# $L^p$ Norm continuity boundedness

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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As it is ($r\to \infty$), the limit is $0$ for every compactly supported function. –  Jose27 Aug 15 '12 at 2:56
Sorry I was supposed to approach 0 not $\infty$! –  Dave Aug 15 '12 at 3:08
Recall the Lebesgue Differentiation Theorem. –  leo Aug 15 '12 at 3:30
@leo Okay so that gives me continuity in $x$. –  Dave Aug 15 '12 at 4:04
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. –  Davide Giraudo Aug 15 '12 at 8:05

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 $r\to 0$...
What is $\mu(A\Delta B)$? The measure of the difference set, $\mu A\B)$ ? If so I don't see how that helps. Are $A$ and $B$ balls of radius $r$ centered at $x4, and$y$? – Dave Aug 15 '12 at 3:01 @Dave it's the symmetric difference.$A,B$are arbitrary (measurable). – tomasz Aug 15 '12 at 3:07 Okay that's what I was thinking but I forgot that notation. So since I can make$x$and$y$arbitrarily close the symmetric difference of$B_{x}(r)$and$B_{y}(r)$can be as small as I want? – Dave Aug 15 '12 at 3:20 Okay that's what I was thinking but I forgot that notation. So since I can make x and y arbitrarily close the symmetric difference of$B_{x}(r)$and$B_{y}(r)$can be as small as I want? – Dave Aug 15 '12 at 3:29 @Dave: Yes. Similarly if$r,r'\$ are close, then the annulus of the symmetric difference will be small. –  tomasz Aug 15 '12 at 11:47