# Average integral for continuous functions with compact support

Let $f$ be a continuous function with compact support in $\mathbb{R}^n$. Show that $$\lim_{r\to 0} \frac{1}{|B_r(x)|} \int_{B_r(x)} f(y)\,dy = f(x),$$ where $B_r(x)$ is the ball of radius $r$ centered at $x$.

In 1d, it looks like the Fundamental Theorem of Calculus. However, I'm not sure how to prove it in higher dimensions. How can I use the "compact support"?

Thank you.

• Continuity plus compact support give uniform continuity, then the claim is trivial. Aug 24 '15 at 22:10
• you should constraint the integral to $B_r(x)$ Aug 24 '15 at 22:19
• Hint: Let $\epsilon > 0$. Then, for $r$ sufficiently small, we have $|f(y) - f(x)|<\epsilon$ for $y\in B_r(x)$. Aug 24 '15 at 22:21
• @JackD'Aurizio: Why do we need $f$ to be with compact support? It seems to me that "continuous" suffices to prove the statement. Since the proof I came up with, does not rely on the compact support (and is still very simple) I wonder why, when assuming $f$ with compact support, "the claim is trivial". math.stackexchange.com/questions/1106426/… Aug 6 '16 at 11:42

Let's make a sketch. Given na $\epsilon > 0$ there is a $\delta >0$ such that $$|\frac{1}{|B_r(X)|}\int_{B_r(x)}f(y)dy - f(x)| < \epsilon$$ if $|x-y| < \delta$. Then $$|\frac{1}{|B_r(X)|}\int_{B_r(x)}f(y)dy - f(x)| \le \frac{1}{|B_r(X)|}\int_{|B_r(X)|}|f(x)-f(y)|dy.$$ Since we have compact support and $f$ is continuous the $f$ is uniformly continuous. Then given $\epsilon >0$ there is $\delta >0$ such that $|f(x) - f(y)| < \epsilon$ if $|x-y| < \delta$. Then $$\frac{1}{|B_r(X)|}\int_{B_r(X)}|f(x)-f(y)|dy < \epsilon.$$
• you should link $r$ to $\delta$ ... somehow Aug 24 '15 at 22:47
• In fact, once you are getting $x,y \in B_r(x)$ you want that this distance do not "ultrapasse" the ball. But $r \to 0$ means $\delta \to 0$. Aug 24 '15 at 22:49
• @LeonardoFranciscoCavenaghi: How exactly is the compact support of $f$ used? Do we really need $f$ to be uniformly continuous? Since $f$ is continuous and $x$ is fixed, we already get $|f(x) - f(y)|< \varepsilon$ for all $y$ with $|x - y| < \delta$ . What am I missing? Aug 6 '16 at 10:13