Sequence of convex functions

If a sequence of convex functions $\{ f_n \}$ on [0,1] converges pointwise to a continuous function f, then is the convergence uniform?

The questions is almost identical to this one, except that the functions are not assumed to be continuous on [0,1]. Now I know that convex functions are continuous on open sets, so we can easily prove that the convergence is uniform on compact subsets of (0,1) as in the link above. But if we try to include the endpoints 0,1 this approach does not work, and I am actually starting to think if maybe there is a counterexample. Could anyone provide any ideas? Thanks?

EDIT:Since the proof in the above link does not seem to work as it is formulated at the moment, could someone please explain in detail how one works to prove the assertion assuming $f_n$ are continuous?

Solution 1:

Let $g_n(x)=f_n(x)$ for $0<x<1$, $g_n(0)=f_n(0+)$, and $g_n(1)=f_n(1-)$. This new function is continuous on $[0,1]$ and is still convex. Also, $g_n\le f_n$.

I claim that $g_n\to f$ pointwise. Suppose not. Then there is $\epsilon>0$ and a subsequence $n_k$ such that $g_{n_k}(0) \le f(0)-\epsilon$ for all $k$. Since $g_{n_k}$ are convex and uniformly bounded from above, it follows that $g_{n_k}\le f(0)-\epsilon/2$ in some interval $[0,\delta]$ with $\delta$ independent of $k$. Passing to the limit we find that $f(x)\le f(0)-\epsilon/2$ on $(0,\delta]$, contradicting the continuity of $f$.

Now the result to which you linked applies to $g_n$ and shows that $g_n\to f$ uniformly. Taking into account that $f_n(0)\to f(0)$ and $f_n(1)\to f(1)$, we conclude that $f_n\to f$ uniformly.