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?


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.

  • $\begingroup$ Thank you, that makes sense... but today when I tried to rigorously formulate the argument in the above link, I run into some trouble... see link above for details. Thank you for your help though. $\endgroup$
    – Bernard
    Sep 11 '12 at 6:41
  • $\begingroup$ Why $g_n$ is continuous? $\endgroup$
    – vesszabo
    Sep 16 '12 at 16:56
  • $\begingroup$ @vesszabo We are talking about functions with real values, not including infinity. For such functions convexity implies continuity in the interior of the interval. At the endpoints $g_n$ is continuous by design. $\endgroup$
    – user31373
    Sep 16 '12 at 17:34
  • 2
    $\begingroup$ A short hint that $f_n(0+)$ exists may be in order: $f_n(\epsilon)$ is bounded from above by the line through $(0,f_n(0))$ and $(1,f_n(1))$ and (if $\epsilon<\frac12$) from below by the line through $(\frac12,f_n(\frac12))$ and $(1,f_n(1))$. If $f_n$ is decreasing for some $x$ then it is decreasing for all $\epsilon<x$, hence $\epsilon\to0$ produces an increasing bounded sequence, hence a limit. If on the other hand $f_n$ is never decreasing, we have a nonincreasing bounded sequence and again a limt. $\endgroup$ Sep 16 '12 at 19:00
  • $\begingroup$ I see. Thanks. Why does $g_{n_k}(0)\leq f(0)-\varepsilon$ hold, why not $|g_{n_k}(0)-f(0)|>\varepsilon$? $\endgroup$
    – vesszabo
    Sep 17 '12 at 10:40

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