Pointwise convergence for a sequence of convex functions implies uniform convergence

Question

I'm trying to prove the following fact.

Let be $(u_k)_{k\in\mathbb{N}}$ a sequence, where $u_k:\mathbb{R}^n\rightarrow (-\infty,+\infty]$ is convex.

I suppose that $u_k$ converges pointwise to a function $u:\ \mathbb{R}^n\rightarrow (-\infty,+\infty]$ such that the interior of the effective domain of $u$ is non empty.

Then I want to prove that $(u_k)$ converges uniformly on every compact set of the effective domain of u.

I recall the definition of effective domain of $u$; it is the set $$dom(u)=\{x\in\mathbb{R}^n\;|\;u(x)<+\infty\}. $$

Attempt

My idea is to use Ascoli-Arzelà theorem. I have proved that, for every $K$ compact set of $dom(u)$, the sequence $\{u_k\}_{k\in\mathbb{N}}$ is equi-bounded and equi-lipschitz on $K$, using the fact that a convex function is continous on the interior of the effective domain and is locally compact.

So, for Ascoli Arzelà, for every $K$ compact in $dom(u)$, I can find a subsequence of $(u_k)$ which converges uniformly on $K$ to $u$.

Then using a diagonalisation argument, I've proved that there is a subsequence $(u_{k_{h}})$ of $(u_k)$ which converges uniformely on every compact set $K$ of $dom(u)$.

My question: How can I prove now that $(u_k)$ converges uniformely on every compact sets $K$ of $(dom)(u)$?

I know that the following fact is true:

if for every $(u_{k_{h}})$ subsequence of $(u_k)$ exists a subsubsequence $(u_{k_{h_{j}}})\subseteq (u_{k_{h}})$ such that $u_{k_{h_{j}}}\rightrightarrows u$ on $K$ then $u_k\rightrightarrows u$ on $K$.

But here I can't use this fact, because I've found only the existence of a subsequence of $(u_k)$ that satisfies the condition of the preceding fact; I, ve not proved that for every subsequence of $(u_k).$

How can I conclude that the whole sequence converges uniformly on $K$?

Can someone help me? Thanks!

Answer 1

How can I conclude that the whole sequence converges uniformly on $K$?

Show that before the diagonalisation argument - which then becomes superfluous because $K$ was an arbitrary compact subset of the effective domain.

You have shown that the full sequence $(u_k)_k$ is equi-Lipschitz and uniformly bounded on $K$. Now let $(u_{k_m})_m$ be an arbitrary subsequence. Clearly that is equi-Lipschitz and uniformly bounded on $K$, hence it has a uniformly convergent subsequence $\bigl(u_{k_{m_n}}\bigr)_n$. Since $u_k \to u$ pointwise, it follows that the limit of the uniformly convergent sub-subsequence is $u$. Thus by the criterion, the full sequence converges to $u$ uniformly on $K$.

Answer 2

I do not know, whether I miss something, but the claim is not true. Define $$ u_n(x) = \begin{cases} 0 & \text{ if } x\le 0\\ x^n & \text{ if } x\in(0,1]\\ +\infty & \text{ if } x>1 \end{cases}, $$ which converges pointwise to the convex function $$ u(x) = \begin{cases} 0 & \text{ if } x < 1\\ 1 & \text{ if } x=1\\ +\infty & \text{ if } x>1 \end{cases}. $$ The convergence cannot be uniform on subsets $[1-\epsilon,1]$, for obvious reasons.