\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 $$\begin{gathered} v(t,x)=\mathbf{E} \Bigl[ \int_t^T f(r,\xi_r (t,x)) e^{-\varphi_r(t,x)} \,dr + e^{-\varphi_T(t,x)} g(\xi_T(t,x))\Bigr],\\ \varphi_s(t,x)=\int_t^s c(r,\xi_r(t,x))\,dr \end{gathered} \label{e1.1}$$ 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 $$\int_0^T \left[ \|\sigma (t)\|^2 +|b(t)|\right]\,dt <\infty. \label{e2.1}$$ 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 $n\to \infty$ $$\mathbf{E} \Delta^1_{h_n,y} g (\xi_T(t,x)) \to \mathbf{E} g_{y}(\xi_T(t,x)).\label{e2.2}$$ 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 $\bar y \in E_d$ $$\frac{v_y(t,x+h\bar y)-v_y(t,x)}h=\mathbf{E} \Delta^1_{h,\bar y}\, g_y(\xi_T(t,x)). \label{e2.*}$$ 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 $$|v_{y \bar y}(t^n,x^n)-v_{y \bar y}(t,x)| \le \mathbf{E} |g_{y \bar y}(\xi_T(t^n,x^n)) - g_{y \bar y}(\xi_T(t,x))|. \label{e2.3}$$ 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{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} 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, \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} 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 $(t,x) \in \bar H_T$ \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} If in addition, $\sigma, b$ satisfy $\sup_{t\le T}(\|\sigma (t)\| +|b(t)|)\le K$, inequality \eqref{e2.6} yields, with $N_1=N_1(d,m,K)$ \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} 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 $t+\epsilon^3 0$ 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