Convolution of an approximate identity with a Schwartz function.

Question

The following was stated in class:

If $f$ is a Schwartz function, and $(\varphi_i)_{i ∈ N}$ is an approximate identity (in other words, $\varphi_i(x) \ge 0$ , $\int_{\mathbb{R}} \varphi_i=1 $ for all $i$, and given fixed $\delta>0$, $\lim_{i \rightarrow \infty} \int_{|X| \ge \delta} \varphi_i=0),$ then the convolution $f * \varphi_i→ f.$

However, there was not a proof given, and I had trouble finding one by searching the web. I was wondering:

1. How is this proven?
2. In what sense does $f * \varphi_i$ converge to $f$? Is this pointwise, or in $L^2$, or uniformly pointwise?

Answer 1

I ended up figuring this out in the Fall, but I forgot to return to answer this question.

The answer is that convergence is at least pointwise.

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Lemma: For $\varphi_n$ an approximate identity and $f$ a Schwartz function, $\lim_{n \to \infty}\int_{\mathbb{R}} \varphi(x) \circ f(x)=f(0)$.

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We know that $$f*\varphi_n(x)=\int_{\mathbb{R}}f(x-y)\varphi_n(y)dy. $$

Let $$F_x(y)=f(x-y)=\int_{\mathbb{R}}F_x(y)\varphi_n(y)dy.$$

Taking limits and using the lemma, at a fixed $x$, we have \begin{align} \lim_{n \to \infty}f * \varphi_n(x) &=\lim_{n \to \infty} \int_{\mathbb R} F_x(y)\varphi_n(y)dy\\ &=\lim_{n \to \infty}F_x(0)\\ &=f(x-0)\\ &=f(x). \end{align}