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A sequence of mollifiers $(\rho_n)_{n \geq 1}$ is any sequence of functions on $\mathbb{R}^d$ such that $$ \rho_n \in C_c^{\infty} (\mathbb{R}^d), \quad \operatorname{supp} \rho_n \subset \overline{B(0, r_n)}, \quad \sup_n \|\rho_n\|_1 < \infty, \quad \int \rho_n \xrightarrow{n \to \infty} 1, $$ where $r_n \to 0^+$ and $B(x, r)$ denotes an open ball centered at $x$ with radius $r$.

I would like to prove a statement mentioned in this comment, i.e.,

Theorem Let $f \in \mathcal C (\mathbb R^d)$. Then $\rho_n * f \xrightarrow{n \to \infty} f$ uniformly on compact sets.

Could you have a check on my below attempt?

Proof WLOG, we assume $\sup_n r_n \le 1$. Fix a compact subset $K$ of $\mathbb{R}^d$. Fix $\varepsilon>0$. Then $f$ is uniformly continuous on $K' :=K - \overline{B(0, 1)}$. There is $\delta>0$ such that if $x,y \in K'$ with $|x-y| < \delta$ then $|f(x)-f(y) |< \varepsilon$. Let $a_n := \int \rho_n$. Then $a_n \to 1$. There is $N \in \mathbb N$ such that $|a_n-1| < \varepsilon$ for all $n \ge N$. Let $\alpha :=\sup_{x\in K} |f(x)| < \infty$. If $x \in K$ and $r_n < \delta$ and $n > N$ then $$ \begin{align} &|\rho_n * f (x)-f(x)| \\ = & \bigg | \int_{\overline{B(0, r_n)}} (f(x-y)-f(x)) \rho_n(y) \, \mathrm d y + (a_n-1) f(x) \bigg |\\ \le & \int_{\overline{B(0, r_n)}} |f(x-y)-f(x)| | \rho_n(y)| \, \mathrm d y + \alpha |a_n-1| \\ \le & \varepsilon \|\rho_n\|_1 + \varepsilon\alpha. \end{align} $$

The claim then follows by taking the limit $\varepsilon \downarrow 0$.

Rhys Steele
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Analyst
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1 Answers1

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I see two minor points.

For the final inequality you need $r_n < \delta$ rather than $r_n \le 1$ (which of course you can arrange by taking $n$ large enough which puts a second constraint on your choice of $N$).

For the final inequality, you should also write $\sup_k \|\rho_k\|_1$ instead of $\|\rho_n\|_1$ since the value of $n$ you can choose needs to depend on $\delta$ and hence on $\varepsilon$ so as written you cannot easily pass $\varepsilon \to 0$.

Otherwise, everything looks right to me.

Rhys Steele
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  • Maybe I misunderstood your comment, but I wrote "If $x \in K$ and $r_n < \delta$...". – Analyst Apr 12 '23 at 07:18
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    Maybe I am making a proof writing point more than a correction, you say "if $r_n < \delta$..." but you never arrange for the condition in that if to be satisfied. A properly written proof would take an $N$ such that $n \ge N$ implies that $|a_n - 1|< \varepsilon$ and $r_n < \delta$ – Rhys Steele Apr 12 '23 at 07:21
  • Ah I got it. Thank you so much for your elaboration! – Analyst Apr 12 '23 at 07:21