Let $f ∈ L^1(loc)$, the space of locally integrable functions. Show that for all $x ∈ \mathbb{R}^n$ we have that $g(r) =\int_{B(x,r)}f(y)dy$ is a continuous function.
I was given a hint to show the result holds for all continuous functions and use this to show the result for a general locally integrable function, close to a continuous one.
First I consider an arbitrary continuous function $h$. I need to show, for $\epsilon>0$, there is a $\delta>0$ such that $|r-r_0|<\delta \implies |k(r)-k(r_0)|<\epsilon$ where $k(r)=\int_{B(x,r)}h(y)dy$.
I did the following $$|\int_{B(x,r)}h(y)dy-\int_{B(x,r_0)}h(y)dy|=|(\frac{r}{r_0})^n\int_{B(x,r_0)}h(x+\frac{r}{r_0}(z-x))dz-\int _{B(x,r_0)}h(z)dz|\\=(\frac{r}{r_0})^n|\int_{B(x,r_0)}h(x+\frac{r}{r_0}(z-x))dz-(\frac{r_0}{r})^n\int _{B(x,r_0)}h(z)dz|$$
Now I am trying to use the continuity of $h$ but I cannot figure how to do this. Any suggestion?
Best Answer
You don't need the dominated convergence theorem for this. Note that (using your notation for $k$) \begin{eqnarray} |k(r) - k(r_0) | &=& \left|\, \int_{B(x,r)} h(y) \, dy - \int_{B(x,r_0)} h(y) \, dy \,\right| \\ &=& \left| \, \int_{A(x; r,r_0)} h(y) \, dy \,\right| \\ &\le& \int_{A(x; r,r_0)} \left| h(y)\right| \, dy \le \text{vol}A(x;r,r_0) ||h||_{\infty,B(x,r)} \end{eqnarray} where $A(x;r,r_0)$ is the $(n-)$ dimensional 'annulus' with center $x$, inner radius $r_0$ and outer radius $r$. This is assuming $r>r_0$ -- if the reverse inequality applies the same statement holds with $A(x;r_0,r)$ instead. The volume of $A(x;r, r_0)$ equals $C_1 (r^n- r_0^n)$ for some constant $C_1 = C(n)$.
If $f$ is continuous, then $|f|\le C_2 $ on some ball $B_R(x)$ with $R> \max{\{r, r_0\}}$, so $$|k(r) - k(r_0)| \le C_1 C_2 |r^n - r_0^n|$$ which tends to $0$ if $r\rightarrow r_0$ due to the continuity of $r\mapsto r^n$. Can you finish the proof now for $f\in L^1_{\text{ loc}}$?