Convergence of semi convex functions

Question

Definition. Let $u:\Omega \rightarrow \mathbb{R} $. A function $u$ is called semiconvex if $u=v+w$ for some $v\in C^{1,1}(\Omega)$ and a convex function $w$.

Note. Saying that $u$ is semiconvex is equivalent to say that there exists a $\lambda$ such that the function $$ z(x)=u(x)+\dfrac{|x|^2}{2\lambda}\text{ is convex}.$$

It's known from the Alexandrov Theorem that a semi-convex function is twice differentiable a.e., that is for almost everywhere $x_0$ there exists a unique $(p,A)$ such that $$ u(x)=u(x_0)+<p,x-x_0>+<A(x-x_0),x-x_0> + o(|x-x_0|^2) $$

Let $u_k = u\star \eta_k $ the mollification of $u$. I would show some kind of convergence of the second and first derivatives of $u$, that is $D_{ij}u_k$ and $D_iu_k$.

Can I say that $D_{ij}u_k-D_{ij}u$ for a.e. $x\in\Omega$?

Can be shown the uniformly convergence?

Can somebody please help me?

Answer 1

You get almost everywhere convergence for both first and second derivatives. In general, no uniform convergence can be expected. Take for example $\Omega=(-1,1)\subset\mathbb{R}$ and $u(x)=|x|$. Then $Du=\frac{x}{|x|}$ is discontinuous and $D^2u=2\delta_0$ in the sense of measures. On the other hand, both $Du_k$ and $D^2u_k$ are continuous functions, hence they can't converge to anything discontinous.

Now to the proof. Let me assume that $u$ is convex: if not, just add $|x|^2/2\lambda$. Let me also assume that $\eta$ is supported in $B_1(0)$ and define $\eta_k(x)=k^n\eta(kx)$; I expect what follows to be true also when $\eta$ doesn't have compact support, but I haven't checked. Since $u$ is locally lipschitz, $Du\in L^\infty_{loc}(\Omega)$. In particular, for every $x$, $$ Du_k(x)=\int_{B_{1/k}}Du(y)\eta_k(x-y)\,dy.$$ Now $$ |Du(x)-Du_k(x)|\le\int_{B_{1/k}}|Du(x)-Du(y)|\eta_k(x-y)\,dy\le \frac{C||\eta||_{\infty}}{|B_{1/k}|}\int_{B_{1/k}}|Du(x)-Du(y)|\,dy $$ which goes to $0$ as $k\to\infty$ whenever $x$ is a Lebesgue point for $Du$, that is almost everyhwere.

The proof for the second derivative goes more or less the same way. Indeed, by convexity, $D^2u$ is a positive-semidefinite Radon measure whose components are $$(D^2u)_{i,j}=D_{ij}u\mathcal{L}^n+\Gamma_{ij}$$ for some positive-semidefinite Radon measure $\Gamma=\{\Gamma_{ij}\}$ which is singular with respect to $\mathcal{L}^n$ and $D_{ij}u\in L^1(\Omega)$ is as in Aleksandrov's theorem. Therefore, as above, we get $$ |D_{ij}u(x)-D_{ij}u_k(x)|\le\int_{B_{1/k}}|D_{ij}u(x)-D_{ij}u(y)|\eta_k(x-y)\,d\mathcal{L}^n(y)+\int_{B_{1/k}}\eta_k(x-y)\,d\Gamma_{ij}(y) $$ which tends to $0$ as $k\to\infty$ for every $x$ which is a Lebesgue point for $D^2u$ and such that $$\lim_{\rho\searrow0}\frac{|\Gamma|(B_\rho(x))}{\mathcal{L}^n(B_\rho(x))}=0, $$ which happens at $\mathcal{L}^n$-a.e. $x\in\Omega$.

I hope I haven't made any stupid mistake and that my answer was useful!