Gambling for Resurrection and the Heat Equation on a Triangle

Abstract

We consider the problem of controlling the diffusion coefficient of a diffusion with constant negative drift rate such that the probability of hitting a given lower barrier up to some finite time horizon is minimized. We assume that the diffusion rate can be chosen in a progressively measurable way with values in the interval [0, 1]. We prove that the value function is regular, concave in the space variable, and that it solves the associated HJB equation. To do so, we show that the heat equation on a right triangle, with a boundary condition that is discontinuous in the corner, possesses a smooth solution.

Appendix. A

In this appendix we prove the continuity and smoothness properties of the function H claimed in Theorem 3.3. To shorten notation, we take $b=1$ and we show the respective properties for

$$\begin{aligned} \overline{H}(t,x) := 1- H(t,x) = P\big (\rho ^{t,x}_u < \rho ^{t,x}_\ell \big ). \end{aligned}$$

Lemma A.1

For every t, $x\in (t,T)\mapsto \overline{H}(t,x)$ is continuous.

Proof

Let $t\in [0,T)$ and $\delta >0$, we prove that for all $x\in [t,T-\delta ]$, the function is continuous in x. For $t<x<T-\delta$, using Proposition 3.1, and a simple change of variables, we get

$$\begin{aligned} \overline{H}(t,x) = \int _0^{T-t} \frac{1}{\sqrt{2\pi }u^{3/2}}\exp \left( -\frac{(T-x)^2}{2u}\right) \sum _{k=-\infty }^{+\infty } h_k(u,x) \, du, \end{aligned}$$

(A.1)

with

$$\begin{aligned} h_k(u,x)&:= [T-x-2k(T-t)]\\&\quad \times \exp \left( -2k^2\frac{(T-u-t)(T-t)}{u}\right) \exp \left( 2k\frac{(T-u-t)(T-x)}{u}\right) . \end{aligned}$$

Note that $\sum _{k=-\infty }^{+\infty } h_k(u,x) \leqslant 1$, since it is a probability.

• 1. For all $x\in [t,T-\delta ]$,

  $$\begin{aligned} \frac{1}{\sqrt{2\pi }u^{3/2}}\exp \left( -\frac{(T-x)^2}{2u}\right) \le \frac{1}{\sqrt{2\pi }u^{3/2}}\exp \left( -\frac{(T-(T-\delta ))^2}{2u}\right) , \end{aligned}$$

  and this function $u\mapsto \frac{1}{\sqrt{2\pi }u^{3/2}}\exp \left( -\frac{(T-(T-\delta ))^2}{2u}\right)$ is integrable on $(0,T-t)$.

• 2. $x \mapsto \frac{1}{\sqrt{2\pi }u^{3/2}}\exp \left( -\frac{(T-x)^2}{2u}\right)$ is continuous on $[t,T-\delta ]$.

• 3. Let us prove that the series $\sum _{k=-\infty }^{+\infty } h_k(u,x)$ is a continuous function of x. Since each $h_k$ is a continuous function of x, it is sufficient to prove that the series converges uniformly, which is implied by the fact that

  $$\begin{aligned}&\sum _{k=-\infty }^{+\infty } \sup _{t\le x\le T} \left| h_k(u,x)\right| <+\infty , \text { since}\\&\quad \left| h_k(u,x)\right| \le (T-t)|2k-1|\exp \left( -2\frac{(T-u-t)(T-t)}{u} k(k-1)\right) . \end{aligned}$$

The point 1. above and the fact that $\sum _{k=-\infty }^{+\infty } h_k(u,x) \leqslant 1$ show that the integrand in (A.1) is bounded from above by an integrable function, which is independent of x. The points 2. and 3. allow to deduce that the integrand in (A.1) is a continuous function of x. Hence $x\mapsto \overline{H}(t,x)$ is continuous, for $t<x<T$.

Let us now prove that $\lim _{x\rightarrow T^-}\overline{H}(t,x)=1$. Recall that

$$\begin{aligned} \overline{H}(t,x)&= \int _t^{T} f(u) \sum _{k=-\infty }^{+\infty } \left( 1-2k\frac{T-t}{T-x}\right) \\&\quad \times \exp \left( \frac{T-u}{u-t}\left[ -2k^2(T-t)+2k(T-x)\right] \right) \, du, \end{aligned}$$

with

$$\begin{aligned} f(u) = \left\{ \begin{array}{ll} \frac{T-x}{\sqrt{2\pi }(u-t)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(u-t)}\right) &{} \text{ if } x<T \\ \mathbf {1}_{\{u=t\}} &{} \text{ if } x=T. \end{array} \right. \end{aligned}$$

Isolating the term $k=0$ in the sum, we get,

$$\begin{aligned}&\overline{H}(t,x) = \int _t^{T} \left\{ f(u) + \frac{1}{\sqrt{2\pi }(u-t)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(u-t)}\right) \right. \\&\quad \left. \times \sum _{k=-\infty , k\ne 0}^{+\infty } \left( T-x-2k(T-t)\right) \exp \left( \frac{T-u}{u-t}\left[ -2k^2(T-t)+2k(T-x)\right] \right) \right\} \, du. \end{aligned}$$

When $x\rightarrow T$, the second term in the previous integral simplifies to

$$\begin{aligned} \frac{1}{\sqrt{2\pi }(u-t)^{3/2}} \sum _{k=-\infty , k\ne 0}^{+\infty } \left( -2k(T-t)\right) \exp \left( \frac{T-u}{u-t}\left[ -2k^2(T-t)\right] \right) =0, \end{aligned}$$

and

$$\begin{aligned} \lim _{x\rightarrow T} \overline{H}(t,x) = \lim _{x\rightarrow T} \int _t^{T} f(u) \, du = 1. \end{aligned}$$

Finally, for $x=t$, we have

$$\begin{aligned} \overline{H}(t,t) = \int _t^{T} f(u) \sum _{k=-\infty }^{+\infty } \left( 1-2k\right) \exp \left( \frac{(T-u)(T-t)}{u-t}2k(1-k) \right) \, du=0, \end{aligned}$$

since

$$\begin{aligned} \sum _{k=-N}^{N} \left( 1-2k\right) \exp \left( Ck(1-k) \right) = \frac{2N+1}{\exp (CN(N+1))} \end{aligned}$$

goes to 0 as N goes to $+\infty$, for any $C>0$. $\square$

The following observations will be useful: we can write $\overline{H}(t,x)$ in the following way:

$$\begin{aligned} \overline{H}(t,x)&= \int _t^{T} \frac{1}{\sqrt{2\pi }(u-t)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(u-t)}\right) \nonumber \\&\quad \times \left\{ T-x-4(T-t)\sum _{k=1}^{+\infty } k \exp (-2k^2C_{t,u,T})\sinh \left( 2k(T-x)\frac{T-u}{u-t}\right) \right. \nonumber \\&\quad \left. +2(T-x)\sum _{k=1}^{+\infty } \exp (-2k^2C_{t,u,T})\cosh \left( 2k(T-x)\frac{T-u}{u-t}\right) \right\} \, du, \end{aligned}$$

(A.2)

with $C_{t,u,T}:= \frac{(T-t)(T-u)}{u-t}$. This can be written

$$\begin{aligned} \overline{H}(t,x) =&\int _t^{T} h(x,u)\, du, \end{aligned}$$

with

$$\begin{aligned} h(x,u):= g(x,u)\times \Big \{T-x-4(T-t)\sum _{k=1}^{+\infty }f^1_k(x,u)+ 2(T-x)\sum _{k=1}^{+\infty }f^2_k(x,u)\Big \}. \end{aligned}$$

Lemma A.2

Both series $\sum f^1_k$ and $\sum f^2_k$ are $\mathcal C^2$ functions of x and

$$\begin{aligned} \frac{\partial ^2}{\partial x^2} \sum f^i_k = \sum \frac{\partial ^2}{\partial x^2}f^i_k, \quad i=1,2. \end{aligned}$$

Proof

We will prove the claim for $\sum f^1_k$, the proof for $\sum f^2_k$ being very similar. For any $k\ge 1$, the map $x\mapsto f^1_k(x,u)$ is $C^2$ and

$$\begin{aligned} \frac{\partial ^2}{\partial x^2} f^1_k(x,u) = 4k^3 \exp (-2k^2C_{t,u,T})\left( \frac{T-u}{u-t}\right) ^2 \sinh \left( 2k(T-x)\frac{T-u}{u-t}\right) . \end{aligned}$$

The function $\sinh$ being increasing,

$$\begin{aligned} \left| \frac{\partial ^2}{\partial x^2} f^1_k(x,u)\right| \le 4k^3 \exp (-2k^2C_{t,u,T})\left( \frac{T-u}{u-t}\right) ^2 \sinh \left( 2k(T-t)\frac{T-u}{u-t}\right) , \end{aligned}$$

which implies that the series $\sum \frac{\partial ^2}{\partial x^2} f^1_k(x,u)$ converges uniformy, and from which we deduce the claim. $\square$

Lemma A.3

For every t, $x\mapsto \overline{H}(t,x)$ is $\mathcal C^2$.

Proof

As previously, let $x\in [t,T-\delta ]$ where $\delta >0$. We have

$$\begin{aligned} \frac{\partial g}{\partial x}(x,u)&= \frac{1}{\sqrt{2\pi }(u-t)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(u-t)}\right) \left[ \frac{T-x}{u-t}\right] ,\\ \frac{\partial ^2g}{\partial x^2}(x,u)&= \frac{1}{\sqrt{2\pi }(u-t)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(u-t)}\right) \left[ \frac{(T-x)^2}{(u-t)^2}-\frac{1}{u-t}\right] . \end{aligned}$$

Using Lemma A.2, we get

$$\begin{aligned} \frac{\partial ^2h}{\partial x^2}(x,u) = \frac{1}{\sqrt{2\pi }(u-t)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(u-t)}\right) (I_1+I_2+I_3), \end{aligned}$$

where

$$\begin{aligned} I_1&:= \left[ \frac{(T-x)^2}{(u-t)^2}-\frac{1}{u-t}\right] \Big \{T-x-4(T-t)\sum _{k=1}^{+\infty }f^1_k(x,u)\\&\quad +2(T-x)\sum _{k=1}^{+\infty }f^2_k(x,u)\Big \}, \\ I_2&:= -16(T-t)\frac{(T-u)^2}{(u-t)^2} \sum _{k=1}^{+\infty } k^3 \exp (-2k^2C_{t,u,T})\sinh \left( 2k(T-x)\frac{T-u}{u-t}\right) \\&\quad +8\frac{T-u}{u-t} \sum _{k=1}^{+\infty } k \exp (-2k^2C_{t,u,T})\sinh \left( 2k(T-x)\frac{T-u}{u-t}\right) \\&\quad +8(T-x)\frac{(T-u)^2}{(u-t)^2} \sum _{k=1}^{+\infty } k^2 \exp (-2k^2C_{t,u,T})\cosh \left( 2k(T-x)\frac{T-u}{u-t}\right) , \end{aligned}$$

and

$$\begin{aligned} I_3&:= \left[ \frac{T-x}{u-t}\right] \\&\quad \times \left\{ -1+8C_{t,u,T} \sum _{k=1}^{+\infty } k^2 \exp (-2k^2C_{t,u,T})\cosh \left( 2k(T-x)\frac{T-u}{u-t}\right) \right. \\&\quad -2 \sum _{k=1}^{+\infty } \exp (-2k^2C_{t,u,T})\cosh \left( 2k(T-x)\frac{T-u}{u-t}\right) \\&\quad \left. - 4\frac{(T-u)(T-x)}{u-t}\sum _{k=1}^{+\infty } k \exp (-2k^2C_{t,u,T})\sinh \left( 2k(T-x)\frac{T-u}{u-t}\right) \right\} . \end{aligned}$$

We have

$$\begin{aligned} \left| \frac{1}{\sqrt{2\pi }(u-t)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(u-t)}\right) I_1\right| \le g(x,u)\left( \frac{T-x}{u-t}\right) ^2+g(x,u)\frac{1}{u-t}. \end{aligned}$$

For any $x\in [t, T-\delta ]$, we have

$$\begin{aligned} g(x,u)\left( \frac{T-x}{u-t}\right) ^2 \le \frac{(T-t)^2}{\sqrt{2\pi }(u-t)^{5/2}}\exp \left( -\frac{(T-(T-\delta )^2}{2(u-t)}\right) , \end{aligned}$$

and

$$\begin{aligned} g(x,u)\frac{1}{u-t}\le \frac{1}{\sqrt{2\pi }(u-t)^{5/2}}\exp \left( -\frac{(T-(T-\delta )^2}{2(u-t)}\right) . \end{aligned}$$

The two functions on the RHS are integrable on $(0,T-t)$.

We can show that all the terms appearing in the quantities $\left| \frac{1}{\sqrt{2\pi }(u-t)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(u-t)}\right) I_2\right|$ and $\left| \frac{1}{\sqrt{2\pi }(u-t)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(u-t)}\right) I_3\right|$ are bounded from above by $\frac{K}{(u-t)^{1/2}}$, for $u\in (t,t+\varepsilon )$, where $\varepsilon >0$ and where K is a positive constant (independent of u and x). Let us for example study in detail the term

$$\begin{aligned} \tilde{I}_2&:= \frac{1}{\sqrt{2\pi }(u-t)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(u-t)}\right) 16(T-t)\frac{(T-u)^2}{(u-t)^2} \\&\quad \times \sum _{k=1}^{+\infty } k^3 \exp (-2k^2C_{t,u,T})\sinh \left( 2k(T-x)\frac{T-u}{u-t}\right) , \end{aligned}$$

the arguments for the other terms are very similar.

Using that $\sinh$ is increasing and that $C_{t,u,T} = \frac{(T-t)^2}{u-t}- (T-t)$, we get

$$\begin{aligned} \left| \tilde{I}_2\right|&\le \frac{1}{\sqrt{2\pi }(u-t)^{3/2}} \exp \left( -\frac{(T-x)^2}{2(u-t)}\right) 16\frac{(T-t)^3}{(u-t)^2}\\&\quad \times \sum _{k=1}^{+\infty } k^3 \exp \left( -k^2\left[ \frac{2(T-t)^2}{u-t}-2(T-t)\right] \right) \sinh \left( 2k\frac{(T-t)^2}{u-t}\right) . \end{aligned}$$

Define

$$\begin{aligned} S(u):= \sum _{k=1}^{+\infty } k^3 \exp \left( -k^2\left[ \frac{c}{u}-c'\right] \right) \sinh \left( k\frac{c}{u}\right) , \quad 0<u\le (T-t) \end{aligned}$$

where $c=2(T-t)^2$ and $c'=2(T-t)$ ($\frac{c}{u}-c'\ge 0$ for $u\le T-t$). We are going to prove that $\lim _{u\rightarrow 0^+} u\log S(u) = 0$. Assuming momentarily that this limit is indeed equal to 0, then there exists $\bar{u}>t$ such that for any $u\le \bar{u}$,

$$\begin{aligned} 2(u-t)\log S(u-t)-2(u-t)\log \left( \frac{(2\pi )^{1/6}(u-t)}{2^{4/3}(T-t)}\right) ^3\le T-(T-\delta ). \end{aligned}$$

This implies for any $x\in [t,T-\delta ]$ and any $u\le \bar{u}$,

$$\begin{aligned} \frac{1}{\sqrt{2\pi }(u-t)^{3/2}} \exp \left( -\frac{(T-x)^2}{2(u-t)}\right) 16\frac{(T-t)^3}{(u-t)^2} S(u-t)\le \frac{1}{(u-t)^{1/2}}. \end{aligned}$$

This implies, by splitting the interval [t, T] into $[t,t+\varepsilon )$ and $[t+\varepsilon , T]$ with $\varepsilon$ small enough, that $\left| \frac{\partial ^2h}{\partial x^2}(x,u)\right|$ is bounded from above by an integrable function which is independent of x, on each of these intervals. Using similar arguments as in the proof of Lemma A.2, we can show that $x \mapsto \frac{\partial ^2h}{\partial x^2}(x,u)$ is a continuous function. From this, we can conclude that $x\mapsto \overline{H}(t,x)$ is $\mathcal C^2$.

It remains to prove that $\lim _{u\rightarrow 0^+} u\log S(u) = 0$, which is the purpose of Lemma A.4. $\square$

Lemma A.4

$\lim _{u\rightarrow 0^+} u\log S(u) = 0$.

Proof

Define,

$$\begin{aligned} \varphi (v,u):= \frac{1}{2} v^3 \exp \left( -v^2\left[ \frac{c}{u}-c'\right] \right) \left( \exp \left( v\frac{c}{u}\right) - \exp \left( -v\frac{c}{u}\right) \right) , \quad v>1, \end{aligned}$$

(A.3)

which is a differentiable function such that

$$\begin{aligned} \frac{\partial \varphi }{\partial v} (v,u)= & {} \exp \left( -v^2\left[ \frac{c}{u}-c'\right] \right) v^2\left[ v \frac{c}{u} \cosh \left( v \frac{c}{u}\right) \right. \\&\left. -\left( 2 v^2\left[ \frac{c}{u}-c'\right] -3\right) \sinh \left( v \frac{c}{u}\right) \right] . \end{aligned}$$

The sign of this derivative is given by the sign of $v \frac{c}{u} \cosh \!\left( \!v \frac{c}{u}\!\right) \!-\!\left( \!2 v^2[\frac{c}{u}\!-\!c'] \!-\!3\!\right) \sinh \!\left( \!v \frac{c}{u}\!\right)$, which is going to be negative for u small enough. Indeed, this quantity is negative if and only if

$$\begin{aligned} \tanh \left( v \frac{c}{u}\right) \left( 2v - 2v \frac{c'}{c}u - \frac{3}{v} \frac{u}{c}\right) >1. \end{aligned}$$

Since $\tanh \left( v \frac{c}{u}\right) \left( 2v - 2v \frac{c'}{c}u - \frac{3}{v} \frac{u}{c}\right)$ converges to $2v>1$ as u goes to 0, we have that $v\mapsto \varphi (v,u)$ is decreasing for u small enough. We can then apply the standard technique of comparison between series and integrals: for $k \ge 1$,

$$\begin{aligned} \int _k^{k+1}\varphi (v,u) dv \le \varphi (k,u) = \int _{k-1}^k \varphi (k,u) dv \le \int _{k-1}^k \varphi (v,u) dv. \end{aligned}$$

Taking the sum from $k=2$ to $k=N$, and then the limit when N goes to $+\infty$, we have

$$\begin{aligned} \varphi (1,u)+\int _2^{+\infty }\varphi (v,u) dv \le \sum _{k=1}^{+\infty } \varphi (k,u) \le \varphi (1,u)+\int _1^{+\infty }\varphi (v,u) dv. \end{aligned}$$

(A.4)

Let us compute the integral on the right hand side, (which is well defined since for v in the neighbourhood of $+\infty$, $\varphi (v,u) = o(\frac{1}{v^p})$ for any $p>1$), that we expand as:

$$\begin{aligned} K&:=\int _1^{+\infty }\varphi (v,u) dv =\frac{1}{2} (K^+-K^-), \quad \text {where}\\ K^+&:= \int _1^{+\infty } v^3 \exp \left( -v^2\left[ \frac{c}{u}-c'\right] +v \frac{c}{u}\right) \, dv \quad \text {and }\\ K^-&:= \int _1^{+\infty } v^3 \exp \left( -v^2\left[ \frac{c}{u}-c'\right] -v \frac{c}{u}\right) \, dv. \end{aligned}$$

Writing $\bar{\Phi }_j(x):= E[Z^j\mathbf {1}_{\{Z>x\}}]$, $j \ge 0$ and $\bar{\Phi }(x):=\bar{\Phi }_0(x)=P(Z>x)$, with Z a r.v. having a standard Gaussian distribution, we have

$$\begin{aligned} K^+&= \sqrt{2\pi }\frac{\exp \left( \frac{c^2}{4u^2[\frac{c}{u}-c']}\right) }{4[\frac{c}{u} -c']^2} \\&\quad \times \left[ \kappa _2^3\bar{\Phi }(\kappa _1-\kappa _2) +3\kappa _2^2\bar{\Phi }_1(\kappa _1-\kappa _2)+3\kappa _2\bar{\Phi }_2(\kappa _1-\kappa _2)+\bar{\Phi }_3(\kappa _1-\kappa _2)\right] , \end{aligned}$$

where $\kappa _1(u):= \sqrt{2(\frac{c}{u}-c')}$ and $\kappa _2(u):= \frac{c}{u\kappa _1(u)}$ (for simplicity, we do not indicate the dependence on u of $\kappa _1$ and $\kappa _2$ when there is no ambiguity). Similarly we have

$$\begin{aligned} K^-&= \sqrt{2\pi }\frac{\exp \left( \frac{c^2}{4u^2[\frac{c}{u}-c']}\right) }{4[\frac{c}{u} -c']^2} \\&\quad \times \left[ \bar{\Phi }_3(\kappa _1+\kappa _2) -3\kappa _2\bar{\Phi }_2(\kappa _1+\kappa _2)+3\kappa _2^2\bar{\Phi }_1(\kappa _1+\kappa _2)-\kappa _2^3 \bar{\Phi }(\kappa _1+\kappa _2)\right] . \end{aligned}$$

By an integration by parts,

$$\begin{aligned} \bar{\Phi }_3(x)= & {} \frac{1}{\sqrt{2\pi }}\exp \left( -\frac{x^2}{2}\right) (x^2+2)\\ \bar{\Phi }_2(x)= & {} \frac{x}{\sqrt{2\pi }}\exp \left( -\frac{x^2}{2}\right) +\bar{\Phi }(x)\\ \bar{\Phi }_1(x)= & {} \frac{1}{\sqrt{2\pi }}\exp \left( -\frac{x^2}{2}\right) , \quad \text {and it is standard that}\\ \bar{\Phi }(x)\sim & {} \frac{1}{x} \frac{1}{\sqrt{2\pi }}\exp \left( -\frac{x^2}{2}\right) . \end{aligned}$$

Using these explicit expressions, the value of $\varphi (1,u)$ given through (A.3) and inequality (A.4), it is straightforward to show that $u\log S(u)$ converge to 0 as u goes to $0^+$. $\square$

Lemma A.5

For every $x\in (0,T)$, $t \in [0,x] \mapsto \overline{H}(t,x)$ is $\mathcal C^1$.

Proof

Recall that (do a change of variable $v=\frac{u-t}{T-t}$)

$$\begin{aligned} \overline{H}(t,x)&= \int _0^{T} \left\{ f(u,t,x) + \frac{1}{\sqrt{2\pi }(T-t)^{1/2}(u)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(T-t)u}\right) \right. \\&\quad \times \sum _{k=-\infty , k\ne 0}^{+\infty } \left( T-x-2k(T-t)\right) \exp \left( \frac{1-u}{u}\left[ -2k^2(T-t)\right. \right. \\&\quad \left. \left. \left. +2k(T-x)\right] \right) \right\} \, du.\\&=\int _0^{T} \Big \{ f(u,t,x)+g(t,x,u)\sum _{k=-\infty , k\ne 0}^{+\infty } h_k(t,x,u) \Big \}\, du, \end{aligned}$$

where

$$\begin{aligned} f(u,t,x) = \left\{ \begin{array}{ll} \frac{T-x}{\sqrt{2\pi }(T-t)^{1/2}(u)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(T-t)u}\right) &{} \text{ if } x<T \\ \mathbf {1}_{\{u=t\}} &{} \text{ if } x=T. \end{array} \right. \end{aligned}$$

Let $x\in [0,T)$, $t_0\le 0$ and $\delta >0$ such that $t_0+\delta \in [0,x)$.

• For all $t\in [0, t_0+\delta ]$

  $$\begin{aligned} \left| \frac{\partial f}{\partial t} \right|&=\frac{T-x}{2\sqrt{2\pi }(T-t)^{3/2}(u)^{3/2}}\exp \left( -\frac{(T-x)^2}{2(T-t)u}\right) \\&\quad +\frac{(T-x)^3}{\sqrt{2\pi }(T-t)^{3/2}(u)^{5/2}}\exp \left( -\frac{(T-x)^2}{2(T-t)u}\right) \\&\le \frac{T-x}{2\sqrt{2\pi }(T-t_0-\delta )^{3/2}(u)^{3/2}}\exp \left( -\frac{(T-x)^2}{2Tu}\right) \\&\quad +\frac{(T-x)^3}{\sqrt{2\pi }(T-t_0-\delta )^{3/2}(u)^{5/2}}\exp \left( -\frac{(T-x)^2}{2Tu}\right) . \end{aligned}$$

  The right hand side is an integrable function on [0, 1] independent of $t\in [0, t_0+\delta ]$. The function $t\mapsto \frac{\partial f}{\partial t}$ is continuous on $[0, t_0+\delta ]$

• Note that $g(t,x,u)=f(t,x,u)/(T-x)$. Hence the functions $t\mapsto g(t,x,u)$ and $t\mapsto \frac{\partial g}{\partial t}$ are continuous on $[0, t_0+\delta ]$ and can be bounded from above by a function independent of f and integrable w.r.t u.

• Let us prove that the $\sum _{k=-\infty , k\ne 0}^{+\infty } h_k(t,x,u)$ is a $\mathcal C^1$ function. Since

  $$\begin{aligned}&\sup _{t\in [0,t_0+\delta ]}\left| \frac{\partial h_k(t,x,u)}{\partial t}\right| \\&\quad =\sup _{t\in [0,t_0+\delta ]}\left| 2k\exp \left( \frac{1-u}{u}\left[ -2k^2(T-t)+2k(T-x)\right] \right) \right. \\&\qquad + \left. 2k^2\frac{1-u}{u}(T-x-2k(T-t))\exp \left( \frac{1-u}{u}\left[ -2k^2(T-t)+2k(T-x)\right] \right) \right| \\&\quad \le 2|k|\exp \left( \frac{1-u}{u}\left[ -2k^2(T-t_0-\delta )+2k(T-x)\right] \right) \\&\qquad +2k^2\frac{1-u}{u}(T-x+2|k|T)\exp \left( \frac{1-u}{u}\left[ -2k^2(T-t_0-\delta )+2k(T-x)\right] \right) \\&\quad \le 2|k|\exp \left( \frac{1-u}{u}\left[ -2k^2(T-x)+2k(T-x)\right] \right) \\&\qquad +2k^2\frac{1-u}{u}(T-x+2|k|T)\exp \left( \frac{1-u}{u}\left[ -2k^2(T-x)+2k(T-x)\right] \right) , \end{aligned}$$

  which is the term of a convergent series, the result follows.

• $|\frac{\partial g(t,x,u)}{\partial t}\sum _{k=-\infty , k\ne 0}^{+\infty } h_k(t,x,u)|\le |\frac{\partial g(t,x,u)}{\partial t}|$, which is bounded from above by an integrable function independent of $t\in [0, t_0+\delta ]$ (see above).

• Define

  $$\begin{aligned} \widetilde{S}_n(u):= \sum _{k=1}^{+\infty } k^n \exp \left( -k^2c\left[ \frac{1-u}{u}\right] \right) \sinh \left( kc\frac{1-u}{u}\right) , \quad 0<u\le (T-t), \end{aligned}$$

  where $c=(T-x)$, $n=1,2,3$. Similar arguments as the ones used in the proof of Lemma A.4 imply that $\lim _{u\rightarrow 0^+} u \log \widetilde{S}_n(u) = 0$. Then there exists $\bar{u}$ such that for any $u\le \bar{u}$

  $$\begin{aligned} \left| g(t,x,u)\sum _{k=-\infty , k\ne 0}^{+\infty }\frac{\partial h_k(t,x,u)}{\partial t}\right| \le C\frac{1}{\sqrt{u}}, \end{aligned}$$

  where C is constant independent of u and t.$\square$