\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 77, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2003 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/77\hfil The Kolmogorov equation]
{The Kolmogorov equation with time-measurable coefficients}

\author[Jay Kovats\hfil EJDE--2003/77\hfilneg]
{Jay Kovats}

\address{Jay Kovats \newline
Department of Mathematical Sciences \\
Florida Institute of Technology \\
Melbourne, FL 32901, USA}
\email{jkovats@zach.fit.edu}

\date{}
\thanks{Submitted March 11, 2003. Published July 13, 2003.}
\subjclass[2000]{35K15, 35B65, 35K15, 60J60}
\keywords{Diffusion processes, Kolmogorov equation, Bellman equation}


\begin{abstract}
 Using both probabilistic and classical analytic techniques,
 we investigate the parabolic Kolmogorov equation
 $$
 L_t v +\tfrac {\partial v}{\partial t}\equiv \frac 12 a^{ij}(t)v_{x^ix^j}
 +b^i(t) v_{x^i} -c(t) v+ f(t) +\tfrac {\partial v}{\partial t}=0
 $$
 in $H_T:=(0,T)  \times E_d$ and its solutions when the coefficients are
 bounded Borel measurable functions of $t$. We show that the probabilistic
 solution $v(t,x)$ defined in $\bar H_T$, is twice differentiable
 with respect to $x$, continuously in $(t,x)$, once differentiable with respect
 to $t$, a.e. $t \in [0,T)$ and satisfies the Kolmogorov equation
 $L_t v +\frac {\partial v}{\partial t}=0$ a.e. in $\bar H_T$. Our main tool
 will be the Aleksandrov-Busemann-Feller Theorem. We also examine the
 probabilistic solution to the fully nonlinear Bellman equation with
 time-measurable coefficients in the simple case $b\equiv 0,\,c\equiv 0$.
 We show that when the terminal data function is a paraboloid, the payoff
 function has a particularly simple form.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}
It is well-known in the theory of diffusion processes \cite{K1,K2}
that when $g\in C^2(E_d)$ and the coefficients $a(t,x)$, $b(t,x)$, $c(t,x)$
and free term $f(t,x)$ are sufficiently smooth in $(t,x)$ and satisfy certain
growth conditions, with $c(t,x)\geq 0$, then the function
\begin{equation}
\begin{gathered}
v(t,x)=\mathbf{E} \Bigl[ \int_t^T f(r,\xi_r (t,x))
e^{-\varphi_r(t,x)} \,dr + e^{-\varphi_T(t,x)} g(\xi_T(t,x))\Bigr],\\
 \varphi_s(t,x)=\int_t^s c(r,\xi_r(t,x))\,dr
\end{gathered} \label{e1.1}
\end{equation}
belongs to $C^{1,2}(H_T)$ and satisfies the Kolmogorov equation
$L v (t,x)+ \frac {\partial v}{\partial t}(t,x) =0, \forall\,(t,x)\in \bar H_T$,
where $L v :=\frac 12 a^{ij}(t,x) v_{x^i x^j} +b^i(t,x) v_{x^i} -c(t,x) v + f(t,x)$,
with $v(T,x)=g(x)$. In \eqref{e1.1}, for fixed $(t,x) \in \bar H_T$, 
$\omega \in \Omega$ and $s\ge t$, $\xi_s(t,x)=\xi_s(\omega,t,x)$ is the 
solution of the stochastic
equation $\xi_s=x +\int_t^s \sigma(r,\xi_r)\,d\mathbf{w}_r +\int_t^s b (r,\xi_r)\,dr$,
where $(\Omega,\mathcal{F}, P)$ is a complete probability space on which
$(\mathbf{w}_t,\mathcal{F}_t)$ is a $d_1$-dimensional Wiener process, defined for $t\ge 0$.
Furthermore, $\sigma (t,x)$ and $b(t,x)$ are assumed continuous in $(t,x)$ and have
values in the set of $d \times d_1$ matrices, $E_d$ respectively,
 with $a=\sigma\sigma^*$. The fact that the probabilistic solution $v$ satisfies
the Kolmogorov equation throughout $\bar H_T$ is proved using It\^o's formula and
relies heavily on the continuity in $t$ of the coefficients to establish the
existence and continuity in $(t,x)$ of $\frac {\partial v}{\partial t}$
\cite[Chapter 5]{K2}. In this paper, we show that if the coefficients are only
bounded Borel measurable functions of $t$, the second derivatives $v_{x^ix^j}(t,x)$
exist and are continuous in $(t,x)$ (Theorem \ref{thm2.1}) but in general,
$\frac {\partial v}{\partial t}$ exists only in the generalized sense
(Theorem \ref{thm2.3})
and the Kolmogorov equation will be satisfied only in the almost everywhere sense
(Theorem \ref{thm2.5}). For example, consider the function
$v(t,x)=|x|^2 +2d (\frac 12 -t)_+$. For $t\neq \frac 12$,
$\frac {\partial v}{\partial t}(t,x)$ exists and equals
$-2d \, I_{0\le t<\frac 12}$ and hence for $t\neq \frac 12$, $v$ is a solution of
the degenerate equation
$I_{0\le t<\frac 12}\Delta v + \frac {\partial v}{\partial t} =0$ in
$[0,1)\times E_d$. Note $\frac {\partial v}{\partial t}(t,x)$ is discontinuous in
$t$.

When the coefficients and free term are independent of $x$, the
right hand side of our stochastic equation is independent of
$\xi_.$ and the probabilistic solution \eqref{e1.1} takes a decidedly more
convenient form (see \eqref{e3.3}). Since the other terms in \eqref{e3.3} are
independent of $x$ and their derivatives with respect to $t$ can
be explicitly calculated (almost everywhere) it suffices to
investigate the function $v(t,x)=\mathbf{E} g(\xi_T(t,x))$.


We do this in two ways. In section 1, we use probabilistic
arguments to show that for $g\in C^2(E_d)$, the function
$v(t,x)=\mathbf{E} g(\xi_T(t,x))$ is twice differentiable with
respect to $x$, continuously in $(t,x)$ and once differentiable
with respect to $t$, a.e. $t\in [0,T)$. We then apply the
Aleksandrov-Busemann-Feller theorem to a variant of $v$ to show
that $v$ satisfies the Kolmogorov equation $\frac 12
a^{ij}(t)v_{x^ix^j} +b^i(t) v_{x^i} +\frac {\partial v}{\partial
t}=0$ a.e. in $H_T$. From this it follows (by our previous remark)
that the simplified version of \eqref{e1.1}, given by \eqref{e3.3} satisfies the
more general Kolmogorov equation a.e. in $H_T$. In section 2, we
use the fact that $\xi_T(t,x)$ is a Gaussian vector to express $v$
as a convolution (in $x$) of $g$ with a kernel $p$ which is the
fundamental solution of the Kolmogorov equation (a.e. $t$). Our
proof that this convolution satisfies the Kolmogorov equation
amounts to showing that we can differentiate the kernel under the
integral sign. Here we assume only that $g$ is continuous and
slowly increasing, that is $|g(x)|\le C_1e^{C_2|x|^2}$. Our
derivative estimates are done under the assumption that the
coefficient matrix $a(t)$ is non-degenerate. This assumption was
not needed in section 1, (due to the assumption $g\in C^2(E_d)$)
yet we do get a slightly more refined result here, namely
$v(t,x)=\mathbf{E} g(\xi_T(t,x))$ satisfies the Kolmogorov equation
for almost every $t\in [0,T)$ and any $x\in E_d$. Finally in
section 4, we examine the payoff function for the fully nonlinear
Bellman equation in the simple case $b\equiv 0,\,c\equiv 0$. It
turns out that when $g$ is a paraboloid, the probabilistic
solution of the Bellman equation has a particularly simple form.


\section{The Probabilistic Approach}
Throughout this section, we assume the following.

Let $g \in C^2(E_d)$ and assume that for all $x,y \in E_d$,
$|g(x)|,\,|g_{(y)}(x)|,\,|g_{(y)(y)}(x)|\le K(1+|x|^m)$, where for
any twice differentiable function $u(x)$ and $l \in E_d$,
$u_{(l)}(x)= |l|^{-1} u_{x}(x)\cdot l, u_{(l)(l)}(x)=|l|^{-2} l^*
u_{xx}(x)l$. For $t\in [0,T]$ and $x\in E_d$, we define, for
$s\in [t,T]$, the diffusion process $\xi_s (t,x)= x +\int_t^s
\sigma(r)\, d\mathbf{w}_r +\int_t^s b(r)\, dr$, where the Borel
measurable coefficients $\sigma(t), \,b(t)$ are defined on
$[0,T]$, independent of $\omega \in \Omega$ and satisfy
\begin{equation}
\int_0^T \left[ \|\sigma (t)\|^2 +|b(t)|\right]\,dt <\infty.
\label{e2.1}
\end{equation}
Under these assumptions, we prove our first theorem.

\begin{theorem} \label{thm2.1}
For $(t,x) \in \bar H_T$, the function $v(t,x)=\mathbf{E} g(\xi_T(t,x))$ is twice
differentiable with respect to $x$, continuously in $(t,x)$ and for any
$y, \bar y \in E_d$, $ v_{y\bar y}(t,x)= \mathbf{E} g_{y\bar y}(\xi_T(t,x))$.
\end{theorem}

\begin{proof}
We show that $v(t,x)$ is differentiable with respect to $x$.
Writing $\xi_T (t,x) = x + \eta_T(t)$, where
$\eta_T(t):= \int_t^T \sigma(r)\, d\mathbf{w}_r +\int_t^T b(r)\, dr$, note that
for any $y \in E_d$ and any sequence $h_n\to 0$ as $n\to \infty$
$$
\Delta^1_{h_n,y}v(t,x):=\frac {v(t,x+h_ny) -v(t,x)}{h_n}
= \mathbf{E} \Delta^1_{h_n,y} g ( x+\eta_T(t))
=\mathbf{E} \Delta^1_{h_n,y} g ( \xi_T(t,x)).
$$
Since $g_{y}$ is continuous, the Mean Value Theorem yields
$$
\Delta^1_{h_n,y} g ( \xi_T(t,x))=\int_{0}^1 g_{y}
(\xi_T(t,x) +rh_ny)\,dr=g_{y} (\xi_T(t,x))+\theta h_ny),
$$
for some $\theta\in [0,1]$.
Since $g \in C^1(E_d)$, \,$\Delta^1_{h_n,y} g ( \xi_T(t,x))\to g_{y}(\xi_T(t,x))$ as $n\to \infty$. Furthermore,
as $n\to \infty$
\begin{equation}
\mathbf{E} \Delta^1_{h_n,y} g (\xi_T(t,x)) \to \mathbf{E}
g_{y}(\xi_T(t,x)).\label{e2.2}
\end{equation}
To see this observe that
\begin{align*}
|\Delta^1_{h_n,y} g (\xi_T(t,x))|
&=|g_{y} (\xi_T(t,x)+\theta h_ny)|\\
& \le |y|K\left(1+|\xi_T(t,x)+\theta h_ny|^m\right) \\
& \le 2^m K|y| \left(1 +|\xi_T(t,x)|^m +|\theta h_ny|^m \right) \\
&\le N |y|\Big(1 +|x|^m +\Big|\int_t^T\sigma (r) d\mathbf{w}_r\Big|^m
+\Big|\int_t^T b (r)\, dr\Big|^m +|y|^m \Big),
\end{align*}
where $N=N(m,K)$. By \eqref{e2.1}, the Burkholder-Davis-Gundy inequalities and the
fact that $\sigma, b$ are independent of $\omega$, the last expression above has
finite expectation. Hence by \cite[Lemma III.6.13 (f)]{K2}, \eqref{e2.2} holds.
Since $\{h_n\}$ was an arbitrary sequence converging to 0 as $n\to \infty$,
we conclude
$$
\lim_{h\to 0} \mathbf{E} \Delta^1_{h,y} g (\xi_T(t,x)) =\mathbf{E}
g_{y}(\xi_T(t,x)).
$$
Thus $v(t,x)$ is differentiable with respect to $x$ and for any $y\in E_d$,
$v_{y}(t,x)=\lim_{h\to 0} \mathbf{E} \Delta^1_{h,y} g (\xi_T(t,x))
=\mathbf{E} g_{y}(\xi_T(t,x))$.
We now show that $v(t,x)$ is twice differentiable with respect to $x$.
By the above expression for $v_{y}(t,x)$, we have, for any $\bar y \in E_d$
\begin{equation}
\frac{v_y(t,x+h\bar y)-v_y(t,x)}h=\mathbf{E} \Delta^1_{h,\bar y}\, g_y(\xi_T(t,x)).
\label{e2.*}
\end{equation}
But since $g_{y\bar y}$ is continuous, for any sequence $h_n\to
0$ as $n\to \infty$, the Mean Value Theorem yields
$$
\Delta^1_{h_n,\bar y} \,g_y ( \xi_T(t,x))=\int_{0}^1 g_{y\bar y}
(\xi_T(t,x) +rh_ny)\,dr=g_{y\bar y} (\xi_T(t,x))+\theta h_ny),
$$
for some $\theta\in [0,1]$.
Since $g \in C^2(E_d)$,
$\Delta^1_{h_n,\bar y} \,g_y ( \xi_T(t,x))\to g_{y\bar y}(\xi_T(t,x))$ as
$n\to \infty$. By the argument immediately following \eqref{e2.2},
except with $|y|^2+|\bar y|^2$ in place of $|y|$ and using the growth condition
on $|g_{(y)(y)}(x)|$, we see that $|\Delta^1_{h_n,\bar y} \,g_y ( \xi_T(t,x))|$
is bounded above (independently of $n$) by a random variable which has finite
expectation. Hence
$$
\mathbf{E} \Delta^1_{h_n,\bar y} g_y (\xi_T(t,x)) \to \mathbf{E} g_{y\bar y}(\xi_T(t,x))
\quad \text { as } n\to \infty.
$$
Since $\{h_n\}$ was an arbitrary sequence converging to 0 as $n\to \infty$,
$$
\lim_{h \to 0}\mathbf{E} \Delta^1_{h,\bar y}\, g_y(\xi_T(t,x))
=\mathbf{E} g_{y\bar y}(\xi_T(t,x)).
$$
Thus by \eqref{e2.*}, $v_{y\bar y}(t,x)$ exists and since $y,\bar y \in E_d$ were
arbitrary, $v(t,x)$ is twice differentiable with respect to $x$ and
$$
v_{y\bar y}(t,x)=\lim_{h \to 0}\mathbf{E} \Delta^1_{h,\bar y}\, g_y(\xi_T(t,x))
=\mathbf{E} g_{y\bar y}(\xi_T(t,x)).
$$
We now show the continuity of $v_{y \bar y}(t,x)$ in $(t,x)$. To this end, fix
$(t,x)$ and let $t^n\to t^+ ,x^n \to x$. It suffices to show
$v_{y \bar y}(t^n,x^n)\to v_{y \bar y}(t,x)$. We have
\begin{equation}
|v_{y \bar y}(t^n,x^n)-v_{y \bar y}(t,x)|
\le \mathbf{E} |g_{y \bar y}(\xi_T(t^n,x^n)) - g_{y \bar y}(\xi_T(t,x))|.
 \label{e2.3}
\end{equation}
Observe that $\xi_T(t^n,x^n) \overset P \to \xi_T(t,x)$ and
since $g_{y \bar y}$ is continuous, $g_{y \bar y}(\xi_T(t^n,x^n)) \overset P
\to  g_{y \bar y}(\xi_T(t,x))$.
Since $|g_{y \bar y}(\xi_T(t^n,x^n))|\le \eta $ with $\mathbf{E} \eta <\infty$,
the right hand side of \eqref{e2.4} tends to zero as $n \to \infty$.
The details are as follows. To see that
$\xi_T(t^n,x^n) \overset P \to \xi_T(t,x)$, observe that
\begin{equation}
\begin{aligned}
&|\xi_T(t^n,x^n) -\xi_T(t,x)|\\
&\le |x^n-x| + \Big|\int_{t^n} ^T\sigma (r) d\mathbf{w}_r
 -\int_t ^T\sigma (r) d\mathbf{w}_r\Big|
 +\Big|\int_{t^n} ^T b (r)\, dr
 -\int_t ^T b (r)\, dr \Big|.
\end{aligned}\label{e2.4}
\end{equation}
The middle summand tends to zero in probability as $n\to \infty$ by
\cite[Theorem III.6.6]{K2} and the fact that
$$
\int_0 ^{T}\| I_{t^n \le r} \sigma (r) -I_{t\le r} \sigma (r)\|^2 \, dr =
\int_0 ^{T}\| \sigma (r)\|^2 I_{t\le r < t^n}  \, dr \to 0
\quad\text{ as } n\to \infty
$$
by  \eqref{e2.1} and the Dominated Convergence Theorem. The third summand on the
right hand side of \eqref{e2.4} tends to zero by the Dominated Convergence Theorem.
Since $x^n\to x$, we have $\xi_T(t^n,x^n) \overset P \to \xi_T(t,x)$.
 Since $g_{y \bar y}(x)$ is continuous,
$$
g_{y \bar y}(\xi_T(t^n,x^n)) \overset P \to  g_{y \bar y}(\xi_T(t,x)),
$$
by \cite[Theorem III.6.13 (c)]{K2}. Finally,
\begin{equation}\label{e2.5}
\begin{aligned}
&|g_{y \bar y}(\xi_T(t^n,x^n))|\\
&\le K(|y|^2 +|\bar y|^2)(1 +|\xi_T(t^n,x^n)|^m)\\
&\le 3^m K (|y|^2 +|\bar y|^2) \Big\{1 + |x^n|^m + \Big|\int_{t^n}^T\sigma (r)\,
d\mathbf{w}_r\Big|^m+\Big|\int_{t^n}^T b (r)\, dr\Big|^m\Big\}.
\end{aligned}
\end{equation}
Since
$$
\Big|\int_{t^n}^T\sigma (r)\, d\mathbf{w}_r\Big|^m
\le 2^m \sup_s  \Big|\int_{0}^{s\wedge T}\sigma (r)\, d\mathbf{w}_r\Big|^m
$$
as $x^n \to x$ and $\big|\int_{t^n}^T b (r)\, dr\big|^m
\le \big(\int_{0}^T |b (r)|\, dr\big)^m$, the right hand side of \eqref{e2.5} is
bounded uniformly in $n$ by a random variable, which, by the Burkholder-Davis-Gundy
inequalities and \eqref{e2.1}, has finite expectation. Hence, by
\cite[Theorem III.6.13 (f)]{K2},
$$
\mathbf{E} |g_{y \bar y}(\xi_T(t^n,x^n)) - g_{y \bar y}(\xi_T(t,x))|\to 0
\quad \text{ as } n\to \infty
$$
and hence by \eqref{e2.3}, $v_{y \bar y}(t^n,x^n)\to v_{y \bar y}(t,x)$.
\end{proof}

The proof that $v(t,x)$ and $v_y(t,x)$ are continuous in $\bar H_T$ follow
same the technique shown here, except we use the respective assumptions
$|g(x)| ,\,|g_{(y)}(x)| \le K(1+|x|^m)$. Observe that by \eqref{e2.5} and the
Burkholder-Davis-Gundy inequalities, we obtain the following estimate,
which holds for $(t,x) \in \bar H_T$
\begin{equation} \label{e2.6}
\begin{aligned}
&\|v_{xx}(t,x)\|\\
&\le N(d,m,K)\Big\{ 1 +|x|^m +\Big(\int_t^T \|\sigma (r)\|^2 \, dr\Big)^{m/2}
+ \Big(\int_{t}^T |b (r)|\, dr\Big)^m \Big\}.
\end{aligned}
\end{equation}
If in addition, $\sigma, b$ satisfy $\sup_{t\le T}(\|\sigma (t)\| +|b(t)|)\le K$,
inequality \eqref{e2.6} yields, with $N_1=N_1(d,m,K)$
\begin{equation} \label{e2.7}
\begin{aligned}
\|v_{xx}(t,x)\|&\le 2N(1\vee K^m)(1+|x|^m)\left\{ 1 +(T-t)^m \right\}\\
&\le  4N(1\vee K^m)(1+|x|^m) e^{(T-t)m}\\
&\le N_1(1+|x|)^me^{N_1(T-t)}.
\end{aligned}
\end{equation}
The following lemma appears in \cite[p.~195]{K2}. We will use this lemma and the
fact that $v,v_x,v_{xx}$ are continuous in $(t,x)$ to show that when
$\sigma(t), b(t)$ are bounded, $v(t,x)$ is differentiable with respect to $t$
for almost every $t \in [0,T]$.

\begin{lemma} \label{lm2.2}
Let $\xi_s (t,x)= x +\int_t^s \sigma(r)\, d\mathbf{w}_r +\int_t^s b(r)\, dr$,
where $\sup_{t\le T}(\|\sigma (t)\| +|b(t)|)\le K$. For $\epsilon >0$ and
$(t,x) \in Q$, let
$$
\tau_{\epsilon} (t,x)=\inf \{s\ge t: (s,\xi_s (t,x)) \notin Q_{\epsilon}(t,x)\},
$$
where $Q_{\epsilon}(t,x) =(t-\epsilon^3,t+\epsilon^3) \times B_{\epsilon}(x)$.
Then for any compact set $\Gamma \subset Q_+:=Q\cap\{t\ge 0\}$,
$$ \epsilon^{-3} P\{\tau_{\epsilon} (t,x)-t<\epsilon^3\}\to 0,
\quad \epsilon^{-3} \mathbf{E}[\tau_{\epsilon} (t,x)-t] \to 1,
$$
uniformly in $(t,x) \in \Gamma$, as $\epsilon \to 0^+$.
\end{lemma}

\begin{theorem} \label{thm2.3}
Under the hypotheses of Theorem \ref{thm2.1} suppose that
$\sup_{t\le T}(\|\sigma (t)\| +|b(t)|)\le K$. Then for any $x \in E_d$, the
 function $v(t,x)=\mathbf{E} g(\xi_T(t,x))$ is differentiable with respect to $t$
for almost every $t \in [0,T)$.
\end{theorem}

\begin{proof}
Fix any $(t, x) \in H_T$ and choose $\epsilon$ so small that $t+\epsilon^3 <T$.
Since absolutely continuous functions of a single real variable are differentiable
almost everywhere, it suffices to show that $v(t,x)$ is Lipschitz in $t$.
By the strong Markov property we can write
\begin{equation} \label{e2.8}
v(t,x)=\mathbf{E}v(\tau_{\epsilon} (t,x), \xi_{\tau_{\epsilon} (t,x)}(t,x)),
\end{equation}
which we henceforth abbreviate as
$\mathbf{E} v(\tau_{\epsilon}, \xi_{\tau_{\epsilon}})$.
By It\^o's formula applied to the $C^2$ function (of $x$) $v(t +\epsilon^3, \cdot)$,
 we have
\begin{equation} \label{e2.9}
\begin{aligned}
&v(t,x)-v(t+\epsilon^3,x)\\
&=\mathbf{E}[v(\tau_{\epsilon}, \xi_{\tau_{\epsilon}}) -v(t +\epsilon^3,
\xi_{\tau_{\epsilon}})] +\mathbf{E} [v(t +\epsilon^3, \xi_{\tau_{\epsilon}})
-v(t +\epsilon^3, \xi_t)]\\
&= \mathbf{E}\,I_{\tau_{\epsilon}<t +\epsilon^3}  [v(\tau_{\epsilon},
\xi_{\tau_{\epsilon}}) -v(t +\epsilon^3, \xi_{\tau_{\epsilon}})]
+ \mathbf{E} \int_t^{\tau_{\epsilon}} L_r v(t +\epsilon^3, \xi_r) \,dr
\end{aligned}
\end{equation}
Certainly $|v(\tau_{\epsilon}, \xi_{\tau_{\epsilon}}) -v(t +\epsilon^3,
\xi_{\tau_{\epsilon}})|\le 2\sup_{[t,t+\epsilon^3]\times
 \overline {B_{\epsilon}(x)}}|v|$.
We recall that $v,v_x,v_{xx}$ are continuous and hence bounded in any compact set.
By definition, $L_r v(t +\epsilon^3, \xi_r)=\frac 12 \mathop{\rm tr}[a(r) \,
v_{xx}(t +\epsilon^3, \xi_r)]+b(r)\cdot  v_{x}(t +\epsilon^3, \xi_r)$.
From the elementary inequality $\left|\mathop{\rm tr} [a\cdot m]\right|\le \|a\|\,\|m\|$
and the fact that $\|\sigma (t)\| +|b(t)|\le K$, we get,
for $r\in [t,\tau_{\epsilon}]$,
\begin{equation} \label{e2.10}
\begin{aligned}
\left| L_r v(t +\epsilon^3, \xi_r)\right|
&\le  \frac {K^2}2  \|v_{xx}(t +\epsilon^3, \xi_r)\|
+K |v_{x}(t +\epsilon^3, \xi_r)|\\
&\le N(K) ( \sup_{B_{\epsilon}(x)} \|v_{xx}(t +\epsilon^3, \cdot)\|
+ \sup_{B_{\epsilon}(x)} |v_x(t +\epsilon^3, \cdot)|).
\end{aligned}
\end{equation}
So in any small closed cylinder
$\widetilde Q \supset [t,t+\epsilon^3]\times \overline {B_{\epsilon}(x)}$,
we have, by \eqref{e2.9} and Lemma \ref{lm2.2}, for sufficiently small $\epsilon$,
\begin{align*}
&|v(t,x)-v(t+\epsilon^3,x)| \\
&\le 2\sup_{\widetilde Q} |v| \cdot P\{\tau_{\epsilon} -t<\epsilon^3\}
+  N(K)( \sup_{\widetilde Q} \|v_{xx}\|
+ \sup_{\widetilde Q} |v_x|)\mathbf{E}[\tau_{\epsilon} -t]\\
&\le N_1(K) (\sup_{\widetilde Q} |v| +\sup_{\widetilde Q} |v_x|
+ \sup_{\widetilde Q} \|v_{xx}\|) \epsilon^3.
\end{align*}
Since $t,\epsilon$ were arbitrary (such that $t+\epsilon^3 <T$), we get,
for any $s,t \in [0,T)$ and any fixed $x \in E_d$,
$$|v(t,x)-v(s,x)|\le N_2 |t-s|,$$
where $N_2$ is independent of $s,t,x$. Hence the generalized derivative
$\frac {\partial v}{\partial t}$ exists and
$\left|\frac {\partial v}{\partial t}(t,x)\right| \le N_2$.
\end{proof}

We will now show that the function $v(t,x)=\mathbf{E} g(\xi_{T}(t,x))$ satisfies
 the Kolmogorov equation almost everywhere in $H_T$, under the assumptions
of Theorem \ref{thm2.3}. Our main tool will be the Aleksandrov-Busemann-Feller
(ABF) theorem (see \cite[Theorem 1.1]{K3}) for continuous functions which are
convex in $x$ and non-increasing in $t$.

\begin{theorem}[Aleksandrov-Busemann-Feller] \label{thm2.4}
Let $u(t,x)$ be convex in $x$, non-increasing in $t$ and continuous in $\bar H_T$.
Let $P(s,x,t,y)= u(s,x) +u_s^{(0)}(s,x)t
+u_x(s,x)\cdot y +\frac 12 y^*u_{xx}^{(0)}(s,x)y$,
where $u_s^{(0)}, u_{x^ix^j}^{(0)}$ denote generalized derivatives.
Then for almost all $(s,x)\in E_{d+1}, \,u(s+t,x+y)=P(s,x,t,y) +o(|t| +|y|^2)$
as $(t,y)\to (0,0)$.
\end{theorem}

Equivalently, for almost all
$(t_0,x_0)\in E_{d+1}, u(t,x)=P_{(t_0,x_0)}(t,x) +o(|t-t_0| +|x-x_0|^2)$ as
$(t,x)\to (t_0,x_0)$, where
$P_{(t_0,x_0)}(t,x)=u(t_0,x_0) +u_t^{(0)}(t_0,x_0)(t-t_0) +u_x(t_0,x_0)\cdot (x-x_0)
+\frac 12 (x-x_0)^*u_{xx}^{(0)}(t_0,x_0)(x-x_0)$.
We want to apply the ABF theorem to a variant of $v$. To this end, note that
by \eqref{e2.5}, for any $l \in E_d$, we have
$$
\left|v_{(l)(l)}(t,x)\right|\le\mathbf{E} \left|g_{(l)(l)}(\xi_T (t,x))\right|
\le N e^{N(T-t)} (1+ |x|)^m,
$$
where $N=N(m,K)$. Direct calculation shows that for any $l ,x\in E_d,\,
(m+2)2^{-\frac m2}(1+|x|)^m\le \left[  (1+|x|^2)^{\frac m2 +1} \right]_{(l)(l)}$.
 Hence
$$\left|v_{(l)(l)}(t,x)\right|\le  \frac{ N e^{N(T-t)}2^{\frac m2}}{m+2}
 \left[  (1+|x|^2)^{\frac m2 +1} \right]_{(l)(l)} \le  N e^{N(T-s)}
  \left[  (1+|x|^2)^{\frac m2 +1} \right]_{(l)(l)}
$$
which yields
$$
0\le \left( v(t,x) +  N e^{N(T-t)} (1+|x|^2)^{\frac m2 +1}  \right)_{(l)(l)}
\quad \forall (t,x) \in H_T,\, l\in E_d.
$$
That is, the function $v(t,x) + N e^{N(T-t)} (1+|x|^2)^{\frac m2 +1}$ is convex in
 $x$. We may also consider this function to be decreasing in $t$ by the following
 argument. By Lemma \ref{lm2.2}, the first summand on the right hand side of 
 \eqref{e2.9} is
$o(\epsilon^3)$ as $\epsilon \to 0$. By the continuity of $v_{xx}(t,x), v_x(t,x)$,
the last factor on the right hand side of \eqref{e2.10} tends to
$\|v_{xx}(t,x)\|+ |v_x(t,x)|$ as $\epsilon \to 0$. Since the estimate
$|v_x(t,x)|\le  N e^{N(T-t)} (1+ |x|)^m$ also holds, dividing \eqref{e2.9} by $\epsilon^3$,
letting $\epsilon \to 0$, using \eqref{e2.7} and applying the second result in
Lemma \ref{lm2.2}, we get for almost every $t \in [0,T)$ and any $x \in E_d$
\begin{equation}
\left|\frac {\partial v}{\partial t}(t,x)\right |\le  N e^{N(T-t)} (1+ |x|)^m
\le N e^{N(T-t)} (1+|x|^2)^{\frac m2 +1},\label{e2.11}
\end{equation}
where $N=N(d,m,K)$. From \eqref{e2.11} it follows, as before, that for some
$N=N(d,m,K)$, $v(t,x) + N e^{N(T-t)} (1+|x|^2)^{\frac m2 +1}:=v+v_0$ is
decreasing in $t$.

\begin{theorem} \label{thm2.5}
Under the assumptions of Theorem \ref{thm2.3}, the function $v(t,x)=\mathbf{E} g(\xi_{T}(t,x))$
satisfies the Kolmogorov equation almost everywhere in $H_T$.
\end{theorem}

\begin{proof}
 Since the ABF theorem holds for the function $v + v_0$ and $v_0$ is smooth,
the ABF theorem also holds for $v$. Since $v$ has continuous second derivatives
(by Theorem \ref{thm2.1}), $v^{(0)}_{xx}=v_{xx}$ almost everywhere. So fix any
$(t,x) \in H_T$ for which the assertion of the ABF theorem holds for
$v$, $v^{(0)}_{xx}(t,x)=v_{xx}(t,x)$ and $t$ is in the Lebesgue set of the
operator $L_s \equiv \frac 12 a^{ij}(s)\frac{\partial^2 }{\partial x^i\partial x^j}
+b^i(s)\frac{\partial }{\partial x^i}$.
By the strong Markov property,
$v(t,x)=\mathbf{E}v(\tau_{\epsilon} , \xi_{\tau_{\epsilon} })$, where
$\tau_{\epsilon}(t,x)$ is as in Lemma \ref{lm2.2}.  By the ABF theorem,
$v(\tau_{\epsilon} , \xi_{\tau_{\epsilon}} )= P_{(t,x)} (\tau_{\epsilon} ,
\xi_{\tau_{\epsilon} })+o(|\tau_{\epsilon}-  t| +|\xi_{\tau_{\epsilon}} -x|^2)$
as $\epsilon \to 0$. Since $\xi_t(t,x)=x$ and $P_{(t,x)} (t,x)=v(t,x)$, applying
It\^o's formula to the parabaloid $P_{(t,x)}$ yields
\begin{equation}
0=\mathbf{E} \int_t^{\tau_{\epsilon}} \big(L_rP +\frac {\partial P}{\partial r}\big)
(r,\xi_r)\, dr  +\mathbf{E} [o(|\tau_{\epsilon}- t| +|\xi_{\tau_{\epsilon}} -x|^2)].
\label{e2.12}
\end{equation}
Since $0\le \tau_{\epsilon}-  t\le \epsilon^3$, the estimates
$\mathbf{E} |\xi_{\tau_{\epsilon}} -x|^p \le N(p,K)\epsilon^{\frac {3p}2}
(1+\epsilon^{\frac {3p}2})$ and $|v(t,x)|\le N(T,m,K)(1+|x|)^m$
imply that the second summand on the right of \eqref{e2.12} is $o(\epsilon^3)$.
Let us write the first summand on the right of \eqref{e2.12} as
\begin{equation}
\mathbf{E} I_{ \tau_{\epsilon}<  t+ \epsilon^3}  \int_t^{\tau_{\epsilon}}
\big(L_rP +\frac {\partial P}{\partial r}\big)(r,\xi_r)\, dr
+\mathbf{E} I_{ \tau_{\epsilon}= t+ \epsilon^3}
\int_t^{t+ \epsilon^3} \big(L_rP +\frac {\partial P}{\partial r}\big)(r,\xi_r)\, dr.
 \label{e2.13}
\end{equation}
Since the coefficients of $L_r$ are uniformly bounded and
$r \in [t,\tau_{\epsilon}]$ implies $|\xi_r-x|<\epsilon$, the integrand
in the first summand of \eqref{e2.13} satisfies
$$
\big|\big( L_rP +\frac {\partial P}{\partial r}\big)(r,\xi_r)\big|
\le (K^2+K\epsilon)\|v_{xx}(t,x)\| +K|v_x(t,x)| +|v^{(0)}_t(t,x)|.
$$
Since $\epsilon \in (0,1)$,
\begin{align*}
&\Big|\mathbf{E} I_{ \tau_{\epsilon}<  t+ \epsilon^3}  \int_t^{\tau_{\epsilon}}
\big(L_rP +\frac {\partial P}{\partial r}\big)(r,\xi_r)\, dr\Big|\\
&\le  N(K)(\|v_{xx}(t,x)\| +|v_x(t,x)| +|v^{(0)}_t(t,x)|)
\cdot P\{\tau_{\epsilon}< t+ \epsilon^3\},
\end{align*}
and hence the first expectation in \eqref{e2.13} is $o(\epsilon^3)$ by Lemma \ref{lm2.2}.
Dividing the second expectation in \eqref{e2.13} by $\epsilon^3$ and evaluating it
explicitly yields
\begin{equation} \label{e2.14}
\begin{aligned}
&\mathbf{E} I_{ \tau_{\epsilon}= t+ \epsilon^3}\frac 1{\epsilon^3}
\int_t^{t+ \epsilon^3} \left(L_r v(t,x) + v^{(0)}_t(t,x)\right)\, dr\\
&+\mathbf{E} I_{ \tau_{\epsilon}= t+ \epsilon^3} \frac 1{\epsilon^3}
\int_t^{t+ \epsilon^3}  b(r)^*v_{xx}(t,x)(\xi_r-x)\, dr.
\end{aligned}
\end{equation}
Since $t$ is a Lebesgue point for $L_s$, we have (almost surely)
$$I
_{ \tau_{\epsilon}= t+ \epsilon^3} \frac 1{\epsilon^3}\int_t^{t+ \epsilon^3}
 \left(L_r v(t,x) + v^{(0)}_t(t,x)\right)\, dr \rightarrow
 L_t v(t,x) + v^{(0)}_t(t,x)\quad \text{as } \epsilon \to 0
$$
and since
\begin{align*}
&\Big|I_{ \tau_{\epsilon}= t+ \epsilon^3} \frac 1{\epsilon^3}\int_t^{t+ \epsilon^3}
\left(L_r v(t,x) + v^{(0)}_t(t,x)\right)\, dr\Big|\\
&\le K^2 \|v_{xx}(t,x)\| +K|v_x(t,x)| +|v^{(0)}_t(t,x)|,
\end{align*}
the first expectation in \eqref{e2.14} converges to $L_t v(t,x) + v^{(0)}_t(t,x)$ as
$\epsilon \to 0$. The second expectation in \eqref{e2.14} converges to 0 as
$\epsilon \to 0$. Recalling that $0\le \tau_{\epsilon}-  t\le \epsilon^3$
and that $r \in [t,\tau_{\epsilon}]$ implies $|\xi_r-x|<\epsilon$,
we immediately get the bound
$$
\Big|  I_{ \tau_{\epsilon}= t+ \epsilon^3} \frac 1{\epsilon^3}
\int_t^{t+ \epsilon^3}  b(r)^*v_{xx}(t,x)(\xi_r-x)\, dr \Big|
\le \frac 1{\epsilon^3}\|v_{xx}(t,x)\| K\epsilon^4= \|v_{xx}(t,x)\| K\epsilon.
$$
Hence dividing \eqref{e2.12} by $\epsilon^3$ and letting $\epsilon \to 0$,
we get $L_t v(t,x) + v^{(0)}_t(t,x)=0$.
\end{proof}

\section{Fundamental solutions of the Kolmogorov equation - the analytic approach}

Even the ``analytic'' proof that $v(t,x)=\mathbf{E} g(\xi_{T}(t,x))$ is a solution of
the Kolmogorov equation relies on the well known probabilistic fact that since
coefficients $\sigma (t),\,b(t)$ are independent of $\omega$, the vector
$\xi_{T}(t,x)$ is a Gaussian vector with parameters
$\big( x + \int_t^T b (r)\,dr, \int_t^T a (r)\,dr\big)$.
Hence, the distribution $P\xi_T (t,x)^{-1}$ has density function
$$
p(T,t,y)=\frac {e^{-\frac 12 \left\langle C^{-1}(t)(y-x-\int_t^T b(r)dr),y-x
-\int_t^T b(r)dr\right\rangle}  }{ (2\pi)^{\frac d2}\sqrt{\det C(t)}},
$$
where $C(t)=C_T(t)=\int_t^T a(r)\,dr$.
From this, it follows that a solution to the problem
\begin{equation}
\left\{\begin{gathered}
\tfrac 12 a^{ij}(t)v_{x^ix^j}(t,x) +b^i(t) v_{x^i}(t,x)
+\tfrac {\partial v}{\partial t}(t,x)=0  \quad\text{a. e. }  t\in [0,T)\\
v(T,x)=g(x)\quad  x\in E_d
\end{gathered}\right.\label{e3.1}
\end{equation}
is given by
\begin{equation}
\begin{aligned}
v(t,x)&=\mathbf{E} g(\xi_{T}(t,x))\\
&=\int_{E_d} g(y) \,P\, \xi_T ^{-1}(t,x)\, (dy) \\
&=  \int_{E_d} g(y) \frac {e^{-\frac 12
\left\langle C^{-1}(t)(y-x-\int_t^T b(r)dr),y-x -\int_t^T b(r)dr\right\rangle }}
{(2\pi)^{\frac d2}  \sqrt{\det C(t)}} \,dy ,
\end{aligned}
\end{equation}
where $a(r)=\sigma(r)\sigma ^*(r)$ is non-degenerate. We prove this in
Theorem \ref{thm3.1} below, for slowly increasing $g\in C^0(E_d)$. Viewed analytically,
since the function
\begin{equation}
p(T,t,x) =\frac {e^{-\frac 12 \left\langle C^{-1}(t)(x+\int_t^T b(r)dr),x
+\int_t^T b(r)dr\right\rangle}  }{ (2\pi)^{\frac d2}\sqrt{\det C(t)}}
\label{e3.2}
\end{equation}
is a fundamental solution (in $x$) of the equation
$L_t p(t,x) +\frac {\partial p}{\partial t}(t,x)=0$ a.e. $t\in [0,T)$,
all $x\neq -\int_t^T b(r)dr$, where
$L_t\equiv \frac 12 a^{ij}(t)\frac{\partial^2 }{\partial x^i\partial x^j}
+b^i(t)\frac{\partial }{\partial x^i}$, a solution to \eqref{e3.1} will be given
by the convolution
\begin{align*}
v(t,x)&=\left[ g * p(T,t,\cdot)\right](x)\\
&=\int_{E_d} g(y) \,p(T,t, x-y)\,dy \\
&=\int_{E_d} g(y) \frac {e^{-\frac 12 \left\langle C^{-1}(t)(x-y
+\int_t^T b(r)dr),x-y+\int_t^T b(r)dr\right\rangle}  }{ (2\pi)^{\frac d2}
\sqrt{\det C(t)}} \,dy,
\end{align*}
providing, of course, we can differentiate under the integral sign.
Regarding notation, by {\it fundamental solution}, we mean that
for all $t\in [0,T)$, $p(T,t,x)$ is infinitely differentiable in $x$ and
$\int_{E_d} p(T,t,x)\, dx=1$.

By Lebesgue's differentiation theorem, $p(T,t,x)$ in (2) is differentiable
with respect to $t$, only in the {\it almost everywhere} sense.
This is in contrast to the case where $a(t)=I_d, b(t)=b \,(const.)$ and
the Kolmogorov equation is simply
$\frac 12\Delta u(t,x) + b\cdot u_x(t,x) +\frac {\partial u}{\partial t}(t,x)=0$
for all $(t,x) \in H_T$. In this case,
$p(T,t,x)= (2\pi(T-t))^{-\frac d2}  e^{\frac{-|x+b(T-t)|^2}{2(T-t)}}$ is
infinitely differentiable in both $t$ and $x$.


\begin{theorem} \label{thm3.1}
For $t\in [0,T]$ and $x\in E_d$ and $s\in [t,T]$, let
$\xi_s (t,x)= x +\int_t^s \sigma(r)\, d\mathbf{w}_r +\int_t^s b(r)\, dr$,
where $\sup_{t\le T}(\|\sigma (t)\| +|b(t)|)\le K$. Assume $\exists \delta>0$
for which $\delta I_d \le a(t)$, for all $t \in [0,T]$, where
$a(t)= \sigma(t)\sigma^*(t)$. Then for $p(T,t,x)$ as in (2) and $g$ continuous
and slowly increasing, the function
$$
v(t,x)=\mathbf{E} g(\xi_{T}(t,x))= \int_{E_d} g(y) \,p(T,t, x-y)\,dy
$$
satisfies the Kolmogorov equation
$\frac 12 a^{ij}(t) v_{x^i x^j}(t,x) + b^i(t) v_{x^i}(t,x)
+\frac{\partial v}{\partial t}(t,x)=0$ a.e. $t \in [0,T)$ and any $x\in E_d$.
\end{theorem}

\begin{proof} Direct calculation shows that for almost every $t \in [0,T)$ and
any $x \neq -\int_t^Tb(r)dr \in E_d$, $p(T,t,x)$ is a solution of the Kolmogorov
equation. Thus we need only show that we can differentiate under the integral sign.
Omitting the constant factor of $(2\pi)^{-d/2}$, direct calculation shows that
for almost every $t \in [0,T)$, with $z=y-x$ and $\eta_t:=\int_t^T b(r)\,dr$,
\begin{align*}
&\frac {\partial p}{\partial t}(T,t, x-y)\\
&=\frac{e^{-\frac 12 \left\langle C^{-1}(t)(z-\eta_t),z-\eta_t\right\rangle} }
{2  \sqrt{\det C(t)}} \Big\{\mathop{\rm tr} [a(t)C^{-1}(t)]
-\langle C^{-1}(t)\,a(t)\,C^{-1}(t)(z-\eta_t),z-\eta_t  \rangle\\
&\quad+2 \langle  C^{-1}(t)(y-x),b(t)\rangle + 2 \langle  C^{-1}(t)\, b(t),
\eta_t\rangle \Big\}
\end{align*}
and hence
\begin{align*}
&\big|\frac {\partial p}{\partial t}(T,t, x-y)\big|\\
&\le \frac{e^{-\frac 12 \left\langle C^{-1}(t)(z-\eta_t),z-\eta_t\right\rangle} }
{ \sqrt{\det C(t)}} \Big\{ \|a(t)\|  \|C^{-1}(t)\| \\
&\quad+ \|C^{-1}(t)\,a(t)\,C^{-1}(t)\| |z-\eta_t|^2 + \|C^{-1}(t)\| |z| |b(t)|
+\|C^{-1}(t)\| |b(t)| |\eta_t| \Big\}.
\end{align*}
Since $\sup_{t\le T}(\|\sigma (t)\| +|b(t)|)\le K$ and
$\|ab\|\le \|a\|\, \|b\|$,  $\|a(t)\|=\|\sigma (t) \sigma^* (t)\|\le K^2$.
 From the estimate $\|C(t)\|\le \sqrt{T-t} \sqrt{\int_t^T \|a(r)\|^2\,dr}$,
we have $\|C(t)\|\le K^2(T-t)$. Moreover, by the uniform non-degeneracy condition
$\delta |\lambda|^2 \le a^{ij}(t)\lambda^i\lambda^j$, which holds for all
$t\in [0,T]$ and all $\lambda \in E_d$, we get
$\|C^{-1}(t)\|\le \frac {\sqrt {d}}{\delta (T-t)}$. We also have
$C^{ij}(t)\lambda^i\lambda^j\ge \delta |\lambda|^2(T-t)$, from which it immediately
follows that $\det C(t)\ge [\delta (T-t)]^d$. Obviously, $|\eta_t| \le K(T-t)$.
This gives
\begin{align*}
&\big|\frac {\partial p}{\partial t}(T,t, x-y)\big|\\
&\le \frac{e^{-\frac 12 \left\langle C^{-1}(t)(z-\eta_t),z-\eta_t\right\rangle} }
{ (\delta (T-t))^{d/2}   }\\
&\times \Big\{ \frac{ K^2 \sqrt {d}}{\delta (T-t)}
 +\frac{2 dK^2}{\delta^2 (T-t)^2}\left(|y-x|^2 +K^2(T-t)^2\right)
+ \frac{K \sqrt{d}}{\delta (T-t)} |x-y| + \frac{ K^2 \sqrt {d}}{\delta}  \Big\}.
\end{align*}
Similarly, the gradient and hessian of $p(T,t,x-y)$ satisfy
$$
p_x(T,t,x-y) =\frac {e^{-\frac 12 \left\langle C^{-1}(t)(z-\eta_t),z-\eta_t
\right\rangle}}{ \sqrt{\det C(t)}} \,\, C^{-1}(t)\cdot (z-\eta_t)
$$
\begin{align*}
&p_{xx}(T,t,x-y)\\
& =\frac {e^{-\frac 12 \left\langle C^{-1}(t)(z-\eta_t),z-\eta_t\right\rangle}}
{ \sqrt{\det C(t)}} \left\{  C^{-1}(t)(z-\eta_t) [C^{-1}(t)(z-\eta_t)]^*
- C^{-1}(t)\right\}.
\end{align*}
Thus
\begin{align*}
|p_x(T,t,x-y)| &\le \frac{e^{-\frac 12 \left\langle C^{-1}(t)(z-\eta_t),
z-\eta_t\right\rangle} }{ (\delta (T-t))^{d/2} }  \|C^{-1}(t)\| \cdot |z-\eta_t|\\
&\le \frac{\sqrt{d}\, e^{-\frac 12 \left\langle C^{-1}(t)(z-\eta_t),z-\eta_t
\right\rangle} }{ (\delta (T-t))^{d/2 +1} } \Bigl\{|y-x| +K(T-t)\Bigr\},
\end{align*}
\begin{align*}
&\|p_{xx}(T,t,x-y)\|\\
&\le  \frac{e^{-\frac 12 \left\langle C^{-1}(t)(z-\eta_t),z-\eta_t\right\rangle} }
{ (\delta (T-t))^{d/2} }  \left\{  \|C^{-1}(t)\|^2 \cdot |z-\eta_t|^2
+ \|C^{-1}(t)\|\right\}\\
&\le  \frac{e^{-\frac 12 \left\langle C^{-1}(t)(z-\eta_t),z-\eta_t\right\rangle} }
{ (\delta (T-t))^{d/2} }  \Big\{ \frac{2 d}{\delta^2 (T-t)^2}\left(|y-x|^2
+K^2(T-t)^2\right) + \frac{\sqrt{d}}{\delta (T-t)}\Big\}.
\end{align*}
To estimate the exponential term in each derivative, we use the inequality
$\frac{|z-\eta_t|^2}{K^2(T-t)} \le \left\langle C^{-1}(t)(z-\eta_t),
z-\eta_t\right\rangle$
and Young's inequality (twice):
$
|z-\eta_t|^2\ge \bigl||z|-|\eta_t| \bigr|^2 \ge \frac 12 |z|^2-|\eta_t|^2
\ge \frac 12 |y-x|^2-K^2(T-t)^2 \ge \frac 14 |y|^2-\frac 12 |x|^2 -K^2(T-t)^2
$
to conclude
$$
e^{-\frac 12 \left\langle C^{-1}(t)(z-\eta_t),z-\eta_t\right\rangle} \le
e^{-\frac{|y|^2}{8K^2(T-t)} +\frac{|x|^2}{4K^2(T-t)} +\frac{T-t}2}.
$$
Denoting any of the derivatives $p_t,p_x,p_{xx}$ by $p'(T,t,x-y)$, we see that
$$
|p'(T,t,x-y)|\le \frac{N\cdot e^{-\frac{|y|^2}{8K^2(T-t)}
+\frac{|x|^2}{4K^2(T-t)} +\frac{T-t}2}}{(T-t)^{\frac d2 +2}} q(T-t,|y-x|),
$$
where $N=N(\delta,d,K)$ and $q(a,b)$ is a paraboloid in $a$ and $b$.
Hence if $(t,x) \in [0,t_0] \times B_R$, where $0\le t_0<T$, then
$$
|p'(T,t,x-y)|\le \frac{N\cdot e^{-\frac{|y|^2}{8K^2 T}
+\frac{R^2}{4K^2(T-t_0)} +T}}{(T-t_0)^{\frac d2 +2}} q(T,|y|+R).
$$
So if we require that $|g(x)|\le Ne^{\frac{|x|^2}{16K^2 T}}$,
we see that the integrals $\int_{E_d} g(y) p'(T,t,x-y) \,dy$ converge
uniformly with respect to $(t,x) \in [0,t_0] \times B_R$.
This implies $v(t,x)$ is twice differentiable with respect to $x$,
once differentiable with respect to $t$ (almost everywhere) and its derivatives
can be evaluated by differentiating under the integral sign. Since $p(T,t,x-y)$
satisfies the Kolmogorov equation for almost every $t\in [0,T)$, so does $v(t,x)$.
\end{proof}

\noindent {\bf Remark.\,}
The above growth condition for $g$ is obviously satisfied when $g$ has polynomial
growth, $|g(x)|\le K(1+|x|^m)$. Furthermore, direct calculation shows that for
any $d$-dimensional multi-index $\alpha$, any derivative of $p(T,t,x-y)$ with
respect to $x$ satisfies
$$
\left|D^{\alpha}_x p(T,t,x-y)\right| \le \frac{N\cdot e^{-\frac{|y-x|^2}{4K^2(T-t)}
 +\frac{T-t}2}} {(T-t)^{\frac d2 +|\alpha|}}\cdot q_{\alpha}(T-t,|y-x|),
$$
where $N=N(\delta,d,K,|\alpha|)$ and $q_{\alpha}(a,b)$ is a polynomial of degree
less than or equal to $|\alpha|$ in $a$ and $b$, from which it follows, as above,
that $v(t,x)$ is infinitely differentiable with respect to $x$.
\smallskip

More generally, if $c(t)\ge 0$ is bounded and measurable in $[0,T]$ and we define
$\phi_s (t)=\int_t^s c(r)dr$, the function
$\widetilde p(T,t,x) := p(T,t,x)e^{-\phi_T (t)}$ is an infinitely differentiable
solution (in $x$) of the equation
$L_t u(t,x) -c(t)u(t,x)+\frac {\partial u}{\partial t}(t,x)=0$
a.e. $t\in [0,T)$. Since
$\left[ g * \widetilde p(T,t,\cdot)\right](x)=e^{-\phi_T (t)}
\left[ g * p(T,t,\cdot)\right](x)$, a solution to the problem
\[
\left\{\begin{gathered}
\tfrac 12 a^{ij}(t)v_{x^ix^j}(t,x) +b^i(t) v_{x^i}(t,x) -c(t) v(t,x)
+\tfrac {\partial v}{\partial t}(t,x)=0  \\
\text{a.e.   $t\in [0,T)$, all $x\in E_d$}\\
v(T,x)=g(x) \quad x\in E_d
\end{gathered} \right.
\]
is given by
$$
v(t,x)=e^{-\phi_T (t)}\mathbf{E} g(\xi_{T}(t,x)),
$$
while if $\int_0^T |f(r)|e^{-\phi_r (t)}\,dr<\infty$, direct calculation shows
that the function
\begin{equation}
v(t,x)=e^{-\phi_T (t)}\mathbf{E} g(\xi_{T}(t,x)) +\int_t^T f(r)e^{-\phi_r (t)}\,dr
\label{e3.3}
\end{equation}
satisfies
\[ \left\{
\begin{gathered}
 \tfrac 12 a^{ij}(t)v_{x^ix^j}(t,x) +b^i(t) v_{x^i}(t,x) -c(t) v(t,x)+
f(t) +\tfrac {\partial v}{\partial t}(t,x)=0  \\
\text{a.e. $t\in [0,T)$ all $x\in E_d$}\\
v(T,x)=g(x) \quad x\in E_d.
\end{gathered} \right.
\]

\section{Paraboloid solutions of the Simplest Time-Measurable Bellman equations}

In this section, we prove a result about the payoff function for the Bellman
equation in the simple case where the equation depends only on second
derivatives and $t$ and the coefficients are Borel measurable functions of $t$.
Let $A$ be a separable metric space, where for $(\alpha, t)\in A\times [0,T]$,
$\sigma (\alpha, t)$ is a $d\times d_1$ matrix and $f^{\alpha}(t)$ is a function,
both continuous in $\alpha$ and Borel measurable in $t$.
Now let $(\Omega,\mathcal{F}, P)$ be a complete probability space on which
$(\mathbf{w}_t,\mathcal{F}_t)$ is a $d_1$-dimensional Wiener process. We consider the
controlled diffusion process $\xi_s(\alpha, t,x)$, defined for $s \in [0,T]$
by $\xi_s(\alpha, t,x)=x +\int_0^s \sigma(\alpha_r, t+r)\,d\mathbf{w}_r$,
where $t \in [0,T], x \in E_d$ are fixed and $\alpha_t$ is a strategy in class $U$,
that is, progressively measurable with values in $A$.

Suppose $g \in C^2(E_d)$ and satisfies
$|g(x)|,\,|g_{(y)}(x)|,\,|g_{(y)(y)}(x)|\le K(1+|x|^m)$, $\forall\,x,y \in E_d$,
where $K,m$ are nonnegative constants. It is known \cite{K3} that if for any
$\alpha \in A$, $f^{\alpha},\sigma (\alpha,\cdot)$ are differentiable with respect
to $t$ with derivatives not exceeding $K$, then the payoff function
\begin{equation} \label{e4.1}
v(t,x)=\sup_{\alpha \in U} \mathbf{E} \Big[\int_0^{T-t} f^{\alpha_r}(r+t)\,dr
+  g(\xi^{\alpha}_{T-t}(t,x))\Big]
\end{equation}
satisfies the Bellman equation
$$
\sup_{\alpha \in A} \{a^{ij}(\alpha,t)v_{x^ix^j}(t,x) +f^\alpha(t)\}
+ \frac{\partial v}{\partial t} (t,x) =0 \quad \text {a. e. } H_T,
\quad v(T,x)=g(x).
$$
In the special case where $g$ is a paraboloid, the payoff function takes a
very convenient form and clearly satisfies the Bellman equation under the weak
assumption that
$\sup_{\alpha \in A} f^{\alpha},\,\sup_{\alpha \in A} \sigma (\alpha,\cdot)
\in L_1([0,T]),  L_2([0,T])$, respectively.

\begin{theorem} \label{thm4.1}
Let $p(x)$ be any paraboloid defined on $E_d$, i.e.
$p(x)=\frac 12 x^*mx +l\cdot x +l_0$, where $m \in E_{d^2},l\in E_d, l_0\in E_1$.
Then the probabilistic solution of the Bellman equation
\[
\left\{\begin{aligned}
&\sup_{\alpha \in A} \{a^{ij}(\alpha,t)v_{x^ix^j}(t,x) +f^\alpha(t)\}
+ \frac{\partial v}{\partial t} (t,x) =0 \quad\text{a.e. } t \in [0,T)\\
&v(T,x)=p(x) \quad x \in E_d.
\end{aligned} \right.
\]
is given by
\begin{equation} \label{e4.2}
v(t,x)=p(x) +\int_t^T  \sup_{\alpha \in A} \left\{\mathop{\rm tr} [a(\alpha,r)m] +f^\alpha(r)
\right\} dr.
\end{equation}
\end{theorem}

\begin{proof}
 From the theory of controlled diffusion processes \cite{K1}, the probabilistic
solution to this Bellman equation is the payoff function \eqref{e4.1} with $g=p$
and $a(\alpha,t)=\frac 12 \sigma(\alpha,t)\sigma(\alpha,t)^*$. It immediately
follows from It\^o's formula that $\forall \alpha \in U,\,t\in [0,T]$ and
$x \in E_d$, we have
\begin{equation}
\mathbf{E} p(\xi^{\alpha}_{T-t}(t,x))=p(x)+ \mathbf{E}\int_0^{T-t}
\mathop{\rm tr} [a(\alpha_r,t+r)m] \, dr.\label{e4.3}
\end{equation}
We give a more direct proof of \eqref{e4.3} using Wald's identity. Writing
$\xi^{\alpha}_{T-t}=\xi^{\alpha}_{T-t}(t,x)$, we have
$p(\xi^{\alpha}_{T-t} (t,x))=\dfrac{\xi_{T-t}^{\alpha \,*} m \xi_{T-t}^{\alpha}}2
+l\cdot \xi_{T-t}^{\alpha} +l_0$
and
$$
\xi_{T-t}^{\alpha \,*}m\xi_{T-t}^{\alpha}
=\langle m \xi_{T-t}^{\alpha},\xi_{T-t}^{\alpha }\rangle = \langle mx,x\rangle
+2 \langle mx,\eta^{\alpha,t}_{T-t}\rangle
+ \langle m\eta^{\alpha,t}_{T-t},\eta^{\alpha,t}_{T-t}\rangle,
$$
where $\eta^{\alpha,t}_{T-t}:=\int_0^{T-t} \sigma(\alpha_r,t+r)\, d\mathbf{w}_r$.
Writing $m=ODO^*$, where $D=(\lambda^i\delta^{ij})$, we get
$$\langle m\eta^{\alpha,t}_{T-t},\eta^{\alpha,t}_{T-t}\rangle=\langle ODO^* \eta^{\alpha,t}_{T-t},\eta^{\alpha,t}_{T-t}\rangle=\langle Dz^{\alpha,t}_{T-t},z^{\alpha,t}_{T-t}\rangle= \sum_{i=1}^d \lambda^i \left(z^{\alpha,t,i}_{T-t}\right)^2$$
where
$$
z^{\alpha,t}_{T-t}:=O^*\cdot\eta^{\alpha,t}_{T-t}=\int_0^{T-t} O^*\cdot
\sigma(\alpha_r,t+r)\, d\mathbf{w}_r :=\int_0^{T-t} \widetilde\sigma(\alpha_r,t+r)\, d
\mathbf{w}_r.
$$
Orthogonality and the Wald identity yield
$$
\mathbf{E} \big(z^{\alpha,t,i}_{T-t}\big)^2
=\mathbf{E} \sum_{k=1}^d \Big(\int_0^{T-t} \widetilde\sigma^{ik}(\alpha_r,t+r)\,
dw^k_r\Big)^2
=\sum_{k=1}^d \mathbf{E} \int_0^{T-t} [\widetilde\sigma^{ik}(\alpha_r,t+r)]^2\, dr,
$$
and hence
\begin{align*}
\mathbf{E}\langle m\eta^{\alpha,t}_{T-t},\eta^{\alpha,t}_{T-t}\rangle
&=\sum_{i=1}^d \lambda^i\sum_{k=1}^d \mathbf{E} \int_0^{T-t} [\widetilde\sigma^{ik}
(\alpha_r,t+r)]^2\, dr \\
&=2\mathbf{E}\int_0^{T-t} \mathop{\rm tr} [a(\alpha_r,t+r)m] \, dr.
\end{align*}
By Wald's identity, we also have $\mathbf{E} \langle mx,\eta^{\alpha,t}_{T-t}\rangle =0$
and $\mathbf{E}[l\cdot \xi_{T-t}^{\alpha} +l_0]
=\mathbf{E}[l\cdot (x +\eta^{\alpha,t}_{T-t})+l_0]=l\cdot x +l_0$.
Thus
\begin{align*}
\mathbf{E} p(\xi^{\alpha}_{T-t}(t,x))
&= \mathbf{E} \Big[\frac{\xi_{T-t}^{\alpha *}m\xi_{T-t}^{\alpha}}2
+l \xi_{T-t}^{\alpha} +l_0\Big] \\
&= p(x)+ \mathbf{E}\int_0^{T-t} \mathop{\rm tr} [a(\alpha_r,t+r)m] \, dr.
\end{align*}
Therefore,
\begin{align*}
v(t,x)&=p(x)+ \sup_{\alpha \in U} \mathbf{E}
\Big[\int_0^{T-t} \mathop{\rm tr} [a(\alpha_r,t+r)m] +f^{\alpha_r}(r+t)\,dr \Big]\\
&=p(x)+ \int_t^{T} \sup_{\alpha \in \mathcal{A}} \left\{\mathop{\rm tr} [a(\alpha,r)m]
+f^{\alpha}(r)\right\}dr.
\end{align*}
\end{proof}

This result is hardly a surprise since the second-order derivatives of any
paraboloid are constant. Hence by Lebesgue's differentiation theorem, for
any operator $F(b,t)$ for which $\int_0^T |F(b,t)|\,dt <\infty$, the function
$$
u(t,x)=p(x) +\int_t^T F(p_{xx}(x),r)\,dr
 $$
satisfies
\[
\left\{\begin{aligned}
&F(u_{xx}(t,x),t) + \frac {\partial u}{\partial t}(t,x)=0  \quad
\text{a.e. }t\in [0,T)\\
&u(T,x)=p(x) \quad x \in E_d\,.
\end{aligned} \right.
\]

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\end{thebibliography}
\end{document}