# Confused about two steps of the smoothness for functions satisfying the mean value property

Evans page 28. He shows that a function satisfying the mean value property is smooth.

He uses mollification, i.e. let $u^\epsilon = \eta_\epsilon * u$

\begin{align*} u^\epsilon(x)&=\int_U \eta_\epsilon(x-y)u(y)dy\\ &= \frac{1}{\epsilon^n}\int_U\eta\left(\frac{|x-y|}{\epsilon}\right)u(y) dy\\ &= \frac{1}{\epsilon^n}\int_0^\epsilon\eta\left(\frac{r}{\epsilon}\right)\left(\int_{\partial B(x,r)} u dS\right) dr\\ \end{align*} How is that last step done. I was thinking mean value theorem here, but this gives for any $B(y,r)\subset U$: \begin{align*} =\frac{1}{\epsilon^n}\int_U\eta\left(\frac{|x-y|}{\epsilon}\right)(-\!\!\!\!\!\!\int_{\partial B(y,r)} u dS) dy&= \end{align*} and I can't get this to equal the above.

Secondly, they then get from the above the following equality: $$\frac{1}{\epsilon^n}u(x)\int_0^\epsilon \eta(r/\epsilon) n\alpha(n)r^{n-1}dr=u(x)\int_{B(0,\epsilon)} \eta_\epsilon dy$$ and I just can't see how transforming from $dr$ to $dy$ destroys the $n\alpha(n) r^{n-1}$ terms.

## 1 Answer

\begin{align}\frac{1}{ε^n} &∫_U\eta(|x-y|/ε ) u(y) \ dy \\ &\overset{1}{=} \frac{1}{ε^n} ∫_{B(x,\epsilon)}\eta(|x-y|/ε ) u(y) \ dy \\ &\overset{2}{=} \frac{1}{ε^n}∫_{r=0}^ε \eta(r/ ε)∫_{\partial B(x,r)}u \ dS \ dr\\ &\overset{}{=} \frac{1}{ε^n}∫_{r=0}^ε n\alpha(n)r^{n-1}\eta(r/ ε)\left(\frac1{n\alpha(n)r^{n-1}}∫_{\partial B(x,r)}u \ dS\right) \ dr \\ &\overset{3}{=} \frac{1}{ε^n}∫_{r=0}^ε n\alpha(n)r^{n-1}\eta(r/ ε) u(x) \ dr \\ &\overset{4}{=} \frac{1}{ε^n}∫_{r=0}^ε \eta(r/ ε) u(x) \ ∫_{\partial B(x,r)}1 \ dS \ dr \\ &\overset{5}{=} u(x)\frac{1}{ε^n}∫_{B(x,ε)}\eta(|x-y|/\varepsilon) \ dy \\ &\overset{}{=} u(x)∫_{B(0,ε)}\eta_ε \ dy \end{align}

1. support of $\eta(\cdot/\varepsilon)$
2. co-area formula(see the first bullet point under "applications")
3. mean value property
4. area of the surface of the $r$-sphere around $x$
5. co-area formula
• That's oddly beautiful. Thank you! – F.White Jul 3 '17 at 3:45