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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.

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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 $$ \left\|Dh_r\right\|_{L^{q}(B)}\leq C \|D^{2}h_{r}\|_{L^{p}(B)} + \hat C \, \left|\int_{B}Dh_r\right|. $$

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  • $\begingroup$ Indeed, I still cannot see how $||Dh_r||_{L^{q}}\rightarrow 0$. Are you talking about Holder's inequality to do more? $\endgroup$ – Hahn Dec 17 '18 at 8:17
  • $\begingroup$ With which term on the right-hand side do you have problems? $\endgroup$ – gerw Dec 17 '18 at 8:32
  • $\begingroup$ 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 $$ \hat C \, \left|\int_{B}Dh_r\right| \leq ||Dh_{r}||_{L^{q}}.$$ $\endgroup$ – Hahn Dec 17 '18 at 8:36
  • $\begingroup$ 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. $\endgroup$ – gerw Dec 17 '18 at 8:39
  • $\begingroup$ No, it should be $\int_B \nabla h_r = r^{-n-1} \, \int_{B_r} \nabla h $. $\endgroup$ – Hahn Dec 17 '18 at 8:51

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