Suppose $f$ is a continuous function, and let the convolution $f_n(x) := f \star \varphi_n(x)$ where $\varphi_n$ are smooth test functions. We know $f_n \in C^\infty$.
We know that if $f$ is uniformly continuous, $f_n \to f$ uniformly.
We only have continuity of $f$… how can I show that $f_n$ (or there is a sequence $f_n$) that converges uniformly to $f$ on compact subsets?
We can consider the convolution of $f|_K$ onto a compact subset $K \subset \mathbb{R}$, but this gives us a different sequence $f_n(K)$ for each $K$. I want a single sequence that converges uniformly on the compact subsets.
Best Answer
Let $f$ be as stated, and let $K$ be a compact subset of $\mathbb{R}$. I'll show that $f\star\varphi_{n}$ converges uniformly to $f$ on $K$.
To do this, define a function $h$ which is continuous on $\mathbb{R}$, is $1$ for all points $x$ for which $\mbox{dist}(x,K) \le 1$, and which tapers to $0$ for all $x$ for which $\mbox{dist}(x,K) \ge 2$. This is not hard to construct because the distance function $d(x)=\mbox{dist}(x,K)$ is continuous. The function $(fh)\star\varphi_{n}$ must equal $f\star\varphi_{n}$ on $K$ whenever $n$ is chosen large enough that $\varphi_{n}$ has its support contained in the closed ball of radius $1$ centered at $0$. $fh$ is compactly supported and continuous and, therefore, $(fh)\star\varphi_{n}$ converges uniformly to $fh$ on $\mathbb{R}$. However, $(fh)\star\varphi_{n}=f\star\varphi_{n}$ on $K$ for large $n$, and, therefore, converges uniformly to $f$ on $K$. So $f\star\varphi_{n}$ converges uniformly on compact subsets to $f$.