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Let $f:\mathbb{R}^3\rightarrow\mathbb{R}$ be a continuous function. Prove that

$f(x) = \lim_{\epsilon\rightarrow 0}\frac{1}{Vol(B_\epsilon(x))}\int_{B_\epsilon(x)}f(y)dV$

where $B_\epsilon(x)$ denotes the ball of radius $\epsilon$ centered at $x$, and $Vol(B_\epsilon(x))$ denotes its volume.

My attempt: Intuitively, this makes sense, as when you decrease the radius of the ball to $0$, you wind up at the point $x$ anyways. However, I'm having difficulties proving this rigorously. Would taking the derivative help at all?

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  • $\begingroup$ $f$ might not be differentiable. $\endgroup$ – Henricus V. Jan 31 '17 at 15:21
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Let $\epsilon'>0$ and choose $\delta$ such that if $y \in B(x,\delta)$ then $f(y) \in B(f(x),\epsilon')$.

Suppose $\epsilon \le \epsilon'$, then $\int_{B(x,\epsilon)} \|f(x)-f(y)\| dy \le \epsilon m B(x,\epsilon)$. It follows that $\lim_{\epsilon \to 0} {1 \over m B(x,\epsilon)} \int_{B(x,\epsilon)} \|f(x)-f(y)\| dy = 0$.

Since \begin{eqnarray} \| {1 \over m B(x,\epsilon)}\int_{B(x,\epsilon)} (f(x)-f(y)) dy \| &=&\| f(x)- {1 \over m B(x,\epsilon)} \int_{B(x,\epsilon)} f(y) dy \| \\ &\le& {1 \over m B(x,\epsilon)} \int_{B(x,\epsilon)} \|f(x)-f(y)\| dy \end{eqnarray} we have the desired result.

Note that this holds in much greater generality, see the Lebesgue differentiation theorem.

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First of all, even though $f$ may take values in $\Bbb R^3$ and this wouldn't change anything significant in the proof, I am almost sure that $f$ is supposed in fact to take values in $\Bbb R$ - it's probably a typo.

Next, consider the change of variable $y = x + \epsilon u$ with $u \in B_1(0)$. This gives us

$$\int \limits _{B_\epsilon(x)} f(y) \ \Bbb dy = \int \limits _{B_1(0)} f(x + \epsilon u) \epsilon^n \ \Bbb du$$

and

$$\textrm {Vol} (B_\epsilon(x)) = \int \limits _{B_\epsilon(x)} 1 \ \Bbb dy = \int \limits _{B_1(0)} 1 \cdot \epsilon^n \ \Bbb du = \epsilon^n \ \textrm {Vol} (B_1(0)) ,$$

which taken together give

$$\lim_{\epsilon \to 0} \frac 1 {\textrm {Vol} (B_\epsilon(x))} \int \limits _{B_\epsilon(x)} f(y) \ \Bbb dy = \lim_{\epsilon \to 0} \frac 1 {\textrm {Vol} (B_1(0))} \int \limits _{B_1(0)} f(x + \epsilon u) \ \Bbb du = \frac 1 {\textrm {Vol} (B_1(0))} \int \limits _{B_1(0)} f(x) \ \Bbb du = \\ \frac 1 {\textrm {Vol} (B_1(0))} \textrm {Vol} (B_1(0)) \ f(x) = f(x) .$$

In the above I have used the fact that $f$ is continuous, therefore $\lim_{\epsilon \to 0} f(x + \epsilon u) = f(x)$ and a variation of Lebesgue's dominated convergence theorem (the functions $x \mapsto f(x + \epsilon u)$ are dominated by $\sup _{y \in B_r(x)} |f(y)|$ for all $0 < \epsilon <r$, and this supremum is finite because $f$ is continuous on $\Bbb R^3$ and $B_r(x)$ is relatively compact.)

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Consider $f\colon\Bbb R^3\to\Bbb R$ (a scalar function)

Fix $x\in\Bbb R^3$ and $\varepsilon>0$. It follows by the intermediate value property that there exists $x_{\varepsilon}\in B_{\varepsilon}(x)$ s.t. $$\frac{1}{Vol(B_\varepsilon(x))}\int_{B_\varepsilon(x)}f(y)dV=f(x_{\varepsilon}).$$ Try to show this. Now tending with $\varepsilon$ to $0$ causes $x_{\varepsilon}\to x$, which finishes our proof.

If, as you claim, $f\colon\Bbb R^3\to\Bbb R^3$ (the vector-valued function), integrate the components separately.

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Hint: Since $f$ is continuous, so are the component functions $f_i;\ 1\le i\le 3.\ $ Then, $m^i_{\epsilon }(vol \overline B_{\epsilon (x)})\le \int_{B_\epsilon(x)}f_i(y)dV\le M^i_{\epsilon }(vol \overline B_{\epsilon (x)}),\ $where $M^i_{\epsilon }$ and $m^i_{\epsilon }$ are the maximum, minimum, respectively, of $f_i$ on the compact set $\overline B_{\epsilon (x)}$.

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  • $\begingroup$ Your can consider components separately. But it does need to be said. Thanks. $\endgroup$ – Matematleta Jan 31 '17 at 15:30
  • $\begingroup$ I deleted my previous comment because of your corrections. $\endgroup$ – szw1710 Jan 31 '17 at 15:51

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