For $t>0$, we define $$ g_t:\mathbb R^d \to \mathbb R_{\ge 0}, x \mapsto e^{- |x|^2 /t}. $$
In solving this question about heat kernel, I have to verify the following lemma, i.e.,
Let $t_n, t >0$ such that $t_n \to t$ as $n \to +\infty$. Then $g_{t_n} \to g_t$ uniformly as $n \to +\infty$.
- Could you have a check on my attempt?
- Are there other approaches?
Thank you so much for your elaboration!
By mean value theorem, for each $x \in \mathbb R^d$, there is $t_{n, x}$ between $t_n$ and $t$ such that $$ \begin{align} g_{t_n} (x) - g_t (x) &= e^{- |x|^2 /t_n} - e^{- |x|^2 /t} \\ &= \frac{|x|^2}{t_{n, x}^2} e^{- |x|^2 /t_{n, x}} (t_n - t) \\ &= \frac{|x|^2}{t_{n, x}} e^{- |x|^2 /t_{n, x}} \frac{t_n - t}{t_{n, x}}. \end{align} $$
We have $\sup_{h \ge 0} he^{-h} \le \frac{1}{e} < 1$. Then $$ |g_{t_n} (x) - g_t (x)| \le \frac{t_n - t}{t_{n, x}} \quad \forall x \in \mathbb R^d. $$
WLOG, we assume $\alpha :=\inf_n t_n >0$. Then $$ |g_{t_n} (x) - g_t (x)| \le \frac{t_n - t}{\alpha} \quad \forall x \in \mathbb R^d. $$
The claim then follows.
With the same strategy, we can proved that $\|g_{t_n} - g_t\|_{L^p} \to 0$ for $p \in [1, \infty)$.
For $t>0$, we define $f_t:\mathbb R^d \to \mathbb R$ by $$ f_t (x) := \frac{|x|^{2}}{t} \exp \bigg ( -\frac{|x|^{2}}{t} \bigg ) $$
Then $$ g_{t_n} (x) - g_t (x) = f_{t_{n, x}} (x) \frac{t_n - t}{t_{n, x}}. $$
WLOG, we assume there are $0<a<b<\infty$ such that $t_n \in [a, b]$ for all $n$. Then $t_{n, x} \in [a, b]$ for all $n \in \mathbb N$ and $x\in \mathbb R^d$. Then $$ \|g_{t_n} - g_t\|_{L^p} \le \frac{|t_n - t|}{a} \sup_{t \in [a, b]} \|f_t\|_{L^p} $$
We can prove that $$ \sup_{t \in [a, b]} \|f_t\|^p_{L^p} =\sup_{t \in [a, b]} \int_{\mathbb R^d} |f_t (x)|^p \, \mathrm d x < \infty. $$
Then claim then follows.