n}$ denotes the Gaussian measure  on $\bsU$ with  covariance form $S_n$.  Then

$$E_n(f)=\int_{\bsW} \bar{f}_n(\bw) \gamma_{C_n}(|d\bw|). \tag{2}\label{2}$$

For $\bw\in \bsW$ and $r\in (0,1]$ we set

$$m(\bw, r) := \sup_{|\bu|\leq r}|f(\bw+u)- f(u)|.$$

Note that

$$\exists L>0: m(\bw,r)\leq Lr,\;\;\forall |\bw|= 1\tag{3}\label{3}$$

In general,   we set $\bar{\bw}:=\frac{1}{|\bw|} \bw$. If $|\bu|\leq r$  and we have
$$\bigl|\;f(\bw+\bu)-g(\bw) \;\bigr|= |\bw|^k \left| f\Bigl(\bar{\bw}+\frac{1}{|\bw|} \bu\Bigr) -f(\bar{\bw})\right| \leq L |\bw|^{k-1} r,$$
so that
$$m(w,r) \leq L|\bw|^{k-1} r,\;\;\forall \bw\in\bsW,\;\;r\in (0,1]. \tag{4}\label{4}$$

To proceed further, we need a vector counterpart for the Chebysev inequality.

Lemma 1.  Suppose $S:\bsU\to \bsU$ is a  symmetric, positive definite operator. We set $R:=S^{-\frac{1}{2}}$ and  denote by $\gamma_{S}$ the associated   Gaussian measure. Then for any $c,\ell>0$  we have $\newcommand{\bsi}{\boldsymbol{\sigma}}$

$$\int_{ |R \bu|\geq c} |\bu|^\ell d\gamma_S(|\bu|) \leq \sqrt{2^{\ell+m-\frac{3}{2}} \Gamma\Bigl(\; \ell+m-\frac{1}{2}\;\Bigr)} \frac{\bsi_{m-1}}{(2\pi)^{\frac{m}{2}}} \Vert S\Vert^{\frac{\ell}{2}}c^{-\frac{1}{2}}e^{-\frac{c^2}{4}} , \tag{5}\label{5}$$

where $m=\dim\bsU$ and  and $\bsi_N$ denote the area of the $N$-dimensional unit sphere.

Proof.   We make the change in variables $\newcommand{\bx}{\boldsymbol{x}}$  $\bx:=R\bu$ and we  deduce
$$\int_{ |R \bu|\geq c} |\bu|^\ell d\gamma_S(|\bu|)\leq \frac{1}{(2\pi)^{\frac{m}{2}}} \int_{|\bx|\geq c} |S^{\frac{1}{2}} \bx|^\ell e^{-\frac{1}{2}|\bx|^2} |d\bx|$$

$$\leq \frac{\Vert|S\Vert^{\frac{\ell}{2}}}{(2\pi)^{\frac{m}{2}}} \int_{|\bx|\geq c} |\bx|^\ell e^{-\frac{1}{2}|\bx|^2} |d\bx|=\frac{\bsi_{m-1}\Vert S\Vert^{\frac{\ell}{2}}}{(2\pi)^{\frac{m}{2}}}\int_{t>c} t^{\ell+m-1} e^{-\frac{1}{2} t^2} dt$$
$$\leq \frac{\bsi_{m-1}\Vert S\Vert^{\frac{\ell}{2}}}{(2\pi)^{\frac{m}{2}}} \left(\int_{t>c} e^{-\frac{1}{2} t^2} dt\right)^{\frac{1}{2}}\left(\int_{t>0} t^{2\ell+2m-2} e^{-\frac{1}{2} t^2} dt\right)^{\frac{1}{2}}$$
Now observe that we have
$$\int_{t>c} e^{-\frac{1}{2} t^2} dt \leq \frac{1}{c} e^{-\frac{c^2}{2}},$$
and  using the change of variables $s=\frac{t^2}{2}$ we  deduce

$$\int_{t>0} t^{2\ell+2m-2} e^{-\frac{1}{2} t^2} dt =2^{\ell+m-\frac{3}{2}}\int_0^\infty s^{\ell+m-\frac{1}{2}-1} e^{-s} ds= 2^{\ell+m-\frac{3}{2}} \Gamma( \ell+m-\frac{1}{2}).$$

This proves the lemma. q.e.d

We now want to compare $\bar{f}_n(\bw)$ and $f(\bw)$ for $\bw\in\bsW$.  We plan to use Lemma  1.   Set $R_n:=S_n^{-\frac{1}{2}}$ and $m:=\dim\bsU$.   Observe that

$$|\bu\|= |S_n^{\frac{1}{2}}R_n\bu|\leq \Vert S_n^{\frac{1}{2}}\Vert\cdot |R_n\bu|.$$

For simplicity   set $s_n:= \Vert S_n^{\frac{1}{2}}\Vert$.   Choose a sequence of positive numbers  $c_n$ such that $c_n\to\infty$ and  $s_n c_n\to 0$.  Later  we will add several requirements to this sequence.

$$\bigl|\;\bar{f}_n(\bw)-f(\bw)\;\bigr|=\left| \int_{\bsU} (\; f(\bw+\bu)- f(\bw)\; ) \gamma_{S_n}(|d\bu|)\right|$$
$$\leq \left| \int_{|R_n\bu|\leq c_n} (\; f(\bw+\bu)- f(\bw)\; ) \gamma_{S_n}(|d\bu|)\right|+\left|\int_{|R_n\bu|\geq c_n} (\; f(\bw+\bu)- f(\bw)\; ) \gamma_{S_n}(|d\bu|)\right|$$
$$\stackrel{(\ref{4})}{\leq} L|\bw|^{k-1}s_n c_n +C \int_{|R_n\bu|\geq c_n}(|\bw|^k+|\bu|^k) \gamma_{S_n}(|d\bu|)$$

$$\stackrel{(\ref{5})}{\leq} L|\bw|^{k-1}s_n c_n + Z(k, m)c_n^{-\frac{1}{2}} e^{-\frac{c_n^2}{4}}(1+s_n^k),$$
where $Z(k,m)$ is a constant that depends only on   $k$ and $m$.

We deduce that there exists a constant $C>0$ independent of   $n,w$  such that  for any sequence $c_n\to \infty$ such that $s_nc_n\to 0$, $s_n:=\Vert S_n\Vert^{\frac{1}{2}}$ we have

$$\bigl|\;\bar{f}_n(\bw)-f(\bw)\;\bigr| \leq C\bigl(\; |\bw|^{k-1}s_nc_n + e^{-\frac{c_n^2}{4}}\;\bigr). \tag{6}\label{6}$$
We deduce that

$$\Bigl|\; E_n(f) -\int_{\bsW} f(\bw) \gamma_{C_n}(|d\bw|)\;\Bigr| \leq C\left(s_nc_n\int_{\bsW} |\bw|^{k-1} \gamma_{C_n}(|d\bw|) + e^{-\frac{c_n^2}{4}}\;\right).\tag{7}\label{7}$$

Finally let us estimate

$$D_n:=\int_{\bsW} f(\bw) \gamma_{C_n}(|d\bw|)-\int_{\bsW} f(\bw) d\gamma_{C}(|d\bw|).$$

We have  $\newcommand{\one}{\boldsymbol{1}}$
$$D_n= \int_{\bsW} \left( f\bigl( C_n^{\frac{1}{2}}\bw\;\bigr)-f\bigl( C^{\frac{1}{2}}\bw\;\bigr) \;\right)\gamma_{\one}(|\bw|)$$
and we conclude that
$$\left|\; \int_{\bsW} f(\bw) d\gamma_{C_n}(|d\bw|)-\int_{\bsW} f(\bw) d\gamma_{C}(|d\bw|)\;\right| \leq L \Bigl\Vert \;C_n^{\frac{1}{2}}-C^\frac{1}{2}\;\Bigr\Vert \int_{\bsW}|\bw|^k \gamma_{\one}(|d\bw|). \tag{8}\label{8}$$

In (\ref{7}) we let $c_n:=s_n^{-\ve}$. If we denote by $A_\infty$ the limit of the covariance matrices $A_n$, $A=\lim_{n\to\infty} A_n =0\oplus C$, then we deduce from the above computations that for any $\ve>0$ there exists a  constant $C_\ve>0$ such that
$$\left|\; \int_{\bsV} f(\bv) \gamma_{A_n} (|d\bw|) -\int_{\bsV} f(\bv) \gamma_{A_\infty} (|d\bw|) \;\right|\leq C_\ve \left(s_n^{1-\ve}+ \Bigl\Vert \;C_n^{\frac{1}{2}}-C^\frac{1}{2}\;\Bigr\Vert\right)\leq C_\ve \Bigl\Vert A_n^{\frac{1}{2}}-A_\infty^{\frac{1}{2}}\Bigr\Vert^{1-\ve}.\tag{9}\label{9}$$
This can be generalized a bit. Suppose that $T_n:\bsU\to \bsU$ is a sequence of orthogonal operators such that  $T_n\to \one_{\bsU}$

Using (\ref{7})  we deduce

$$\left|\;\int_{\bsV} T^*_nf(\bv) \gamma_{A_n}(|d\bv|)-\int_{\bsV} f(\bv) \gamma_{A_n}(|d\bv|)\right|= \left| \int_{\bsV} f(T_n A_n^{\frac{1}{2}}\bx)- f( A_n^{\frac{1}{2}}\bx)\gamma_{\one}(|d\bx|) \right| \leq L \Bigl\Vert A_n^{\frac{1}{2}}\Bigr\Vert \Vert T_n-\one\Vert.$$
Observe that
$$\int_{\bsV} T^*_nf(\bv) \gamma_{A_n}(|d\bv|)=\int_{\bsV} f(\bv) \gamma_{B_n}(|d\bv|),$$
where
$$B_n= T_nA_nT_n^*.$$

Suppose that we are in the fortunate case when $f|_{\bsW}=0$.    Then

$$\int_{\bsW} f(\bw) d\gamma_{C_n}(|d\bw|)=\int_{\bsW} f(\bw) d\gamma_{C}(|d\bw|)=0$$

and (\ref{9})  can be improved to

$$\left|\; \int_{\bsV} f(\bv) \gamma_{A_n} (|d\bw|)\right|\leq C_\ve s_n^{1-\ve}.$$

### On the 11/8-conjecture

Stefan Bauer has just posted a proof for the 11/8- conjecture for simply connected $4$-manifolds

## Tuesday, November 27, 2012

### FRANK FEST: Workshop on High Dimensional Topology @ ND 2012

This is a conference in honor of Frank Connolly who is retiring this semester.

Workshop on High Dimensional Topology @ ND 2012

## Wednesday, November 21, 2012

### Sharp nondegeneracy estimates for a family of random Fourier series

$\newcommand{\bR}{\mathbb{R}}$  $\newcommand{\ve}{{\varepsilon}}$ $\newcommand{\eS}{\mathscr{S}}$ $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\bZ}{\mathbb{Z}}$

Suppose that $w\in \eS(\bR)$ is an even, nonnegative  Schwartz  function.  Assume that $w\not\equiv 0$. $\newcommand{\hw}{\widehat{w}}$ We denote by $\hw(t)$   its Fourier transform

$$\hw(t)=\int_{\bR}e^{-\ii t x} w(x) dx.$$

For $n\in \bZ$ we  set $\newcommand{\be}{\boldsymbol{e}}$

$$\be_n(\theta) :=\frac{1}{\sqrt{\pi}}\begin{cases} \frac{1}{\sqrt{2}}, & n=0,\\ \sin n \theta , & n<0,\\ \cos n\theta , & n>0. \end{cases}$$

Observe that  the collection $\lbrace \be_n(\theta)\rbrace_{n\in\bZ}$ is an orthonormal  basis of $L^2(\bR/2\pi\bZ)$.   $\newcommand{\bT}{\mathbb{T}}$      For any positive integer $N$ we  denote by $\bT^N$ the $N$-dimensional torus

$$\bT^N:= (\bR/2\pi\bZ)^N$$.

Consider the random Fourier series

$$f_\ve(\theta)=\sum_{n\in \bZ} \sqrt{w(\ve n) } c_n \be_n(\theta),$$

where $(c_n)_{n\in\bZ}$ are i.i.d. Gaussian  random variables with mean zero and variance $1$. $\newcommand{\eE}{\mathscr{E}}$  The   correlation kernel of this random function is  $\newcommand{\bsE}{\boldsymbol{E}}$ $\newcommand{\vfi}{\varphi}$

$$\eE^{\ve}:\bT^1\times \bT^1\to \bR,\;\;\eE^\ve (\theta,\vfi)=\bsE\bigl( f_\ve(\theta)\cdot f_\ve(\vfi)) =\sum_{n\in \bZ} w(\ve n) \be_n(\theta)\be_n(\vfi)$$

$$=\frac{1}{2\pi} w(0)+\frac{1}{\pi}\sum_{n>0}w(\ve n)\cos n(\theta-\vfi)=\frac{1}{2\pi}\sum_{n\in\bZ}w(\ve n) e^{\ii n(\theta-\vfi)}=W_\ve(\theta-\vfi). \tag{1}\label{1}$$

Poisson formula.   For any $\phi\in \eS(\bR)$  and any $c\in\bR\setminus 0$ we have

$$\frac{2\pi}{c}\sum_{n\in\bZ} \phi\Bigl(\frac{2\pi n}{c}\Bigr)= \sum_{\nu\in\bZ} \widehat{\phi}(n c).$$

Suppose $\phi\in\eS(\bR)$ and $c$ are such that

$$\phi\Bigl(\;\frac{2\pi n}{c}\;\Bigr)= w(\ve n) e^{\ii n(\theta-\vfi)} .$$

If we formally replace $n =\frac{c x}{2\pi}$ we deduce from the above equality that

$$\phi(x)= w\Bigl(\frac{\ve c x}{2\pi}\Bigr) e^{\ii\frac{c(\theta-\vfi)x}{2\pi}}=w(ax)e^{\ii b x}, \;\; a:=\frac{\ve c}{2\pi},\;\;b :=\frac{c(\theta-\vfi)}{2\pi}.$$

Then

$$\widehat{\phi}(t) =\int_{\bR} e^{-\ii tx} w(ax) e^{\ii bx} dx = \frac{1}{a}\int_{\bR} e^{-\ii \frac{t-b}{a}x} w(y) dy = \frac{1}{a}\hw\Bigl( \frac{t-b}{a}\Bigr).$$

We now set $c:=2\pi$ so that $a=\ve$, $b=(\theta-\vfi)$. Using The Poisson formula in (\ref{1}) we deduce

$$W_\ve(\theta-\vfi)=\eE^\ve(\theta,\vfi) =\frac{1}{2\pi\ve} \sum_{n\in\bZ} \hw\Bigl(\frac{2\pi n-(\theta-\vfi)}{\ve}\Bigr) . \tag{2}\label{2}$$

Now consider the  random function

$$F_\ve:\bT^N\to \bR,\;\; F_\ve(\vec{\theta}) = \sum_{j=1}^n f_\ve(\theta_j).$$

The  correlation kernel of this  random function is

$$\eE_N^\ve(\vec{\theta},\vec{\vfi}) =\sum_{1\leq j,k\leq N} \eE^\ve(\theta_j-\vfi_k).$$

The differential of $F_\ve$ at a point $\vec{t}\in\bT^N$ is a Gaussian  random vector with covariance matrix $\newcommand{\pa}{\partial}$

$$S^\ve(\vec{t})= \Bigl( S^\ve_{jk}(\vec{t})\;\Bigr)_{1\leq j,k\leq N},\;\; S^\ve_{jk}(\vec{t})= \frac{\pa^2}{\pa\theta_j\pa \vfi_k} \eE_N^\ve\bigl(\;\vec{\theta},\vec{\vfi}\;\bigr)|_{\vec{\theta}=\vec{\vfi}=\vec{t}}=-W_\ve''(t_j-t_k)=\frac{1}{2\pi}\sum_{n\in\bZ} n^2w(\ve n) e^{\ii n(t_j-t_k)}\tag{3}\label{3}.$$

Definition. We say that $\vec{t}\in\bT^n$ is  nondegenerate if $t_j-t_k\in\bR\setminus 2\pi\bZ$, $\forall j\neq k$. We denote by $\bT^N_*$ the collection of nondgenerate  points in $\bT^N$.

$$\ast\ast\ast$$

We have the following result similar to the one in our   previous post.

Proposition 1.  There exists $\ve_0=\ve_0(w,N)>0$ such that if $\ve \in (0,\ve_0)$ and  $\vec{t}\in \bT^N$ is nondegenerate, then the  matrix $S^\ve(\vec{t})$ is positive  definite.

Proof.     Set

$$Z_\ve:=\bigl\{ n\in\bZ;\;\;w(\ve n)\neq 0\;\bigr\}.$$

Consider the space $H_\ve$ consisting of functions $\newcommand{\bC}{\mathbb{C}}$

$$u: Z_\ve \to \bC,\;\;\sum_{n\in Z_\ve} |u(n)|^2 n^2 w(\ve n) <\infty.$$

This is a separable  Hilbert space with inner product

$$(u,v)_\ve= \frac{1}{2\pi} \sum_{n\in\bZ} u(n)\cdot \overline{v(n)}\; n^2w(\ve n).$$

We denote by $\Vert-\Vert_\ve$ the associated norm.

For $t\in\bT^1$ consider the truncated character $\chi^\ve_t:Z_\ve\to \bT^1$, $\chi^\ve_t(n)=e^{\ii tn}$.  For $\vec{z}\in \bC^N$ $\newcommand{\vez}{{\vec{z}}}$  and $\vec{t}\in \bT^N$ consider $T_{\vez,\vec{t}}\in H_\ve$

$$T_{\vez,\vec{t}}(n)=\sum_{j=1}^n z_j \chi^\ve _{t_j}(n)=\sum_{j=1}^N z_j e^{\ii t_j n},\;\;n\in Z_\ve.$$

From the equality (\ref{3}) we deduce that

$$\sum_{j,k=1}^n S^\ve_{jk}(\vec{t}) z_j\bar{z}_k = \Vert T_{\vez,\vec{t}}\Vert_\ve^2.$$

Thus, the matrix $S^\ve(\vec{t})$ has a kernel if and only if the truncated characters $\chi^\ve_{t_1},\dotsc, \chi^\ve_{t_N}$ are linearly dependent.  We show that this is not possible if  $\vec{t}$ is  nondegenerate and  $\ve$ is sufficiently small.

Fix  $\ve_0=\ve_0(N,w)$ such that  if $\ve<\ve_0$ the support   of $x\mapsto w(\ve x)$ contains a long  interval of the form $[\nu_\ve, \nu_\ve+N-1]$, for some integer $\nu_\ve>0$. (Recall that $w$ is even.) In other words $\nu_\ve,\nu_\ve+1,\cdots,\nu_\ve+N-1\in Z_\ve$.

Let $\ve\in (0,\ve_0)$  and suppose that  $\vez\in\bC^N\setminus 0$ and $\vec{t}\in\bT^N$ are such that  such that  $T_{\vez,\vec{t}}=0$.   Thus

$$(T_\vez, u)_\ve =0,\;\;\forall u\in H_\ve$$

For any $m\in Z_\ve$ consider the Dirac function $\delta_m: Z_\ve\to \bC$,  $\delta_m(n)=\delta_{mn}=$ the Kronecker  delta.

We deduce that for any $m=\nu_\ve,\nu_\ve+1,\dotsc, \nu_\ve+N-1$ we have

$$0 = (T_\vez, \delta_m)_\ve=m^2w(\ve m) \sum_{j=1}^N z_j e^{\theta_j m} .$$

This can happen if and only if

$$0= \det\left[ \begin{array}{cccc} e^{\ii \nu_\ve t_1} & e^{\ii \nu_\ve t_2} & \cdots & e^{\nu_\ve t_N}\\ e^{\ii(\nu_\ve+1)t_1} & e^{\ii(\nu_\ve+1)t_2} & \cdots & e^{\ii (\nu_\ve+1) t_N}\\ \vdots & \vdots &\vdots &\vdots\\ e^{\ii(\nu_\ve+N-1)t_1} & e^{\ii(\nu_\ve+N-1)t_2} &\cdots & e^{\ii(\nu_\ve+N-1)t_N} \end{array} \right] = e^{\ii\nu_\ve(t_1+\cdots +t_N)} \prod_{j<k} \Bigl( e^{\ii t_k}-e^{\ii t_j}\Bigr) .$$

This shows that  $S^\ve(\vec{t})$ has a kernel if and only of $\vec{t}$ is degenerate.    Q.E.D.

Remark.    Here is an alternate proof of  Proposition 1 that yields a bit more. The above proof shows that

$$S^\ve_{jk}(\vec{t})= (\chi_{t_j},\chi_{t_k})_\ve.$$

Suppose for simplicity that $0\in Z_\ve$, i.e.,    $w(0)>0$. Then for $\ve>0$ sufficiently small we have $1,\dotsc, N\in Z_\ve$.  Observe that

$$(\delta_j,\delta_k)_\ve= \frac{k^2w(\ve k)}{2\pi}\delta_{jk},\;\;j,k=1,\dotsc, N.$$

We have a Cauchy-Schwartz inequality

$$\Bigl|\; \bigl( \chi_{t_1}\wedge\cdots \chi_{t_n}, \delta_1\wedge \cdots \wedge \delta_N\;\bigr)_\ve\;\Bigr| \leq \bigl|\; \chi_{t_1}\wedge\cdots \wedge\chi_{t_n}\;\bigr|_\ve\cdot \bigl|\;\delta_1\wedge \cdots \wedge \delta_N\;\bigr|_\ve.$$

This translates to

$$\Bigl| \det\Bigl(\; (\chi_{t_j},\delta_k)_\ve\;\Bigr)_{1\leq i,j\leq N}\;\Bigr| \leq \sqrt{\det\Bigr( \; (\chi_{t_j},\chi_{t_k})_\ve\;\Bigr)_{1\leq i,j\leq N} } \cdot \sqrt{\det\Bigr( \; (\delta_j,\delta_k)_\ve\;\Bigr)_{1\leq i,j\leq N} },$$
or, equivalently

$$\prod_{j<k} \Bigl| e^{\ii t_j}-e^{\ii t_k}\Bigr|^2 \leq \frac{1}{(2\pi)^N}\Bigl(\prod_{j=1}^N j^2w(\ve j)\Bigr) \det S^\ve(\vec{t}). \tag{4} \label{4}$$

$$\ast\ast\ast$$

The basic question that interests me is the following: what  happens to $S^\ve(\vec{t})$ as $\ve\to 0$, and $\vec{t}$ is nondegenerate.

Observe that (\ref{2}) implies that

$$W_\ve''(t)=\frac{1}{2\pi\ve^3}\sum_{n\in\bZ}\hw''\Bigl(\frac{2\pi n-t}{\ve}\Bigr).$$

We make the change in variables $t=\ve \tau$, we set

$$C^\ve(\vec{\tau}) := S^\ve(\ve\vec{\tau})$$

and we deduce

$$C^\ve_{jk}(\vec{\tau})=\frac{1}{2\pi}\sum_{n\in\bZ}n^2w(\ve n) e^{\ii\ve n(\tau_j-\tau_k)} =-\frac{1}{2\pi\ve^3}\sum_{n\in\bZ}\hw''\Bigl(\tau_j-\tau_k-\frac{2\pi n}{\ve}\Bigr),\;\;0\leq \tau_j <\frac{2\pi}{\ve},\;\;j=1,\dotsc, N. \tag{5}\label{5}$$

For $\vec{t}\in\bT^N$ nondegenerate  we denote by $X(\vec{t})\subset H_\ve$ the vector space spanned by the characters $\chi_{t_j}$, $j=1,\dotsc, N$. $\newcommand{\bsD}{\boldsymbol{D}}$ Denote  by $\bsD_N\subset H_\ve$ the space spanned by $\delta_1,\dotsc, \delta_n$.   For $k=1,\dotsc, N$ we  set $\newcommand{\bde}{\check{\delta}}$

$$\bde_k=\bde_k^\ve :=\frac{\sqrt{2\pi}}{k\sqrt{w(\ve k)}}\delta_k.$$

By construction,  the collection $\bde_1,\dotsc,\bde_N$ is an $(-,-)_\ve$-orthonormal basis of $\bsD_N$.

The  $(-,-)_\ve$-orthogonal projection $P_\ve=P_\ve(\vec{t}): X(\vec{t})\to \bsD_n$ is given by

$$P_\ve \chi_{t_j} =\sum_{k=1}^n (\chi_{t_j},\bde_k)_\ve \bde_k = \sum_{k=1}^N e^{\ii kt_j} \delta_k.$$

With respect to the natural bases $\chi_{t_1},\dotsc,\chi_{t_N}$ of $X(\vec{t})$ and $\delta_1,\dotsc, \delta_N$ of $\bsD_N$ the    projection is therefore given by the  Vandermonde matrix

$$V= V(\vec{t}),\;\; V_{kj}= e^{\ii k t_j},\;\; V=\left[\begin{array}{cccc} e^{\ii t_1} & e^{\ii t_2} &\cdots & e^{\ii t_N}\\ e^{2\ii t_1} & e^{2\ii t_2} & \cdots & e^{2\ii t_N}\\ \vdots &\vdots &\vdots &\vdots\\ e^{N\ii t_1} & e^{N\ii t_2} &\cdots & e^{N\ii t_N} \end{array} \right].$$

## Wednesday, November 14, 2012

### On a family of symmetric matrices

$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\eS}{\mathscr{S}}$   Suppose that $w: \bR\to[0,\infty)$ is an integrable function.  Consider its   Fourier transform $\newcommand{\ii}{\boldsymbol{i}}$

$$\widehat{w}(\theta)=\int_{\bR} e^{-\ii x\theta}w(\theta) dx.$$

For any $\vec{\theta}\in\bR^n$ we form the  complex Hermitian  $n\times n$ matrix

$$A_w(\vec{\theta})= \bigl(\; a_{ij}(\vec{\theta})\;)_{1\leq i,j\leq n},\;\; a_{ij}(\vec{\theta})=\widehat{w}(\theta_i-\theta_j) .$$

Observe that  for any $\vec{z}\in\mathbb{C}^n$  we have $\newcommand{\bC}{\mathbb{C}}$

$$\bigl(\; A_w(\vec{\theta})\vec{z},\vec{z}\;\bigr)=\sum_{i,j} \widehat{w}(\theta_i-\theta_j) z_i\bar{z}_j =\int_{\bR} | T_{\vec{z}}(x,\vec{\theta})|^2 w(x) dx,$$

where  $T_{\vec{z}}( x)$ is  is the trigonometric polynomial $\newcommand{\vez}{\vec{z}}$

$$T_{\vez}(x,\vec{\theta})= \sum_j z_j e^{\ii \theta_j x}.$$

We denote by $(-,-)_w$ the inner product

$$(f,g)_w=\int_{\bR} f(x) \bar{g(x)} w(x) dx,\;\;f,g:\bR\to \bC.$$

We see that $A_w(\vec{\theta})$  is the Gramm-Schmidt matrix

$$a_{ij}(\vec{\theta})= (E_{\theta_i}, E_{\theta_j})_w,\;\; E_\theta(x)=e^{\ii\theta x}.$$

We see that $\sqrt{\;\det A_w(\vec{\theta})\;}$ is equal to the $n$-dimensional volume   of the parallelepiped   $P(\vec{\theta})=L^2(\bR, wdx)$ spanned  by the  functions $E_{\theta_1},\dotsc, E_{\theta_n}$. We observe  that if these exponentials are  linearly  dependent,  then this volume is  zero.  Here is a first elementary result.

Lemma  1.   The exponentials  $E_{\theta_1},\dotsc, E_{\theta_n}$ are linearly dependent  (over $\bC$) if and only if $\theta_j=\theta_k$ for  some $j\neq k$.

Proof.    Suppose that

$$\sum_{j=1}^n z_j E_{\theta_j}(x)=0,\;\;\forall x\in \bR.$$

Then for any $f\in \eS(\bR)$ we have

$$\sum_{j=1}^n z_j E_{\theta_j}(x)f(x)=0,\;\;\forall x\in \bR.$$

By taking the Fourier Transform of the last equality we deduce

$$\sum_{j=1}^n z_j \widehat{f}(\theta-\theta_j) =0. \label{1}\tag{1}$$

If we now choose $\newcommand{\ve}{{\varepsilon}}$ a  family $f_\ve(x)\in\eS(\bR)$ such that, as $\ve\searrow 0$,  $\widehat{f}_\ve(\theta)\to\delta(\theta)=$ the Dirac  delta function concentrated at $0$, we  deduce  from (\ref{1})  that

$$\sum_{j=1}^n z_j\delta(\theta-\theta_j)=0. \tag{2}\label{2}$$

Clearly this can happen if and only if  $\theta_j=\theta_k$ for  some $j\neq k$.  q.e.d.

If we set

$$\Delta(\vec{\theta}) :=\prod_{1\leq j<k\leq n} (\theta_k-\theta_j),$$

then we deduce from the above lemma that

$$\det A_w(\vec{\theta})= 0 \Leftrightarrow \Delta(\vec{\theta})=0.$$

A more precise statement is true.

Theorem 2.  For any integrable weight $w:\bR\to [0,\infty)$  such that $\int_{\bR} w(x) dx >0$ there  exists a  constant $C=C(w)>0$ such that for any $\theta_1,\dotsc, \theta_n\in [-1,1]$ we have

$$\frac{1}{C}|\Delta(\vec{\theta})|^2 \leq \det A_w(\vec{\theta}). \tag{E}\label{E}$$

Proof.          We regard $A_w(\vec{\theta})$ as a hermitian operator

$$A_w(\vec{\theta}):\bC^n\to \bC^n.$$

We denote by $\lambda_1(\vec{\theta})\leq \cdots \leq \lambda_n(\vec{\theta})$ its eigenvalues so that

$$\det A_w(\vec{\theta})=\prod_{j=1}^n \lambda_j(\vec{\theta}) \tag{Det}\label{D}.$$

Observe that $\newcommand{\Lra}{\Leftrightarrow}$ $\newcommand{\eO}{\mathscr{O}}$

$$\vec{z}\in \ker A(\vec{\theta}) \Lra \sum_{j=1}^n z_j E_{\theta_j}(x) =0,\;\;\forall x\in{\rm supp}\; w \Lra \sum_{j=1}^n z_j E_{\theta_j}(x) =0,\;\;\forall x\in\bR. \tag{Ker}\label{K}$$

We want to give a  more precise description    of $\ker A_w(\vec{\theta})$.    Set

$$I_n:=\{1,\dotsc, n\},\;\; \Phi_{\vec{\theta}}=\{ \theta_1,\dotsc,\theta_n\}\subset \bR.$$

We want to emphasize that $\Phi{\vec{\theta}}$ is not a multi-set so that $\#\Phi(\vec{\theta})\leq n.$  $\newcommand{\vet}{{\vec{\theta}}}$.

Example 3.  For example  with $n=6$ and $\vet=(1,2,3,2,2,4)$ we have

$$\Phi_\vet=\Phi_{(1,2,3,2,2,4)}=\{1,2,3,4\}.$$

For $\newcommand{\vfi}{{\varphi}}$ $\vfi\in\Phi_\vet$ we set

$$J_\vfi=\bigl\{ j\in I_n;\;\; \theta_j=\vfi\;\bigr\}.$$

In the example above for $\vet=(1,2,3,2,2,4)$ and $\vfi=2$ we have $J_\vfi=\{2,4,5\}$.   $\newcommand{\vez}{\vec{z}}$  For $J\subset I_n$ we set

$$S_J:\bC^n\to \bC,\;\;S_J(\vez)=\sum_{j\in J} z_.$$

In particular, for  any $\vfi\in\Phi_\vet$ we define

$$S_\vfi:\bC^n\to \bC,\;\; S_{\vfi}(\vec{z})=S_{J_\vfi}(\vez)=\sum_{j\in J_\vfi} z_j.$$

We deduce

$$\sum_{j\in I_n} z_jE_{\theta_j}=\sum_{\vfi\in \Phi_\vet} S_\vfi(\vec{z}) E_\vfi.$$

Using (\ref{K}) we deduce

$$\vez\in\ker A(\vet)\Lra \sum_{\vfi\in \Phi_\vet} S_\vfi(\vec{z}) E_\vfi\Lra S_\vfi(\vez)=0,\;\;\forall \vfi\in \Phi_\vet . \tag{3}\label{3}$$

In particular we deduce

$$\dim \ker A(\vet)=n-\#\Phi_\vet.$$

Step 1.   Assume  that $w$ has compact support so that $\widehat{w}(\theta)$  is real analytic over $\bR$.   We will show  that we have the two-sided estimate

$$\frac{1}{C}|\Delta(\vec{\theta})|^2 \leq \det A_w(\vec{\theta}) \leq C |\Delta(\vec{\theta})|^2. \tag{E_*}\label{Es}$$

In this case  $\det A_w(\vet)$ is real analytic and symmetric in the variables $\theta_1,\dotsc, \theta_n$  and vanishes   if and only if $\theta_j=\theta_k$ for some $j=k$.  Thus $\det A_w(\vet)$ has a  Taylor series expansion (near $\vet=0$)

$$\det A_w(\vet)= \sum_{\ell\geq 0} P_\ell(\vet),$$

where $P_\ell(\vet)$ is a  symmetric polynomial  in $\vet$ that vanishes   when $\theta_j=\theta_k$ for some $j\neq k$. Symmetric  polynomials of this type  have the form,

$$\Delta(\vet)^{2N} \cdot Q(\vet)$$

where $N$ is some positive integer  and $Q$ is a symmetric polynomial.  We deduce from  the \Lojasewicz inequality  for  subanalytic functions that there exists  $C=C(w)>0$, a positive integer $N$  and a rational number and $r>0$ such that

$$\frac{1}{C} |\Delta(\vet)|^{r}\leq \det A_w(\vet) \leq C \Delta(\vet)^{2N},\;\;\forall |\vet|\leq 2\pi. \tag{4} \label{4}$$

We want to show  that in (\ref{4})  we have $2N=r=2$.   We argue by contradiction, namely we assume that $r\neq 2$ or $N\neq 1$.      Let

$$\vet(t)= (0, t, \theta_3, \dotsc, \theta_n), \;\; 0\leq |t| < \theta_3<\cdots < \theta_n.$$

Set $A_w(t)=A_w\bigl(\,\vet(t)\;\bigr)$.   Denote its eigenvalues by

$$0\leq \lambda_1(t)\leq \lambda_2(t)\cdots \leq \lambda_n(t).$$

The eigenvalues are so arranged so that the functions $\lambda_k(t)$ are real analytic for $t$ in a neighborhood of $0$. We deduce from (\ref{3}) that $\ker A_w(0)$ is one dimensional so that $\lambda_1(0) =0$,  $\lambda_k(0)>0$, $\forall k>1$. Hence

$$\det A_w(t) \sim \lambda_1(t) \prod_{k=2}^n \lambda_k(0)\;\;\mbox{as t\searrow 0}. \tag{5}\label{5}$$

On the other hand

$$\Delta(\vet(t))^2 \sim Zt^2 \;\;\mbox{as}\;\; t\searrow 0$$

for some positive constant $Z$.  Using this estimate in (\ref{4}) we deduce $r=2N$.  On the other hand, using the above estimate in (\ref{5}) we deduce

$$\lambda_1(t) \sim Z_1 t^{2N} \;\;\mbox{as}\;\;t\searrow 0. \tag{6}\label{6},$$

for another positive constant $Z_1$.

The   kernel of $A_w(0)$ is spanned by the  unit vector

$$\vez(0)= (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0,\dotsc 0).$$

We can find a real analytic family of vectors $t\mapsto \vec{z}(t)$ satisfying

$$|\vez(t)|=1,\;\; A_w(t) \vez(t)=\lambda_1(t)\vez(t),\;\;\lim_{t\to 0}\vez(t)=\vez(0).$$

In particular, we deduce

$$\dot{A}_w(0)\vez(0)+A_w(0)\dot{\vez}(0)=\dot{\lambda}_1(0)\vez(0)+\lambda_1(0)\dot{\vez}(0)=0.$$

A simple computation  shows that $\dot{A}_w(0) \vez(0)=0$ so we deduce  $A_w(0)\dot{\vez}(0)=0$.  This shows that

$$\dot{z}_1(0)+\dot{z}_2(0)=0,\;\;\dot{z}_k(0)=0,\;\;\forall k>2.$$

$$\lambda_1(t)= (A_w(t) \vez(t),\vez(t))= \int_{\bR} \Bigl| \;\underbrace{\sum_{j=1}^n z_j(t) e^{\theta_j(t) x}}_{=:f_t(x)}\;\Bigr|^2 w(x) dx.$$

Observe that

$$f_t(x):= \sum_{j=1}^n z_j(t) e^{\theta_j(t) x}= \frac{1}{\sqrt{2}}(1-e^{\ii t x}) +\sum_{j=1}^k \ve_j(t) e^{\ii\theta_j(t) x},\;\;\ve_j(t)=z_j(t)-z_j(0).$$

We   deduce that

$$\lim_{t\to 0} \frac{1}{t}f_t(x) = -\frac{\ii x}{\sqrt{2}} + \sum_{k=1}^n \dot{z}_k(0)= -\frac{\ii x}{\sqrt{2}}\tag{7}\label{7}$$

uniformly  for  $x$   on compacts. Since  $w$ has compact support  we deduce that (\ref{7}) holds for uniformly for $x$ in the support of $w$.  We deduce that

$$\lambda_1(t)\sim \frac{1}{2}\;\underbrace{\left(\int_{\bR} x^2 w(x)dx \right)}_{=\widehat{w}''(0)}\;t^2\;\;\mbox{as t\to 0}.$$

Using the last equality in (\ref{6}) we obtain $2N=2$ which proves  (\ref{Es}) .

Step 2.    We will show that if (\ref{E}) holds for $w_0$ and $w_1(x) \geq w_0(x)$,  $\forall x$,  then (\ref{E}) holds for  $w_1$ as well.       For any weight $w$  and any $\vet$ such that the $\Delta(\vet)\neq 0$ consider the ellipsoid

$$\Sigma_w:=\bigl\{\vez\in\bC^n;\;\; (A_w\vez,\vez)\leq 1\bigr\}.$$

Then

$${\rm vol}\, \bigl(\;\Sigma_w(\vet)\;\bigr)=\frac{\pi^n}{n!\det A_w(\vet)}.$$

Observe that if $w_0\leq w_1$ then $\Sigma_{w_0}(\vet)\subset \Sigma_{w_1}(\vet)$ and we  deduce

$$\det A_{w_0}(\vet) \leq \det A_{w_1}(\vet).$$

This proves our claim.

Step 3. We show that (\ref{E}) holds for any integrable weight.   At  least one of the level sets $\{w\geq \ve\}$, $\ve>0$ is nonempty.  We can find a compact set of nonzero measure  $K \subset \{w\geq \ve \}$. Now define $w_0=I_{K}$. Clearly $I_K\leq w$.   From  Step 1 we know that (\ref{E}) holds for $w_0$. Invoking Step 2 we deduce that (\ref{E}) holds for  $w$.  Q.E.D.

### The half-life of a theorem, or Arnold's principle at work - MathOverflow

This is a very interesting thread.

The half-life of a theorem, or Arnold's principle at work - MathOverflow

## Tuesday, November 13, 2012

### Supersymmetry in doubt

About a dozen of years ago, at a Great Lakes Conference dinner at Northwestern  I asked  Witten  what parts of high energy physics he thinks will be  confirmed  experimentally in our lifetime.  Super-symmetry was one of the first things he mentioned. Now comes  this news from the Large Hadron Collider casting some doubt on the supersymmetry premise.

BBC News - Popular physics theory running out of hiding places

A word of caution though.   About a year ago people thought that the Large Hadron Collider detected a particle traveling faster than the speed of light.

News - Popular physics theory running out of hiding places

### Influence By Degree, Episode 1

An interesting BBC documentary on how private donors  influence academic life.

BBC World Service - The Documentary, Influence By Degree, Episode 1

## Friday, November 9, 2012

### Tim Gowers recovering after a heart intervention

I'll let the Great Man tell the story.

## Thursday, November 1, 2012

### Analele Stiintifice Iasi

I thought I ought to do some advertising for the math journal published by my Alma Mater.  I am talking of course of the  Analele Stiintifice ale Universitatii  "Al.I. Cuza" Iasi. Sectia Matematica.

## Wednesday, October 31, 2012

### Separable random functions

$\newcommand{\si}{\sigma}$ $\newcommand{\es}{\mathscr{S}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bsT}{\boldsymbol{T}}$ First, what is a random function? To  define it we need a parameter space $\bsT$, $\newcommand{\eS}{\mathscr{S}}$ a  probability space $(\Omega, \eS, P)$, and a target space $X$. Roughly speaking  a  random function $\bsT\to X$ is defined to be a choice of probability measure (and underlying $\si$-algebra of events)  on $X^{\bsT}=$ the space of functions $\bsT\to X$.

In applications $X$ is a metric space and $\bsT$ is a  locally closed subset of some Euclidean space $\bR^N$. (Example to keep in mind:  $\bsT$ an open subset of $\bR^N$ or $\bsT$ a properly embedded submanifold of $\bR^N$. Often $X$ is a vector space.)

A random function on $\bsT$ is then a function

$$f:\bsT\times\Omega\to X,\;\; \bsT\times\Omega\ni (t,\omega)\mapsto f(t,\omega) \in X,$$

such that, for any $t\in\bsT$,   the   correspondence

$$\Omega\ni \omega\mapsto f_t(\omega) :=f(t,\omega)\in X$$

is measurable  with respected to the $\si$-algebra of Borel subsets  of $X$. In other words,  a random function on $\bsT$ is a family of random variables (on the same probability space) parameterized by $\bsT$.

Observe that we have a natural  map $\Phi: \Omega\to X^{\bsT}$,

$$\Omega\ni \omega\mapsto f_\omega\in X^\bsT,\;\;f_\omega(t)=f(t,\omega).$$

The pushforward via $\Phi$ of $(\eS,P)$ induces  structure of probability space on $X^{\bsT}$.  The functions $f_\omega$, $\omega\in \Omega$ are called the   sample functions  of the given random function.

Let us observe that  there are certain  properties of functions which a priori may not  measurable subsets of $\Omega$.  For example the set of $\omega$'s such that $f_\omega$ is continuous on $\bsT$   may not be measurable if $\bsT$   is uncountable.  To deal  with such issues  we will restrict our attention to certain  classes of  random functions, namely the  separable ones.

Definition 1.   Suppose that  $\bsT$ is a locally closed subset  of $\bR^N$ and $X$ is a Polish space, i.e., a  complete, separable  metric space. Fix a countable, dense subset $S\subset \bsT$. A random function $f:\bsT\times\Omega\to X$ is called $S$-separable    if  there exists a negligible subset $N\subset \Omega$, with the following property: for any closed subset $F\subset X$, any open subset $U\subset \bsT$ the  symmetric  difference of the sets

$$\Omega(U,F):=\Bigl\{ \omega\in \Omega;\;\; f_\omega(t)\in F,\forall t\in U\;\Bigr\},\;\;\Omega_S(U, F):=\Bigl\{ \omega\in \Omega;\;\; f_\omega(t)\in F,\;\;\forall t\in U\cap S\;\Bigr\} \tag{1}\label{1}$$

is a subset of $N$, i.e.,

$$\Omega(U,S)\setminus \Omega_S(U,S),\;\;\Omega_S(U,F)\setminus \Omega(U,F)\subset N.$$

Definition 2.  Let $\bsT$ and $X$ be as in  Definition 1.  A random function  $g: \bsT\times \Omega\to X$ is called a  version of the  random function $f:\bsT\times \Omega\to X$  if

$$P(g_t=f_t)=1,\;\;\forall t\in\bsT.$$

Let me   give an application of separability.      We say that a random function $g:\bsT\times \Omega\to X$ is  a.s.  continuous if

$$P\bigl(\;\lbrace \omega;\;\; f_\omega: \bsT\to X\;\;\mbox{is continuous} \rbrace\;\bigr)=1.$$

Proposition 3.   Suppose that $f$ is an $S$-separable  random function $\bsT\times \Omega\to \bR$ and $g$ is a version of $f$.  If $g$ is a.s. continuous, then
$P\bigl(\lbrace \omega; \;\;g_\omega=f_\omega\rbrace\bigr)=1.$
In particular,  $f$ is a.s. continuous.

Proof.    Consider the set $N\subset \Omega$ in the definition of $S$-separability of $f$.   Define

$$\Omega_*:=\bigl\lbrace \;\omega\in \Omega\setminus N;\;\;g_\omega\;\mbox{is continuous},\;\;g(s,\omega)=f(s,\omega),\;\;\forall s\in S\;\bigr\rbrace.$$

Observe that $P(\Omega_*)=1$.   We will prove that
$g_\omega(t)=f_\omega(t),\;\;\forall \omega\in \Omega_*,\;\; t\in\bsT.\tag{\ast}\label{ast}$

Fix an open set $U\subset \bsT$. For any $\omega\in\Omega_*$ set

$M_\omega(U, S):= \sup_{t\in S\cap U} f_\omega(t).$
Invoking the definition of separability with $F=(-\infty, M_\omega(U, S)]$ we deduce  that
$f_\omega(t)\leq M_\omega(U,S),\;\;\forall t\in U,$
so that
$\sup_{t\in U} f_\omega(t)\leq \sup_{t\in S\cap U} f_\omega(t)\leq \sup_{t\in U} f_\omega(t).$
In other words, for any open set $U\subset\bsT$ we have
$\sup_{t\in U}f_\omega(t)=\sup_{t\in S\cap U} f_\omega(t),\;\;\forall \omega\in\Omega_*\tag{2}\label{2}.$
A variation of the above argument shows
$\inf_{t\in U}f_\omega(t)=\inf_{t\in S\cap U} f_\omega(t),\;\;\forall \omega\in\Omega_*\tag{3}\label{3}.$
Now let $\omega\in\Omega_*$,  $t_0\in T$. Given   $\newcommand{\ve}{\varepsilon}$ $\ve>0$, choose  an neighborhood $U=U(\ve \omega)$ of $t_0$ such that
$g_\omega(t_0)-\ve\leq g_\omega(t)\leq g_\omega(t_0)+\ve,\;\;\forall t\in U(\ve,\omega).$
Since $g_\omega(t)=f_\omega(t)$ for $t\in S\cap U$ we deduce from (\ref{2}) and (\ref{3}) that
$g_\omega(t_0)-\ve \leq \inf_{t\in U(\ve,\omega)} f(t) \leq \sup_{t\in U(\ve,\omega)} f_\omega(t)\leq g_\omega(t_0)+\ve.$
In particular, we deduce
$g_\omega(t_0)-\ve \leq f_\omega(t_0)\leq g_\omega(t_0)+\ve,\;\;\forall \ve>0.$
This proves (\ref{ast}).     Q.E.D.

We have the following result.

Theorem 4.  Suppose that $f:\bsT\times \Omega\to X$ is a random function, where $\bsT$ are Polish spaces. If $X$ is compact, then  $f$ admits a separable version.

Proof.     We follow the approach in Gikhman-Skhorohod. $\DeclareMathOperator{\cl}{\mathbf{cl}}$  Fix a countable dense subset $S\subset \bsT$. $\newcommand{\eV}{\mathscr{V}}$ Denote by $\eV$ the collection of open balls in $\bsT$ centered at points in $S$ and with rational radii. For any  $\omega\in\Omega$  and any  open set $U\subset \bsT$ we set

$$R(U,\omega):=\cl\bigl\lbrace f_\omega(t);\;\;t\in S\cap U\,\bigr\rbrace,$$

$$R(t,\omega)=\bigcap_{t\in V\in\eV} R(V,\omega),$$

where $\cl$ stands for the closure of a set.   Observe that $R(t,\omega)\neq\emptyset$ because it is the intersection of a family of compact sets such that any finitely many sets in the family have nonempty intersection.

Lemma 5.  The following statements are equivalent.

(a) The random function  $f$ is $S$-separable.

(b) There exists $N\subset \Omega$ such that $P(N)=0$ and for any $\omega\in\Omega\setminus N$ and any $t\in\bsT$ we have $f_\omega(t)\in R(t,\omega)$.

Proof of the lemma.  (a) $\Rightarrow$ (b)   We know that $f$ is $S$-separable. Choose $N$ as in the definition of $S$-separability.   Fix $\omega_0\in \Omega\setminus N$ and $t_0\in \bsT$.   For any ball $V\in \eV$ that contains $t_0$  we have

$\bigl\lbrace \omega\in \Omega\setminus N;\;\;f_\omega(t)\in R(V,\omega_0)\;\;\forall t\in S\cap B\;\bigr\}=\bigl\lbrace\; \omega\in \Omega\setminus N;\;\;f_\omega(t)\in R(V,\omega_0)\;\;\forall t\in B\;\bigr\rbrace.$
Observe that $\omega_0$ belongs to the set in the left-hand-side of the above equality and so it must belong to the set in the right-hand-side.  Hence

$f_{\omega_0}(t)\in R(B,\omega_0),\;\;\forall t\in B$
and therefore $f_{\omega_0}(t_0)\in R(B,\omega_0)$ for any $B\in\eV$ that contains $t_0$. Thus $f_{\omega_0}(t_0)\in R(t_0,\omega_0)$ which finishes the proof of the implication (a) $\Rightarrow$ (b).

(b) $\Rightarrow$ (a)   Set $\Omega_*=\Omega\setminus N$.  Suppose that $F\subset X$ is closed.  For any $B\in\eV$  and $\omega\in \Omega_*$ we have

$$f_\omega(t)\in F\;\;\forall t\in S\cap B \Leftrightarrow F\supset R(B,\omega).$$

Since $R(t,\omega)\subset R(B,\omega)$ for any $t\in B$ we deduce  that

$$\Omega(B,F):=\bigl\{ \omega\in \Omega_*;\;\; f_\omega(t)\in F\;\;\forall t\in B\;\bigr\} =\bigl\{ \omega\in \Omega_*;\;\; f+\omega(t)\in F\;\;\forall t\in S\cap B\;\bigr\}=\Omega_S(B,F).$$

If $U$  an open set then we can write $U$ as a countable union of balls in $\eV$

$$U=\bigcup_n B_n.$$

Then

$$\Omega(U,F)=\bigcap_n \Omega(B_n, F)=\bigcap_n\Omega_S(B_n, F)= \Omega_S(U,F).$$

This finishes the proof of the lemma.  q.e.d.

Lemma 6.  For any  Borel set $B\subset X$ there exists a countable subset $C_B\subset \bsT$ such that  for any $t\in\bsT$ the set

$$N(t, B):=\bigl\{ \omega\in \Omega; \;\;f_\omega(\tau)\in B,\;\;\forall \tau\in C_B,\;\;f_\omega(t)\in \bsT\setminus B\;\bigr\}$$

has probability $0$.

Proof of the lemma.   We construct $C_B$ recursively.  Choose $\tau_1\in\bsT$ arbitrarily and set $C_B^1:=\{\tau_1\}$.   Suppose that we have  constructed  $C_B^k=\{\tau_1,\dotsc,\tau_k\}$.    Set

$$N_k(t):=\bigl\{\omega;\;\; g_\omega(\tau)\in B\;\;\forall \tau\in C_B^k,\;\;f_\omega(t)\in\Omega\setminus B\;\bigl\},\;\;p_k=\sup_{t\in\bsT} P\bigl(\;N_k(t)\;\bigr).$$

Observe that  $p_1\geq p_2\geq \cdots \geq p_k$. If $p_k=0$   we stop and we set $C_B:=C_B^k$.

If this is not the case, there exists $\tau_{k+1}\in \bsT$ such that

$$P\bigl(N_k(\tau_{k+1}\bigr)\geq \frac{1}{2}p_k.$$

Set $C_B^{k+1}:=C_B^k\cup\{\tau_{k+1}\}$.  Observe that the events $N_1(\tau_2),\dotsc , N_k(\tau_{k+1})$ are mutually  exclusive  and thus

$$1\geq \sum_{j=1}^k P(N_j(\tau_{j+1})) \geq \frac{1}{2}\sum_{j=1}^k p_{j+1}.$$

Hence $\lim_{n\to\infty} p_n=0$. Now  set

$$N(t, B):=\bigcap_{k\geq 1} N_k(t).$$
q.e.d.

Lemma 7.   $\newcommand{\eB}{\mathscr{B}}$ Suppose that $\eB_0$ is a countable family  of Borel subsets of $X$ and $\eB$  is the family obtained by taking the intersections of  all the subfamilies of $\eB_0$.  Then there exists a countable subset $C\subset \bsT$, and for each $t$ a subset $N(t)$ of probability zero  such that for any $B\in \eB$ we have

$$\bigl\{ \omega;\;\; f_\omega(\tau)\in B\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B\;\bigr\}\subset N(t).$$

Proof.   For any $t\in \bsT$ we define

$$C:=\bigcup_{B\in\eB_0}C_B, \;\; N(t) :=\bigcup_{B\in\eB_0} N(t,B),$$

where $C_B$  and $N(t,B)$ are  constructed as in Lemma 6. Clearly $C$ is countable.

If $B'\in\eB$ and $B\in\eB_0$ are such that $B\supset B'$, then

$$\bigl\{ \omega;\;\; f_\omega(\tau)\in B'\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B\;\bigr\}\subset \bigl\{ \omega;\;\; f_\omega(\tau)\in B\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B\;\bigr\}\subset N(t,B)\subset N(t).$$

If

$$B'=\bigcap_{k\geq 1} B_k,\;\;B_k\in\eB_0\;\;\forall k,$$

then

$$\bigl\{ \omega;\;\; f_\omega(\tau)\in B'\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B'\;\bigr\}\subset \bigcup_{k\geq 1} \bigl\{ \omega;\;\; f_\omega(\tau)\in B'\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B_k\;\bigr\}\subset \bigcup_{k\geq 1} N(t, B_k)\subset N(t).$$
q.e.d.

The proof of Theorem 4 is now within reach.  Suppose that $S$ is a countable and dense  set of points in $\bsT$ and $D$ is a countable dense subset of $X$.  Denote by $\eV$ the collection of open balls in $\bsT$ centered at points in $S$. Denote by $\eB_0=\eB_0(D)$ the collection of open balls with  in $X$ of rational radii centered at points in $D$.  As in Lemma 7, denote by $\eB$ the collection of sets obtained by taking intersections of arbitrary families in   $\eB_0$. Clearly $\eB$ contains all the closed subsets of $X$.

Fix a ball $V\in \eV$.  Lemma 7  applied to the restriction of $f$ to $V$ implies the existence of a countable set

$$C(V)\subset V$$

and  of a family of  negligible sets

$$N_V(t)\subset \Omega,\;\;t\in V$$

such that  for any $B\in eB$

$$\{ \omega;\;\; f_\omega(\tau)\in B,\;\;\forall \tau\in C,\;\;f_\omega(t)\in V\setminus B\;\bigr\}\subset N_V(t).$$

Set

$$C=\bigcup_{V\in\eV}C(V),$$

while for $t\in \bsT$ we set

$$N(t):=\bigcup_{\eV\niV\ni t}N_V(t). Clearly C is both countable and dense in \bsT. We can now construct a C-separable version \tilde{f} of f. Define • \tilde{f}_\omega(t)= f_\omega(t) if t\in C or \omega\not\in N(t) • If \omega\in N(t) and t\in \bsT\setminus C we assign \tilde{f}^V_\omega(t) an arbitrary value in R(t,\omega). By construction \tilde{f} is a version of f because for any t\in \bsT$$\{\omega;\;\;\tilde{f}_\omega(t)\neq f_\omega(t)\;\}\subset N(t). $$Since f_\omega(\tau)=\tilde{f}_\omega(\tau) for any \tau\in C, \omega\in \Omega sets R(t,\omega), defined as as in Lemma 5, are the same for both functions \tilde{f} and f. By construction \tilde{f}_\omega(t)\in R(t,\omega), \forall t,\omega. Q.E.D. ## Saturday, October 27, 2012 ### On convolutions Somebody on MathOverlow asked for some intuition behind the operation of convolution of two functions. Here is my take on this. \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} Suppose we are given a function f:\bR\to \bR. Discretize the real axis and think of it as the collection of point \Lambda_\hbar:=\hbar \bZ, where \hbar>0 is a small number. We can then approximate f with its restriction f^\hbar:=f|_{\Lambda_\hbar}. This is determined by its generating function, i.e., the formal power series \newcommand{\ii}{\boldsymbol{i}}$$G^\hbar_f(t)=\sum_{n\in\bZ}f(n\hbar)t^n\in \bR[[t,t^{-1}]]. $$Then$$G^\hbar_{f_0\ast f_1}(t)= G^\hbar_{f_0}(t)\cdot G^\hbar_{f_1}(t).\tag{1} \label{1} $$Observe that if we set t=e^{-\ii\xi \hbar}, then$$G^\hbar_f(t)=\sum_{x\in\Lambda_\hbar} f(x) e^{-\ii \xi x}. $$Moreover$$ \hbar G^\hbar_f(e^{-\ii\xi \hbar})=\sum _{n\in \bZ} \hbar f(n\hbar) e^{-\ii\xi(n\hbar)}, \tag{2}\label{2}$$and the expression in the right hand sum is a "Riemann sum" approximating$$\int_{\bR} f(x)^{-\ii\xi x} dx. $$Above we recognize the Fourier transform of f. If we let \hbar\to 0 in (\ref{2}) and we use (\ref{1}) we obtain the wellknown fact that the Fourier transform maps the convolution to the usual pointwise product of functions. (The fact that this rather careless passing to the limit can be rigorous is what the Poisson formula is all about.) The above argument shows that we can regard \hbar G_f^\hbar(1) as an approximation for \int_{\bR} f(x) dx. Denote by \delta(x) the Delta function concentrated at 0. The Delta function concentrated at x_0 is then \delta(x-x_0). What could be the generating function of \delta(x), G_\delta^\hbar? First, we know that \delta(x)=0, \forall x\neq 0 so that$$G_\delta^\hbar(t) =ct^0=c. $$The constant c can be determined from the equality$$ 1= \int_{\bR} \delta(x) dx=\hbar G_\delta^\hbar(1)=\hbar c$$Hence \hbar G_\delta^\hbar(1)=1. Similarly$$ G^\hbar_{\delta(\cdot-n\hbar)} =\frac{1}{\hbar} t^n. $$In particular, the discretization \delta^\hbar(x-n\hbar) of \delta(x-n\hbar) is the function \Lambda_\hbar\to \bR with value \frac{1}{\hbar} at x=n\hbar and 0 elsewhere. Putting together all of the above we obtain an equivalemn description for the generating functon af a function f:\Lambda_\hbar\to\bR. More precisely$$ G^\hbar_f(t)=\hbar\sum_{\lambda\in\Lambda_\hbar}f(\lambda) G^\hbar_{\delta(\cdot-\lambda)}(t). $$In other words$$f^\hbar= \hbar\sum_{\lambda\in\Lambda_\hbar} f(\lambda)\delta^\hbar_\lambda,\;\;\delta^\hbar_\lambda(\cdot):=\delta^\hbar(\cdot-\lambda). \tag{3}\label{3}$$The last equality suggests an interpretation for the generating function as an algebraic encoding of the fact that f:\Lambda_\hbar\to\bR is a superposition of \delta functions concentrated along the points of the lattice \Lambda_\hbar. The factor \hbar in (\ref{3}) is a discretization of the infinitesimal dx, which indicates that \hbar\delta^\hbar_\lambda should be viewed as a measure. Observe that$$(\hbar\delta^\hbar_\lambda)\ast (\hbar\delta^\hbar_\mu)=\hbar\delta^\hbar_{\lambda+\mu}. \tag{4}\label{4}$$## Tuesday, October 23, 2012 ### Midwest Dynamical Systems Seminar - Department of Mathematics - University of Notre Dame ## Monday, October 22, 2012 ### Mathgen paper accepted! | That's Mathematics! I've just found out from a colleague of this site MathGen which randomly generates math paper. Apparently one such paper has recently been accepted for publication by an Open Access journal, you know, the kind where you pay to have your paper publishe. More details at this site Mathgen paper accepted! | That's Mathematics! Here is the paper MathGen produced on my behalf. ## Friday, October 19, 2012 ### Journal of Gokova Geometry and Topology I thought that you, yes you the guy reading these lines, should have a look at this journal of geometry and topology, Journal of Gokova Geometry Topology. It has a great editorial board and it looks for great papers to publish. ## Thursday, October 18, 2012 ### GmailTeX If you wanted to e-mail math formulas and did not know how, try the link below. GmailTeX ### On an integral geometric formula \newcommand{\bR}{\mathbb{R}} \newcommand{\bsV}{{\boldsymbol{V}}} \DeclareMathOperator{\Graffr}{\mathbf{Graff}^c} \newcommand{\be}{\boldsymbol{e}} \newcommand{\bv}{\boldsymbol{v}} \DeclareMathOperator{\Grr}{\mathbf{Gr}^c} \newcommand{\Gr}{\mathbf{Gr}} \newcommand{\Graff}{\mathbf{Graff}} Suppose that \bsV is a finite dimensional real Euclidean space, M\subset \bsV is a smooth compact submanifold of dimension m and codimension r and we set$$ N:=\dim \bsV=m+r. $$For any nonnegative integer c\leq \dim \bsV we denote by \Graff^c(\bsV) the Grassmannian of affine subspaces of \bsV of codimension c, by \Gr^c(\bsV) the Grassmannian of codimension c vector subspaces of \bsV. We set \Gr_k(\bsV):=\Gr^{N-k}(\bsV). The codimension c Radon transform of a smooth function f: M\to \bR is a function$$ \widehat{f}:\Graff^c(\bsV)\to\bR , $$such that \newcommand{\eH}{\mathfrak{H}}$$\widehat{f}(S) =\int_{S\cap M} f(x) d\eH^{m-c}(x),  \;\; \forall S\in \Graff^c(M), \label{r}\tag{R}$$where d\eH^{m-c} denotes the (m-c)-dimensional Hausdorff measure. If c\leq \dim M then a generic affine plane S\in\Graff^c(\bsV) intersects M transversally in which case the Hausdorff measure in (\ref{r}) is the usual Lebesgue measure induced my the natural Riemann metric on S\cap M. I want to explain how to recover the integral of f over M from its Radon transform. Observe that we have an incidence set \newcommand{\eI}{\mathscr{I}}$$\eI^c(\bsV) :=\Bigl\{ (\bv, S)\in \bsV\times \Graffr(\bsV);\;\; \bv\in S\;\Bigr\} $$equipped with natural projections$$ \bsV\stackrel{\lambda}{\leftarrow}\eI^c(\bsV)\stackrel{\rho}{\to}\Graffr(\bsV).\label{F}\tag{F} $$For any subset X\subset \bsV we define$$\eI^c(X):=\lambda^{-1}(M)\subset \eI^r(X),\;\; \Graffr(X)=\rho\Bigl(\;\eI^r(X)\;\Bigr). $$Note that$$ \Graffr(X)=\Bigl\{ S\in \Graffr(\bsV);\;\; S\cap X\neq \emptyset\;\Bigr\} $$and for any \bv\in\bsV we have$$ \lambda^{-1}(\bv) =\bigl\{ \bv+S;\;\;S\in \Grr(\bsV)\;\bigr\}=\Graffr(\bv)\subset \Graffr(\bsV). $$Observe that \eI^c(V)\to \bsV is a smooth fiber bundle with fiber \Gr^r(\bsV). In particular, \eI^c(M)\to M is the bundle obtained by restricting to the submanifold M. Its fiber is also \Gr^c(\bsV). At this point I need to recall some basic facts described in great detail in Sections 9.1.2, 9.1.3 of Lectures on the Geometry of Manifolds. The Grassmannain \Gr^c(\bsV) is equipped with a canonical O(\bsV)-invariant metric with volume density |d\gamma^c_\bsV| with total volume \newcommand{\sbinom}[2]{\genfrac{[}{]}{0pt}{}{#1}{#2}}$$\int_{\Gr^c(\bsV)} |d\gamma_\bsV^c(L)|=\sbinom{N}{c}, $$where \sbinom{N}{c} is defined in equation (9.1.66) of the Lectures. Now observe that we have a natural projection \pi: \Graff^c(\bsV)\to \Gr^c(\bsV) that associates to each affine plane its translate through the origin. A plane S\in\Graff^c(\bsV) intersects the orthogonal complement of \pi(S) in a unique point C(S)=S\cap \pi(S)^\perp. We obtain a an embeding$$ \Gr^c(\bsV)\ni S\mapsto \bigl(\;C(S), \pi(S)\;\bigr)\in \bsV\times\Gr^c(\bsV),\;\;C(S)\perp \pi(S), $$\newcommand{\eQ}{\mathfrak{Q}} and we will regard \Graff^c(\bsV) as a submanifold of \bsV\times \Gr^c(\bsV). As such, it becomes the total space of a vector bundle \eQ_c\to\Gr^c(\bsV), in fact a subbundle of the trivial bundle \bsV\times \Gr^c(\bsV)\to\Gr^c(\bsV). The orthogonal complement \eQ_c^\perp of this bundle is the tautological vector bundle \newcommand{\eU}{\mathscr{U}} {\eU}^c\to\Gr^c(\bsV). In particular$$\dim\Gr^c(\bsV)= c(N-c)+  c. $$Along \Graff^c(\bsV) we have a canonical vector bundle, the vertical bundle VT\Graff^c(\bsV)\subset T\Graff^c(\bsV) consisting of the kernels of d\pi, i.e., vectors tangent to the fibers of \pi. The vertical bundle is equipped with a natural density |d\bv|_c which when restricted to a fiber of \pi^{-1}(L) induces the natural volume form on the fiber L^\perp viewed as a vector subspace of \bsV. As in Section 9.1.3 of the Lectures we define a product density |d\tilde{\gamma}^c|=|d\tilde{\gamma}_\bsV^c| on  \Graff^c(\bsV),$$|d\tilde{\gamma}_\bsV^c|= |d\bv|_c\times \pi^*|d\gamma_\bsV^{c}| $$Alternatively, the vector bundle \eQ_c, as a subbundle of the trivial bundle \bsV\times \Gr^c(\bsV)\to\Gr^c(\bsV) is equipped with a natural metric connection. The horizontal subbundle HT\eQ_c\subset T\eQ_c is isomorphic to \pi^* T\Gr^c(\bsV) and thus comes equipped with a natural metric. The vertical subbundle VT\eQ_c=VT\Graff^c(\bsV) is also equipped with a natural metric and in this fashion we obtain a metric on \Graff^c(\bsV)=\eQ_c. The density |d\tilde{\gamma}^c_\bsV| is the volume density defined by this metric. Suppose now that c\leq m=\dim M. We denote by \Graff^c_*(M) the subset of \Graff^c(M) consisting of affine planes that intersect M transversally. This is an open subset of \Graff^c(M). The condition c\leq m implies that this set is nonempty. (For c=1 this follows from the fact that the restriction to M of a generic linear function is a Morse function. Then look at iterated slicing by hyperplanes.) Set$$ \eI^c_*(M)= \rho^{-1}\bigl(\;\Graff^c_*(M)\;\bigr)\subset \eI_M $$The fiber of \rho:\eI_*^c(M)\to \Graff^c_*(M) over S\in \Graff^c_*(M) is the submanifold S\cap M which is equipped with a metric density. We obtain a density on \eI^c_*(M)$$ |d\nu^c_M|= |dV_{S\cap M}|\times \rho^*|d\tilde{\gamma}^c|. \tag{$\nu^c$}\label{nu}$$If f: M\to\bR is a smooth function, then$$\int_{\eI^c_*(M)}\lambda^*(f) |d\nu^c_M|=\int_{\Graff^c_*(M)}\left(\int_{S\cap M} f|dV_{S\cap M}\right) |d\tilde{\gamma}^c(S)|. \label{1}\tag{1} $$For any vector subspace U\subset \bsV) we denote by \Gr^c(\bsV)_U the set consisiting of subspaces L\in\Gr^c(\bsV) that intersect U transversely. We now want to integrate \lambda^*(f) along the fibers of \lambda :\eI^c_*(M)\to M. For any vector subspace U\subset \bsV) we denote by \Gr^c(\bsV)_U the set consisting of subspaces L\in\Gr^c(\bsV) that intersect U transversely. The fiber of this map over a point x\in M is an open subset of x+\Gr^c(\bsV)_{T_xM}\subset \Graff^c(\bsV) with negligible complement. The density |d\nu^c_M| on \eI^c_*(M) induces a density$$ |d\nu^c_x|=|d\nu^M|/\lambda^*|dV_M| $$on each fiber \lambda^{-1}(x) and we deduce$$ \int_{\eI^c_*(M)} \lambda^* f|d\nu^c(M)|= \int_M\left(\int_{\lambda^{-1}(x)}|d\nu^c_x|\right) f(x)|dV_N(x)|. \label{2}\tag{2} $$The density |d\nu^c(x)| is the restriction of a density |d\bar{\nu}^c_x| on \Gr^c(\bsV)_{T_xM}. In fact, a reasoning similar to the one in the proof of Lemma 9.3.21 in the Lectures implies that for any U\in\Gr_m(\bsV) there exists a canonical density |d\bar{\nu}^c_U| on \Gr^c(\bsV)_U such that$$ T_*|d\nu^c_U|=|d\bar{\nu^c}_{T(U)}|,\;\;\forall T\in O(\bsV),\;\;U\in \Gr_m(\bsV), \label{3}\tag{3}  |d\bar{\nu}^c_x|=|d\bar{\nu}^c_{T_xM}|,\;\;\forall x\in M. \label{4}\tag{4}$$Using (\ref{3}) (\ref{4}) in (\ref{2}) we deduce that there exists a constant Z=Z(N,m,c) that depends only on N,m,c such that$$ Z(N,m,c)=\int_{\nu^{-1}(x)} |d\bar{\nu}^c_x|,\;\;forall x\in M. $$Using this in (\ref{2}) we conclude from (\ref{1}) that$$ Z(N,m,c)\int_{M}f(x)\; |dV_M(x)| =\int_{\Graff^c(\bsV)}\left(\int_{S\cap M} f(x)|dV_{S\cap M}|\right) |d\tilde{\gamma}^c_\bsV(S)|.\label{5}\tag{5} $$To find the constant Z(N,m,c) we choose M and f judiciously. We let M=\Sigma^m, the unit m-dimensional sphere contained in some (m+1)-dimensional subspace of \bsV. Then, we let f\equiv 1. We deduce from (\ref{5}) that$$ Z(N,m,c)=\frac{1}{{\rm vol}\;(\Sigma^m)} \int_{\Graff^c(\bsV)} {\rm vol}\,(S\cap \Sigma^m)\;|d\tilde{\gamma}^c_\bsV(S)|.\label{6}\tag{6} $$Using the Crofton formula in Theorem 9.3.34 in the Lectures in the special case p=m-c we deduce$$Z(N,m,c)=\sbinom{m}{c}. $$Remark. 1 Consider the Radon transform$$ C_0^\infty(\bsV)\ni f\mapsto  \widehat{f}\in C^\infty\bigl(\;\Graff^c(\bsV)\;\bigr), \;\; \widehat{f}(S)=\int_S f(x)|dV_S(x)|,\;\;\forall S\in \Graff^c(\bsV). $$Observe that \widehat{f} has compact support. Indeed, if the support of f is contained in a ball of radius R, then for any affine plane S\in \Graff^c(\bsV) such that {\rm dist}\,(0,S)>R we have \widehat{f}(S)=0. Consider the dual Radon transform \newcommand{\vfi}{\varphi}$$ C^\infty\bigl(\;\Graff^c(\bsV)\;\bigr)\ni \vfi\mapsto \check{\vfi}\in C^\infty(\bsV),\;\;\check{\vfi}(x)=\int_{\Gr^c(\bsV)} \vfi(x+L)\;|d\gamma^c(L)|,\;\;\forall x\in \bsV. $$Consider the fundamental double fibration (\ref{F}). Given f\in C_0^\infty(\bsV), \vfi\in C^\infty_0\bigl(\;\Graff^c(\bsV)\;\bigr) we obtain a function$$ \Phi=\lambda^*(f)\cdot \rho^*(\vfi)\in C_0^\infty(\bsV) $$Arguing as above, with M=\bsV we observe that \Graff^c_*(\bsV)=\Graff^c(\bsV) and we obtain as in (\ref{nu}) a density |\nu^c_\bsV| on \eI^c_*(\bsV)=\eI^c(\bsV). Denote by \rho_*\Phi |d\nu^c_\bsV| the pushfoward of the density \Phi|d\nu^c_\bsV. It is a density on \Graff^c(\bsV) and we have the Fubini formula (coarea formula)$$ \int_{\eI^c(\bsV)} \Phi(x,S) |d\nu^c_\bsV(x,S)|=\int_{\Graff^c(\bsV)}\rho_*\Phi|d\ni^c_\bsV|\label{7}\tag{7} $$Similarly, we obtain$$ \int_{\eI^c(\bsV)} \Phi(x,S) |d\nu^c_\bsV(x,S)|=\int_{\bsV} \lambda_*\Phi |d\nu^c_\bsV|(x).\label{8}\tag{8} $$From the construction of |d\nu^c_\bsV| we deduce immediately that$$ \rho_*\Phi|d\nu^c_\bsV|(S)=  \widehat{f}(S) \vfi(S) |d\tilde{\gamma}^c|(S). $$From the definitions of |d\nu^c_\bsV|, |d\gamma^c_\bsV| and |d\tilde{\gamma}^c_\bsV| it follows easily that$$ \lambda_*\Phi |d\nu^c_\bsV|(x) =  f(x)\check{\vfi}(x)|dx| $$Using the last equalities in (\ref{7}) and (\ref{8}) we deduce$$ \int_{\bsV} f(x)\check{\vfi}(x)=\int_{\Graff^c(\bsV)} \widehat{f}(S)\vfi(S) |d\tilde{\Gamma}^c(S)|. \tag{D}  \label{d} $$The equality (\ref{d}) shows that the operations f\mapsto \widehat{f} and \vfi\mapsto \widehat{\vfi} are indeed dual to each other. Note also that if we set \vfi\equiv 1  in (\ref{d}) then$$\check{\vfi}(x)={\rm vol}\,\bigl(\;\Gr^c(\bsV)\;\bigr)=\sbinom{N}{c} $$and in this case we reobtain (\ref{5}) in the special case M=\bsV. The equality (\ref{d}) is important for another reason. Denote by C_0^{-\infty}(\bsV) the space of generalized functions with compact supports, then we can extend the Radon transform to such objects. If u\in C_0^{-\infty}(\bsV) then we define its Radon transform \widehat{u} to be the compactly supported generalized density on \Graff^c(\bsV) defined by the equality$$ \langle \widehat{u},\vfi\rangle=\langle u,\check{\vfi}\rangle ,\;\;\forall \vfi\in C^\infty\bigl(\;\Graff^c(\bsV)\;\bigr). $$If M is a compact submanifold of \bsV, then we get a Dirac-type generalized function \delta_M on \bsV defined by integration along M with respect to the volume density on M determined by the induced metric. Then$$ \langle\widehat{\delta}_M,\vfi\rangle =\int_M  \check{\vfi}(x) |dV_M(x)|,\;\;\forall  C^\infty\bigl(\;\Graff^c(\bsV)\;\bigr). $$The generalized function \widehat{\delta}_M is represented by a locally integrable function$$\widehat{\delta}_M(S) =\eH^{m-c}(M\cap S),\;\;\forall S\in \Graff^c(M). $$## Saturday, October 13, 2012 ### An introduction to the Laplace method in the asymptotics of integrals \newcommand{\bR}{\mathbb{R}} There are many sources explaining this old technique of determining the asymptotic behavior of certain integrals depending on a small parameter \hbar. However, in applications, the integrals do not quite fit the setup described in most books I have consulted and I thought it would be nice to present the general strategy. What follows is folklore, and even not the most general possible result, but I took great pain to highlight the key features one should look for when attempting to use the Laplace technique in a concrete case. Consider an interval (a,b)\subset \bR and a family of C^2-functions$$\phi_\hbar: (a,b)\to \bR,\;\;\hbar>0,  $$where the interval (a,b) could be finite, or infinite. We are interested in the behavior of the integral$$I_\hbar:=\int_a^b e^{-\phi_\hbar(t)} dt $$as \hbar \searrow 0 given that the functions \phi_h satisfy certain conditions \mathbf{C}_1 For any \hbar>0 the function \phi_\hbar has a unique critical point \tau=\tau(\hbar)\in (a,b). Moreover, \phi_\hbar''(\tau)>0. In other words, \tau is a nondegenerate local minimum, and the uniqueness assumption implies that \phi_\hbar achieves its absolute minimum at \tau. We set \newcommand{\si}{\sigma}$$ \si=\si(\hbar):=    \frac{1}{\sqrt{\phi_\hbar''(\tau)}}. $$\mathbf{C}_2 The numers \tau(\hbar) and \si(\hbar) satisfy the conditions$$ \lim_{\hbar\to 0}\frac{\tau(\hbar)-a}{\si(\hbar)}=\lim_{\hbar\to 0}\frac{ b-\tau(\hbar)}{\si(\hbar)}=\infty. $$\mathbf{C}_3$$\lim_{\hbar\to 0}\bigl(\,\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)\,\bigr)=\frac{x^2}{2},\;\;\forall x\in\bR $$\mathbf{C}_4 There exists u:\bR\to \bR such that$$\int_{\bR}e^{-u(x)} dx <\infty\;\;\mbox{and}\;\;\phi_\hbar(\tau+ \si x)-\phi_\hbar(\tau)\geq u(x),\;\;\forall \hbar,\;\;x\in J(\hbar). $$Then, under the assumptions \mathbf{C}_1,\dotsc,\mathbf{C}_4 we have$$ I_\hbar \sim\sqrt{2\pi} \si e^{-\phi_\hbar(\tau)}\;\;\mbox{as $\hbar\to 0$}. \label{A}\tag{A}  $$Proof of (\ref{A}). We make the change of variables t=\tau+\si x in the integral I_\hbar to conclude that$$I_\hbar=\si e^{-\phi_\hbar(\tau)}\int_{J(\hbar)} e^{-\bigl(\;\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)\;\bigr)} dx, $$where$$ J(\hbar)= \Bigl[\frac{a-\tau}{\si},\frac{b-\tau}{\si}\Bigr]. $$The condition \mathbf{C}_2 implies that the intervals J(\hbar) expand to \bR as \hbar \to 0 Set \newcommand{\vfih}{\varphi_\hbar}$$\vfih(x): =\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau). $$\mathbf{C}_3 implies that$$\vfih(x)\to\frac{x^2}{2}\;\;\mbox{as $\hbar\to 0$} ,\;\;\forall x\in\bR. $$We can now use \mathbf{C}_4 to invoke the dominated convergence theorem and conclude that$$\lim_{\hbar\to 0} \int_{J(\hbar)} e^{-\vfih(x)} dx=\int_{\bR}e^{-\frac{x^2}{2}}=\sqrt{2\pi}. $$This completes the proof of (\ref{A}). Remark 1. (a) Often in applications each of the functions \phi_\hbar is convex. In such cases the bound \mathbf{C}_4 is a consequence of the bound$$ \bigl|\;\phi'_\hbar(\tau\pm \si)\;\bigr|= O\Bigl(\frac{1}{\si}\Bigr) \;\;\mbox{as $\hbar\to 0$}\label{B}\tag{B}. $$Indeed, \vfih is convex and thus its graph its situated above either of the tangent lines at x=\pm 1. Thus$$ \vfih(x) \geq \max\Bigl\{ \vfih'(1)(x-1)+\vfih(1),\;\;\vfih'(-1)(x+1) +\vfih(-1)\Bigr\}. $$Observing that$$\vfih(\pm 1)>\vfih(0)=0,\;\;0< \pm \vfih'(\pm 1)= \pm \si\phi_\hbar(\tau\pm \si)=O(1), $$we deduce that in \mathbf{C}_4 we can choose u(x) of the form u(x)=C(|x|-1) for some positive constant C. (b) Both conditions \mathbf{C}_3 and \mathbf{C}_4 would follow immediately if one can prove that there exists a C^1-function$$\Psi:[0,\infty)\times \bR\to \bR,\;\;(\hbar,x)\mapsto \Psi(\hbar,x), $$such that$$\Psi(0,x)=\frac{x^2}{2},\;\;\Psi(\hbar,x)=\vfih(x),\;\;\forall x\in J(\hbar),\;\;\forall \hbar >0. $$(c) The esence of the above results is that, under appropriate assumptions, we can replace \phi_\hbar(t) with its 2-nd order jet at \tau$$ \phi_\hbar(t)\approx \phi_\hbar(\tau)+\frac{(t-\tau)^2}{2\si^2} $$and deduce that$$ I_\hbar\sim \int_a^b e^{-\phi_\hbar(\tau)-\frac{(t-\tau)^2}{2\si^2}} dt,\;\;\mbox{as $\hbar \to 0$} $$Example 1. Let me illustrate how the above strategy works in the classical situation described in all the books on asymptotics of integrals. Consider a C^2 convex function \phi:(-a, a)\to\bR with a unique minimum at \tau =0 and such that \phi''(0)>0. Set \phi_\hbar(t)=\frac{1}{\hbar}\phi(t) so that$$I_\hbar=\int_{-a}^ae^{-\frac{1}{\hbar}\phi(t)} dt. $$In this case$$\tau=0,\;\; \si =\sqrt{\frac{\hbar}{\phi''(0)}}. $$The conditions \mathbf{C}_1,\mathbf{C}_2 are obviously satisfied. As for \mathbf{C}_3 we observe that in this case we have$$\vfih(x)=\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)=\frac{1}{\hbar}\bigl(\,\phi(\si x)-\phi(0)\;\bigr) = \frac{1}{\hbar}\Bigl(\;\phi'(0)\si x+ \frac{\phi''(0)}{2}\si^2 x^2+ o(\si^2)\;\Bigr)= \frac{x^2}{2}+ o(1). $$Condition (\ref{B}) is also satisfied since$$\vfih(\pm 1)=\frac{1}{\sqrt{\hbar\phi''(0)}}\phi'\Bigl(\pm \sqrt{\frac{\hbar}{\phi''(0)}}\Bigr) \to \pm 1\;\;\mbox{as $\hbar\to 0$}. $$Hence we conclude that$$\int_{-a}^a e^{-\frac{1}{\hbar}\phi(t)} dt \sim\sqrt{\frac{2\pi\hbar}{\phi''(0)}}\;\;\mbox{as $\hbar\to 0$}. $$Example 2. Consider the integral$$ \Gamma(\lambda +1)=\int_0^\infty  t^\lambda e^{-t}  dt,\;\;\lambda \to  \infty. $$Observing that it has the form I_\hbar where a=0, b=\infty, \hbar=\frac{1}{\lambda} and$$\phi_\hbar(t)= t-\lambda \log t. $$In this case we have$$\phi'_\hbar(t)=1-\frac{\lambda}{t},\;\; \phi_\hbar(t)=\frac{\lambda}{t^2}, \;\; \tau(\lambda)=\lambda,\;\;\si(\lambda)=\frac{1}{\sqrt{\lambda}}. $$thus, the conditions (\mathbf{C}_1) and (\mathbf{C}_2) are satisfied. To verify (\mathbf{C}_3) observe that$$\vfih(x)=\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)= (\lambda +\sqrt{\lambda}x)-\lambda\log(\lambda+\sqrt{\lambda}x)- \lambda -\lambda \log \lambda =\sqrt{\lambda} x-\lambda\log\Bigl(1+\frac{x}{\sqrt{\lambda}}\Bigr). $$The condition \mathbf{C}_3 now follows from the Taylor expansion of \log(1+s) at s=0. To prove \mathbf{C}_4 we observe that in this case \phi_\hbar is continuous, so it suffices to check (\ref{B}), i.e., \vfih(\pm 1)=O(1). In this case we have$$\vfih'(\pm1)=\sqrt{\lambda}-\lambda\log\Bigl(1\pm \frac{1}{\sqrt{\lambda}}\Bigr), $$and (\ref{B}) follows by using the Taylor expansion of \log(1+s) at s=0. In this case$$e^{-\phi_\hbar(\tau)}=\lambda^\lambda e^{-\lambda}$$and we deduce$$ \Gamma(\lambda+1)\sim \sqrt{2\pi}\lambda^{\lambda-\frac{1}{2}}e^{-\lambda}\;\;\mbox{as $\lambda\to\infty$}. $$Example 3. Suppose that w:[0,\infty)\to\bR is a smooth function such that$$w(t), \;w'(t),\;\;w''(t) >0,\;\;\forall t>T>1. $$Then$$\mu_\lambda=\int_0^\infty t^\lambda e^{-w(t)} dt <\infty,\;\;\forall \lambda >0$$and I would like to investigate the behavior of \mu_\lambda as \lambda\to \infty. The quantitites \mu_k, k\in\mathbb{Z}_{\geq 0}, are the moments of the measure e^{-w(t)}dt on the positive semiaxis. Note that$$\mu_\lambda=\int_0^T t^\lambda e^{-w(t)} dt+\int_T^\infty t^\lambda  e^{-w(t)} dt. $$Observe that$$ \int_T^\infty t^\lambda e^{-w(t)} dt \geq T^\lambda\int_T^\infty e^{-w(t)} dt, $$while$$ T^{-\lambda} \int_0^T t^\lambda e^{-w(t)} dt= \int_0^T \left(\frac{t}{T}\right)^\lambda e^{-w(t)} dt \to 0\;\;\mbox{as $\lambda\to \infty$}. $$Thus, as \lambda \to \infty we have$$\mu_\lambda\sim I_\lambda:=\int_T^\infty t^\lambda e^{-w(t)}  dt. $$Observe that$$I_\lambda =\int_T^\infty e^{-\phi_\lambda(t)} dt,\;\;\phi_\lambda(t)=w(t)-\lambda\log t. $$We will show that the Laplace method is applicable in this case if we assume that we have an asymptotic estimate \newcommand{\bZ}{\mathbb{Z}}$$ tw'(t)\sim A t^\alpha(\log t)^p\;\;\mbox{as $t\to\infty$},\;\; A>0\;\;,\alpha>1,\;\;p\in {\bZ}_{\geq 0}\tag{$\ast$} \label{ast} $$which is twice differentiable. The critical points of \phi_\lambda are solutions of the equality$$\lambda=tw'(t). $$Since tw'(t) is increasing and tw'(t)\to\infty as t\to \infty we deduce that the above equation has a unique solution \tau=\tau(\lambda) for \lambda \gg 0. The correspondence \lambda\to \tau(\lambda) is smooth and \tau(\lambda)\to \infty as \lambda\to \infty. This proves \mathbf{C_1}. In view of (\ref{ast}) we deduce that$$\lambda(\tau)\sim At^\alpha(\log \tau)^p \;\;\mbox{as $\tau\to \infty$}. $$Observe that$$\phi_\lambda''(\tau)=w''(\tau)+\frac{\lambda}{\tau^2} = w''(\tau)+\frac{w'(\tau)}{\tau}=\frac{\tau w''(\tau)+w'(\tau)}{\tau}. $$Hence$$\si =\sqrt{\frac{\tau}{\tau w''(\tau)+w'(\tau)}}=\sqrt{\frac{\tau}{\lambda'(\tau)},}\;\;\frac{\tau}{\si}= \sqrt{\tau \lambda'(\tau)}=\sqrt{\tau^2w''(\tau) +\tau w'(\tau)}\to \infty. $$This proves \mathbf{C_2}. Observe$$ \phi_\lambda(\tau+\si x)-\phi_\lambda(\tau)= w(\tau+ \si x)-w(\tau) -\lambda\log \Bigl(1+\frac{\si}{\tau}x \Bigr) = w(\tau+ \si x)-w(\tau) -\tau w'(\tau)\log \Bigl(1+\frac{\si}{\tau}x \Bigr) =\si^2\left( w''(\tau) +\frac{w'(\tau)}{\tau}\right)\frac{x^2}{2} +\frac{\si^3x^3}{3!}w^{(3)}(\theta)  + O\left(\frac{\si^3w'(\tau) x^3}{\tau^2}\right),$$for some \theta=\theta(\tau,x)\in [\tau,\tau+\si x]. Now observe that$$ \frac{\si^3 w'(\tau)}{\tau^2}=\frac{\si}{\tau}  \frac{\si^2w'(\tau)}{\tau}=\frac{\si}{\tau}\frac{w'(\tau)}{\tau w''(\tau)+ w'(\tau)}\leq \frac{\si}{\tau}\to 0,\;\;\mbox{as $\tau\to \infty$}. \tag{1}\label{2}$$To verify \mathbf{C}_3 we need to prove that$$\lim_{t\to \infty}\si^3w^{(3)}(\theta)=0.\tag{2} \label{1} $$It is time to use (\ref{ast}). We have$$w'(t)+ tw''(t) \sim   A\alpha t^{\alpha-1}(\log t)^p+ Apt^{\alpha-1} (\log t)^{p-1}, 2w''(t) +tw^{(3)}(t)\sim A\alpha(\alpha-1)t^{\alpha-2}(\log t)^p+A\alpha pt^{\alpha-2}(\log t)^{p-1}+Ap(\alpha-1) t^{\alpha-2} (\log t)^{p-1}+ Ap(p-1)t^{\alpha-2}(\log t)^{p-2}. $$Now observe that$$ t^3w^{(3)}(t)\sim -2t^2 w''(t)+A\alpha(\alpha-1)t^{\alpha}(\log t)^p+Ap t^\alpha(2\alpha-1) (\log t)^{p-1}+ Ap(p-1)t^{\alpha}(\log t)^{p-2}, $$and$$ -2t^2w''(t)\sim  2tw'(t) -2A\alpha t^{\alpha}(\log t)^p-2 Apt^{\alpha} (\log t)^{p-1} = -2A(\alpha-1)t^\alpha(\log t)^p -2Apt^\alpha (\log t)^{p-1}.\tag{3}\label{3}$$Hence$$t^3w^{(3)} (t) \sim  At^{\alpha}\Bigl(\;(\alpha-1)(\alpha-2)(\log t)^p + p(2\alpha-3)(\log t)^{p-1}+ p(p-1)(\log t)^{p-2}\;\Bigr).$$Also$$\frac{\si}{\tau}=\frac{1}{\sqrt{\tau\lambda'(\tau)}} \sim \frac{1}{\sqrt{A\alpha}}\tau^{-\frac{\alpha}{2}}(\log \tau )^{-\frac{p}{2}},\;\;\si \sim \frac{1}{\sqrt{A\alpha}}\tau^{1-\frac{\alpha}{2}}(\log \tau )^{-\frac{p}{2}}. \tag{4}\label{4}$$We distinguish three cases. Case 1. \alpha \neq 2 In this case \si\to 0 and \newcommand{\ve}{\varepsilon}$$\theta= \tau(1 +c\frac{\si}{\tau}x) w^{3}(\theta) =Ct^{\alpha-3} (\log t)^p\bigl( \;1+\ve(t)\;\bigr), $$where C=A(\alpha-1)(\alpha-2)\neq 0 and \ve(t)\to 0 as t\to \infty. Then$$ \si^3w^{(3)}(\theta) \sim\frac{C}{\sqrt{A\alpha}} t^{-\frac{\alpha}{2}} (\log t)^{-\frac{p}{2}} \to 0. $$This proves (\ref{1}). Case 2. \alpha=2 p=1 In this case (\ref{ast}) implies$$\lim_{t\to\infty}w^{(3)}(t)=0,\;\;\si(t)=O(1)\;\;\mbox{as $t\to\infty$}. $$which clearly implies (\ref{1}). Case 3. \alpha=2, p>0 Proceed as in Case 1. Finally, we want to check \mathbf{C}_4. In our case \phi_h is convex and we will verify (\ref{B}). We have$$\vfih(x) = \frac{d}{dx}\Bigl( w(\tau+\si x) -w(\tau)-\tau w'(\tau)\log\Bigl(1+\frac{\si}{\tau}x\Bigr)\;\Bigr) =\si w'(\tau+\si x)-\frac{\si w'(\tau)}{1+\frac{\si}{\tau}x}=\si w'(\tau)\left(1-\frac{1}{1+\frac{\si}{\tau}x}\right)+\frac{1}{2}\si^2x^2w''(\theta)=\frac{\si^2 w'(\tau)}{\tau}x^2+ \frac{1}{2}\si^2x^2w''(\theta)+ O\left( \frac{\si^3w'(\tau)}{\tau^2} x^2\right), $$for some \theta\in [\tau,\tau+\si]. As in the proof of (\ref{2}) we deduce that$$ \frac{\si^2w'(\tau)}{\tau}=O(1),\;\; \frac{\si^3w'(\tau)}{\tau^2}=o(1). $$We only need to verify that$$ \si^2w''(\theta)=O(1). 

This follows from  (\ref{3}) and (\ref{4}}).