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I'm studying for a Qualifying exam and can't figure out this problem. I see that the limit must be $f(x)$ and can get the boundedness but had trouble with continuity. Any suggestions?

Let $f\in L^{\infty}(\mathbb{R}^{d})$ and let $\phi:\mathbb{R}^{d} \times (0,\infty)\rightarrow \mathbb{R}$ be the following map:

$\phi(x,r)=\frac{1}{r}\int_{B_{r}(x)}f(y)dy.$

Prove that $\phi$ is continuous in $x$, and in $r$, and is uniformly bounded. What can you say about $\lim_{r\rightarrow 0}\phi(x,r)$?

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    Are you sure that it's not $\phi(x,r)=\frac 1{r^d}\int_{B_r(x)}f(y)dy$. Otherwise, taking $f=1$ I don't see why \phi would be bounded.2012-08-15

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I assume $\mathfrak R$ means reals (first time I see someone write them using fraktur, I think :) ).

Notice that if $f\in L^{\infty}$, then for any $A,B$, $\lvert \int_A f-\int_B f\rvert\leq \lVert f\rVert\cdot \mu(A\triangle B)$.

That should make it easy to show that it is continuous in $x$.

For $r$ it is much the same: write the map as a composition of $r\mapsto (\int_{B_r(x)} f,r)$ and $(a,r)\mapsto a/r$. The former map is continuous -- you can show it the same way as you've shown that $\phi$ is continuous in $x$, and the second is obviously continuous.

On the other hand, I don't see why would it be bounded. It also seems to me that the limit is actually $0$ if $d>1$, for example if $f\equiv 1$, then, up to a multiplicative constant, $\phi(x,r)\cong r^{d-1}$, which approaches $0$ as $r\to 0$...

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    @Dave: Yes. Similarly if $r,r'$ are close, then the annulus of the symmetric difference will be small.2012-08-15