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 $\eV$

$$U=\bigcup_n B_n. $$


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

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

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

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

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


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


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

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 $B\in eB$

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



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 $\newcommand{\ii}{\boldsymbol{i}}$

$$G^\hbar_f(t)=\sum_{n\in\bZ}f(n\hbar)t^n\in \bR[[t,t^{-1}]]. $$


$$G^\hbar_{f_0\ast f_1}(t)= G^\hbar_{f_0}(t)\cdot G^\hbar_{f_1}(t).\tag{1} \label{1} $$

Observe that if  we   set $t=e^{-\ii\xi \hbar}$, then

$$G^\hbar_f(t)=\sum_{x\in\Lambda_\hbar} f(x) e^{-\ii \xi x}. $$


$$ \hbar G^\hbar_f(e^{-\ii\xi \hbar})=\sum _{n\in \bZ} \hbar f(n\hbar) e^{-\ii\xi(n\hbar)}, \tag{2}\label{2}$$

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

$$(\hbar\delta^\hbar_\lambda)\ast (\hbar\delta^\hbar_\mu)=\hbar\delta^\hbar_{\lambda+\mu}. \tag{4}\label{4}$$

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


 If you wanted to e-mail math formulas and did  not know how, try the link below.


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

$$ \Gr^c(\bsV)\ni S\mapsto \bigl(\;C(S), \pi(S)\;\bigr)\in \bsV\times\Gr^c(\bsV),\;\;C(S)\perp \pi(S), $$
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.)


$$ \eI^c_*(M)= \rho^{-1}\bigl(\;\Graff^c_*(M)\;\bigr)\subset \eI_M $$

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

$$ \lim_{\hbar\to 0}\frac{\tau(\hbar)-a}{\si(\hbar)}=\lim_{\hbar\to 0}\frac{ b-\tau(\hbar)}{\si(\hbar)}=\infty. $$


$$\lim_{\hbar\to 0}\bigl(\,\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)\,\bigr)=\frac{x^2}{2},\;\;\forall x\in\bR $$

$\mathbf{C}_4$ 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

$$I_\hbar=\si e^{-\phi_\hbar(\tau)}\int_{J(\hbar)} e^{-\bigl(\;\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)\;\bigr)} dx, $$


$$ J(\hbar)= \Bigl[\frac{a-\tau}{\si},\frac{b-\tau}{\si}\Bigr]. $$

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

$$w(t), \;w'(t),\;\;w''(t) >0,\;\;\forall t>T>1. $$


$$\mu_\lambda=\int_0^\infty t^\lambda e^{-w(t)} dt <\infty,\;\;\forall \lambda >0$$

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

$$ \int_T^\infty t^\lambda e^{-w(t)} dt \geq T^\lambda\int_T^\infty e^{-w(t)} dt, $$


$$ T^{-\lambda} \int_0^T t^\lambda e^{-w(t)} dt= \int_0^T \left(\frac{t}{T}\right)^\lambda e^{-w(t)} dt \to 0\;\;\mbox{as $\lambda\to \infty$}. $$

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

$$\phi_\lambda''(\tau)=w''(\tau)+\frac{\lambda}{\tau^2} = w''(\tau)+\frac{w'(\tau)}{\tau}=\frac{\tau w''(\tau)+w'(\tau)}{\tau}. $$


$$\si =\sqrt{\frac{\tau}{\tau w''(\tau)+w'(\tau)}}=\sqrt{\frac{\tau}{\lambda'(\tau)},}\;\;\frac{\tau}{\si}= \sqrt{\tau \lambda'(\tau)}=\sqrt{\tau^2w''(\tau) +\tau w'(\tau)}\to \infty. $$

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

$$ t^3w^{(3)}(t)\sim -2t^2 w''(t)+A\alpha(\alpha-1)t^{\alpha}(\log t)^p+Ap t^\alpha(2\alpha-1) (\log t)^{p-1}+ Ap(p-1)t^{\alpha}(\log t)^{p-2}, $$


$$ -2t^2w''(t)\sim  2tw'(t) -2A\alpha t^{\alpha}(\log t)^p-2 Apt^{\alpha} (\log t)^{p-1} = -2A(\alpha-1)t^\alpha(\log t)^p -2Apt^\alpha (\log t)^{p-1}.\tag{3}\label{3}$$

$$t^3w^{(3)} (t) \sim  At^{\alpha}\Bigl(\;(\alpha-1)(\alpha-2)(\log t)^p + p(2\alpha-3)(\log t)^{p-1}+ p(p-1)(\log t)^{p-2}\;\Bigr).$$

$$\frac{\si}{\tau}=\frac{1}{\sqrt{\tau\lambda'(\tau)}} \sim \frac{1}{\sqrt{A\alpha}}\tau^{-\frac{\alpha}{2}}(\log \tau )^{-\frac{p}{2}},\;\;\si \sim \frac{1}{\sqrt{A\alpha}}\tau^{1-\frac{\alpha}{2}}(\log \tau )^{-\frac{p}{2}}. \tag{4}\label{4}$$

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}}).