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why is for given $f\in C^0(\mathbb{R}^n, \mathbb{R})$ $$\lim\limits_{\varepsilon \rightarrow 0}\frac{1}{\operatorname{vol}(B_\varepsilon(x))}\int\limits_{B_\varepsilon(x)}\left|f-\frac{1}{\operatorname{vol}(B_\varepsilon(x))}\int\limits_{B_\varepsilon(x)}f(y)dy \right|=0$$

Where $B_\varepsilon(x)\subset \mathbb{R}^n$ ist the closed ball around $x$ with radius $\varepsilon$. I tried to use, that $f$ have a supremum on $B_\varepsilon(x)$, but as I have the average of $f$ and not the integral, this seems not to work.

Thank you for your help!

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    epsilon-delta-definition for continous functions. (change the roles of epsilon and delta in standard notation, in your case for every positive delta there is a positive epsilon such that...) – 2017-01-18
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    Thank you for your answer! My problem here is, that $\frac{1}{vol (B_\varepsilon (x))}\rightarrow \infty$. I see that $\int\limits_{B_\varepsilon(x)}|...|<\delta$ but not the first term. – 2017-01-18
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    What is the integration variable in the outer integral? – 2017-01-18
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    may call it z, then you have $\int | f(z)- c|dz$ – 2017-01-18
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    not my point. it's much easier than the answer below: $\int_{B_{\epsilon}(x)}\left|\cdots\right|<\delta\cdot \mathrm{vol}(B_{\epsilon}(x))$ – 2017-01-18

1 Answers 1

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First note that $f$ is uniformly continuous when restricted to $\bar{B}(x,1)$ since the latter is compact. Let $\delta >0$. According to the uniform continuity, we can choose $\kappa >0$ such that $$ x,y \in \bar{B}(x,1) \text{ and } \vert x-y \vert < \kappa \Rightarrow \vert f(x)-f(y) \vert < \delta. $$

Now consider $\epsilon < 1$ and consider $z \in B(x,\epsilon)$. Then $$ f(z) - \frac{1}{\text{vol}( B(x,\epsilon) )} \int_{B(x,\epsilon)} f(y) dy = \frac{1}{\text{vol}( B(x,\epsilon) )} \int_{B(x,\epsilon)}(f(z) - f(y)) dy. $$

Suppose that $\epsilon < \kappa/2$. Then for any $y,z \in B(x,\epsilon)$ we have $$ |y-z| \le |y-x| + |z-x| < 2 \epsilon \le \kappa. $$ We can then use the uniform continuity to estimate $$ \left\vert f(z) - \frac{1}{\text{vol}( B(x,\epsilon) )} \int_{B(x,\epsilon)} f(y) dy \right\vert \le \frac{1}{\text{vol}( B(x,\epsilon) )} \int_{B(x,\epsilon)}|f(z) - f(y)| dy \\ < \frac{1}{\text{vol}( B(x,\epsilon) )} \int_{B(x,\epsilon)} \delta dy =\frac{1}{\text{vol}( B(x,\epsilon) )} \text{vol}( B(x,\epsilon) ) \delta =\delta $$ for $\epsilon < \kappa/2$. Thus for $\epsilon < \kappa/2$ we have that $$ \frac{1}{\text{vol}( B(x,\epsilon) )} \int_{B(x,\epsilon)} \left\vert f(z) - \frac{1}{\text{vol}( B(x,\epsilon) )} \int_{B(x,\epsilon)} f(y) dy \right\vert dz \\ < \frac{1}{\text{vol}( B(x,\epsilon) )} \int_{B(x,\epsilon)} \delta = \delta. $$ Hence $$ \lim_{\epsilon \to 0}\frac{1}{\text{vol}( B(x,\epsilon) )} \int_{B(x,\epsilon)} \left\vert f(z) - \frac{1}{\text{vol}( B(x,\epsilon) )} \int_{B(x,\epsilon)} f(y) dy \right\vert dz =0. $$