How to prove $\sup_{t\in [a, b]} \int_{\mathbb R^d} |f_t (x)|^p \, \mathrm d x<\infty$ in higher dimension?

Question

For $t>0$, we define $$ g_t:\mathbb R^d \to \mathbb R_{\ge 0}, x \mapsto e^{- |x|^2 /t}. $$

Let $t_n, t >0$ such that $t_n \to t$ as $n \to +\infty$. I have proved that

Lemma $g_{t_n} \to g_t$ uniformly as $n \to +\infty$.

Let $p \in [1, \infty)$. I would like to prove that

$g_{t_n} \to g_t$ in $L^p$ as $n \to +\infty$.

For $t>0$, we define $f_t:\mathbb R^d \to \mathbb R$ by $$ f_t (x) := \frac{|x|^{2}}{t} \exp \bigg ( -\frac{|x|^{2}}{t} \bigg ) $$

Let $0<a<b < \infty$. It turns out it suffices to prove $$ \sup_{t\in [a, b]} \int_{\mathbb R^d} |f_t (x)|^p \, \mathrm d x = \sup_{t\in [a, b]} \int_{\mathbb R^d} \frac{|x|^{2p}}{t^p} \exp \bigg ( -\frac{p|x|^{2}}{t} \bigg ) \, \mathrm d x < \infty. \tag{1} $$

In case $d=1$, the result is available here.

Could you please elaborate how to prove (1) in higher dimension?

Answer 1

Actually, it's not too hard to obtain (1). Let $y :=\frac{x}{\sqrt{t/(2p)}}$. Then $\mathrm d x = \frac{t^{d/2}}{(2p)^{d/2}}\mathrm d y$ and $|x|^2 = \frac{t}{2p} |y|^2$. Then $$ \begin{align} & \int_{\mathbb R^d} \frac{|x|^{2p}}{t^p} \exp \bigg ( -\frac{p|x|^{2}}{t} \bigg ) \, \mathrm d x \\ ={} & \frac{t^{d/2}}{(2p)^{p+(d/2)}} \int_{\mathbb R^d} |y|^{2p} \exp \bigg ( -\frac{|y|^{2}}{2} \bigg ) \, \mathrm d y. \end{align} $$

The claim then follows easily.