Mollified by heat kernel: is $f_t \to f$ in $L^1$?

Question

We consider the heat kernel $$ g :\mathbb R_{>0} \times \mathbb R^d \to \mathbb R, (t, x) \mapsto \frac{1}{(4\pi t)^{d/2}} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ). $$

Then $\int_{\mathbb R^d} g(t, x) \, \mathrm dx=1$ for all $t>0$. Fix $f \in C^\infty_c (\mathbb R^d)$. Let $f_t := g(t, \cdot) * f$ for all $t>0$. It is mentioned in this Wikipedia page that $f_t \to f$ pointwise everywhere, i.e., $\lim_{t \to 0^+} f_t (x) = f(x)$ for all $x \in \mathbb R^d$.

1. Is $f_t \to f$ in $L^1$?
2. Is $f_t \to f$ uniformly on compact sets?
3. Is $f_t \to f$ uniformly?

Any reference is appreciated. Thank you so much for your elaboration?

Answer 1

It seems I have found a proof that $f_t \to f$ uniformly. Please let me know I you found some logical mistakes in my below attempt.

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We have $g(t, \cdot)$ is the p.d.f. of a $d$-dimensional Gaussian random variable with mean $0$ and covariance matrix $2t I_d$. Here $I_d$ is the $d$-dimensional identity matrix. We have the Gaussian tail bound $$ \int_{\mathbb R^d \setminus B(0, t^{1/4})} g(t, x) \, \mathrm dx \le 2\exp \bigg ( -\frac{\sqrt t}{2\operatorname{tr} (2tI_d)} \bigg ) = 2\exp \bigg ( -\frac{1}{4\sqrt{t}d} \bigg ). $$

We have $f$ is uniformly continuous. Fix $\varepsilon >0$. There is $\delta>0$ such that $$ |f(x-y) - f(x)| < \varepsilon \quad \forall x\in \mathbb R^d, \forall y \in B(0, \delta). $$

Then for all $x \in \mathbb R^d$, $$ \begin{align} |f_t (x) - f(x)| &= \bigg | \int_{\mathbb R^d} g(t, y)(f(x-y)-f(x)) \, \mathrm dy \bigg | \\ &\le \int_{\mathbb R^d \setminus B(0, t^{1/4})} g(t, y) |f(x-y)-f(x)| \, \mathrm dy + \int_{B(0, t^{1/4})} g(t, y) |f(x-y)-f(x)| \, \mathrm dy \\ &\le 2 \|f\|_\infty \int_{\mathbb R^d \setminus B(0, t^{1/4})} g(t, y) \, \mathrm dy + \int_{B(0, t^{1/4})} g(t, y) |f(x-y)-f(x)| \, \mathrm dy \\ &\le 4 \|f\|_\infty \exp \bigg ( -\frac{1}{4\sqrt{t}d} \bigg ) + \int_{B(0, t^{1/4})} g(t, y) |f(x-y)-f(x)| \, \mathrm dy. \end{align} $$

If $t^{1/4} < \delta$, then $$ \int_{B(0, t^{1/4})} g(t, y) |f(x-y)-f(x)| \, \mathrm dy \le \varepsilon \int_{B(0, t^{1/4})} g(t, y) \, \mathrm dy \le \varepsilon. $$

The claim then follows.

Remark It is enough that $f$ is bounded and uniformly continuous.