# show that this function is continuously differentiable ( application to Lebesgue theorem ?)

Consider $R>0$ and $u \in C_{0}^{\infty}(B(0,R))$ (this is the set of smooth functions with compact support contained in $B(0,R)$).

Let $|\cdot|$ be the Lebesgue measure in $\mathbb R^n$.

Fix $y \in \mathbb R^n$ and define the function :

$$f(t) = \frac{\displaystyle\int_{B(y,t)} u(x) \ dx}{\left|B(y,t)\right|}$$

for $t>0$ .

Show that $f$ is continuously diferentiable in the interval $(0,\infty)$.

My idea :

With some calculus i obtain this :

$$\frac{f(t) - f(t_0)}{t - t_0} = \frac{1}{t-t_0}\left( \frac{{t_0}^n \displaystyle\int_{B(y,t)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n} - \frac{{t}^n \displaystyle\int_{B(y,t_0)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n}\right)$$

where $t_0$ is fixed in the interval above. I am trying to aply the Lebesgue diferentiation theorem in the quotient above ...

Someone can help me? Thanks

• By $B(0,R)$, do you mean an open ball of center $0$ and radius $R$? Jul 24 '13 at 19:54
• @Omnomnomnom yes ^^ Jul 24 '13 at 19:56

The function $t\mapsto \vert B(y,t)\vert$ is positive and $\mathcal C^1$ (in fact, it is $c\, t^n$ for some constant $c$); so you just need to show that $\varphi(t)=\int_{B(y,t)} u(x)\, dx$ is $\mathcal C^1$. If you put $x=y+t\xi$, then the change of variable formula gives $\varphi (t)=t^n\int_{B(0,1)} u(y+t\xi )\, d\xi$, so you have to show that $\psi(t)=\int_{B(0,1)} u(y+t\xi)\, d\xi$ is $\mathcal C^1$. But this follows from the usual differentiation theorems under the integral sign.
Let's begin with the expression $$\frac{f(t) - f(t_0)}{t - t_0} = \frac{1}{t-t_0}\left( \frac{{t_0}^n \displaystyle\int_{B(y,t)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n} - \frac{{t}^n \displaystyle\int_{B(y,t_0)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n}\right)$$ And rewrite that, saying $t=t_0+\Delta t$, as $$\frac{f(t_0+\Delta t) - f(t_0)}{\Delta t} = \frac{1}{\Delta t}\left( \frac{{t_0}^n \displaystyle\int_{B(y,t)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n} - \frac{(t_0+\Delta t)^n \displaystyle\int_{B(y,t_0)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n}\right)$$
Expand the second term and you can cancel the ${t_0}^n$ term, then use Lebesgue's theorem to evaluate the limit.