## 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}}).