We consider the heat kernel $$ g :\mathbb R_{>0} \times \mathbb R^d \to \mathbb R, (t, x) \mapsto \frac{1}{(4\pi t)^{d/2}} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ). $$
Let $t_n, \bar t >0$ such that $t_n \to \bar t$ as $n \to +\infty$. In order to solve this question about heat kernel, I have to verify
$\Delta g(t, \cdot) \to \Delta g (\bar t, \cdot)$ uniformly as $n \to +\infty$.
- Could you have a check on my attempt?
- Are there other approaches?
Thank you so much for your elaboration!
We have $$ \Delta g(t, x) = \left(\frac{|x|^2-2 d t}{4 t^2}\right) g(t, x). $$
I have already proved that $g (t_n, \cdot) \to g(\bar t, \cdot)$ uniformly as $n \to +\infty$. It suffices to prove that $$ \frac{|\cdot|^2}{t_n^2} g(t_n, \cdot) \xrightarrow{n \to +\infty} \frac{|\cdot|^2}{\bar t^2} g(\bar t, \cdot) \quad \text{uniformly}. $$
We have $$ \frac{|x|^2}{t^2} g(t, x) = \frac{|x|^2}{t^2 (4\pi t)^{d/2}} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ). $$
We define $$ h :\mathbb R_{>0} \times \mathbb R^d \to \mathbb R, (t, x) \mapsto \frac{|x|^2}{4t} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ). $$
It suffices to prove that $$ h(t_n, \cdot) \xrightarrow{n \to +\infty} h(\bar t, \cdot) \quad \text{uniformly}. $$
We have $$ \frac{\partial}{\partial t} h(t, x) = \frac{|x|^4}{16t^3} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ) - \frac{|x|^2}{4t^2} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ). $$
WLOG, we assume there are $\alpha, \beta>0$ such that $\alpha \le t_n \le \beta$ for all $n$. Any polynomial function is dominated by exponential function, so $$ \gamma :=\sup_{t \in [\alpha, \beta], x \in \mathbb R^d} \bigg | \frac{\partial}{\partial t} h(t, x) \bigg | <+\infty. $$
By mean value theorem, for each $x \in \mathbb R^d$ there is $t_{n,x}$ between $t_n$ and $t$ such that $$ h(t_n, x) -h(\bar t, x) = \frac{\partial}{\partial t} h(t, x) \bigg |_{t=t_{n, x}} (t_n-\bar t). $$
Then $$ |h(t_n, x) -h(\bar t, x)| \le \gamma|t_n-\bar t| \quad \forall x \in \mathbb R^d. $$
The claim then follows.
With the same strategy, we can proved that $\|\Delta g(t, \cdot) - \Delta g (\bar t, \cdot)\|_{L^p} \to 0$ for $p \in [1, \infty)$.
We have $$ \Delta g(t, x) = \left(\frac{|x|^2-2 d t}{4 t^2}\right) g(t, x). $$
I have already proved that $g (t_n, \cdot) \to g(\bar t, \cdot)$ in $L^p$ as $n \to +\infty$. It suffices to prove that $$ \frac{|\cdot|^2}{t_n^2} g(t_n, \cdot) \xrightarrow{n \to +\infty} \frac{|\cdot|^2}{\bar t^2} g(\bar t, \cdot) \quad \text{in} \quad L^p. $$
We have $$ \frac{|x|^2}{t^2} g(t, x) = \frac{|x|^2}{t^2 (4\pi t)^{d/2}} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ). $$
We define $$ h :\mathbb R_{>0} \times \mathbb R^d \to \mathbb R, (t, x) \mapsto \frac{|x|^2}{4t} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ). $$
It suffices to prove that $$ h(t_n, \cdot) \xrightarrow{n \to +\infty} h(\bar t, \cdot) \quad \text{in} \quad L^p. $$
By mean value theorem, for each $x \in \mathbb R^d$ there is $t_{n,x}$ between $t_n$ and $t$ such that $$ h(t_n, x) -h(\bar t, x) = \frac{\partial}{\partial t} h(t, x) \bigg |_{t=t_{n, x}} (t_n-\bar t). $$
Then $$ |h(t_n, x) -h(\bar t, x)| \le \bigg | \frac{\partial}{\partial t} h(t_{n, x}, x) \bigg| |t_n-\bar t| \quad \forall x \in \mathbb R^d. $$
WLOG, we assume there are $0 < a <b<\infty$ such that $t_n \in [a, b]$ for all $n$. $$ \gamma :=\sup_{t \in [\alpha, \beta], x \in \mathbb R^d} \bigg | \frac{\partial}{\partial t} h(t, x) \bigg | <+\infty. $$
Then $$ \| h(t_n, \cdot) - h(\bar t, \cdot) \|_{L^p} \le |t_n-\bar t| \underbrace{\sup_{t \in [a, b]} \bigg \| \frac{\partial}{\partial t} h(t, \cdot) \bigg \|_{L^p}}_{\gamma}. $$
It suffices to prove that $\gamma<\infty$. We have $$ \frac{\partial}{\partial t} h(t, x) = \frac{|x|^4}{16t^3} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ) - \frac{|x|^2}{4t^2} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ). $$
Let $$ \begin{align} \varphi_t (x) &:= \frac{|x|^4}{16t^3} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ), \\ \psi_t (x) &:= \frac{|x|^2}{4t^2} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ). \end{align} $$
We have $$ \bigg \| \frac{\partial}{\partial t} h(t, \cdot) \bigg \|_{L^p} \le \|\varphi_t\|_{L^p} + \|\psi_t\|_{L^p}. $$
Then $$ \gamma \le \underbrace{\sup_{t \in [a, b]} \|\varphi_t\|_{L^p}}_{\gamma_1} + \underbrace{\sup_{t \in [a, b]} \|\psi_t\|_{L^p}}_{\gamma_2}. $$
Just as in this question, we can prove that $\gamma_1, \gamma_2 <\infty$. The claim then follows.