# A function in $W^{2,p}$ for $p>n/2$ is a.e. second differentiable

Let $$B$$ be the unit ball in $$\mathbb{R}^{n}$$ and $$u\in W^{2,p}(B)$$, with $$p>\dfrac{n}{2}$$. How can we see that $$u$$ is second differentiable almost everywhere in $$B$$?

This result is claimed in Page 25 but I cannot prove it. I can understand the proof of: $$v\in W^{1,p}(B)$$, with $$p>n$$ then $$v$$ is differentiable almost everywhere in $$B$$ (in the same page of the link).

My attempt so far:

I try to follow the same method as in Page 25.

Assume that $$0$$ is a Lebesgue point of $$D^2u\in L^p$$, i.e. $$|B_r|^{-1}\int_{B_r}|D^2u(x)-D^2u(0)|^p\rightarrow 0, \text{as }r\rightarrow 0. \quad (*)$$ Here $$B_r$$ denotes the ball of radius $$r$$ centered at the origin. Now, our aim is to show that $$u$$ is classically second diffentiable at $$0$$. Set $$h(x)= u(x)-u(0)-Du(0)x-D^2u(0)\dfrac{x^2}{2},$$
and set $$h_r(x):=h(rx)/r^{2}$$. It is clear to see that it suffices to show $$||h_r||_{L^{\infty}(B)}\rightarrow 0, \text{as }r\rightarrow 0.\quad \textbf{(1)}$$

Since $$u\in W^{2,p}(B)$$, with $$p>\dfrac{n}{2}$$, it is not hard to see that $$u\in W^{1,q}(B)$$, for some $$q>n$$. For example, assume for simplicity that $$n>p>\frac{n}{2}$$ then we can choose $$q=np/(n-p)$$.

By Morrey's inequality (since $$q>n$$), $$||h_r-h_r(0)||_{L^{\infty}(B)}\leq C ||Dh_r||_{L^q(B)}.$$

Note that $$h_r(0)=0$$, and (*) is equivalent to $$||D^2h_r||_{L^p(B)}\rightarrow 0$$. Therefore, in order to finish the proof (i.e. to show $$\textbf{(1)}$$ is true), we really hope to have the following inequality $$||Dh_r||_{L^{q}(B)}\leq C ||D^{2}h_{r}||_{L^{p}(B)}.$$ However, this cannot be seen by the Sobolev embedding theorem since the Sobolev embedding on a bounded domain should be $$||Dh_r-|B|^{-1}\int_{B}Dh_r||_{L^{q}(B)}\leq C ||D^{2}h_{r}||_{L^{p}(B)}.$$

How can I resolve this problem?

Thanks for any suggestion.

Hint: From $$\left\|Dh_r-|B|^{-1}\int_{B}Dh_r\right\|_{L^{q}(B)}\leq C \|D^{2}h_{r}\|_{L^{p}(B)}.$$ and the triangle inequality, you have $$\left\|Dh_r\right\|_{L^{q}(B)}\leq C \|D^{2}h_{r}\|_{L^{p}(B)} + \hat C \, \left|\int_{B}Dh_r\right|.$$
• Indeed, I still cannot see how $||Dh_r||_{L^{q}}\rightarrow 0$. Are you talking about Holder's inequality to do more? – Hahn Dec 17 '18 at 8:17
• Since $||D^{2}h_{r}||_{L^{p}} \rightarrow 0$, we need something like $\hat C \, \left|\int_{B}Dh_r\right|< (1-\epsilon) ||Dh_{r}||_{L^{q}}$ to finish, right? But it's not clear. Note that, by applying Holder's inq, you would get $$\hat C \, \left|\int_{B}Dh_r\right| \leq ||Dh_{r}||_{L^{q}}.$$ – Hahn Dec 17 '18 at 8:36
• If I am not mistaken, you have $\int_B \nabla h_r = r^{-n} \, \int_{B_r} \nabla h = r^{-n} \, \int_{B_r} \nabla u(x) - \nabla u(0)$. If $0$ is a Lebesgue point of $\nabla u$, this goes to zero. – gerw Dec 17 '18 at 8:39
• No, it should be $\int_B \nabla h_r = r^{-n-1} \, \int_{B_r} \nabla h$. – Hahn Dec 17 '18 at 8:51