Suppose that $X$ is a subset of a Euclidean space and $f_n:X\to\mathbb{R}$ converges pointwise to $f:X\to\mathbb{R}$. A book I'm reading seems to say that: if $X$ is convex and compact and each $f_n$ is continuous and concave, then the convergence is uniform.

Main question: is the above true and, if so, could you please provide a proof (or a reference to a proof)? I know that if $f_n$ is continuous, then it is uniformly continuous but I can't proceed further.

Edit 2: as Nate's answer below demonstrates, the claim is false given the premises as they are. What if we further assume that $f$ is continuous?

Sub-question if you have time: can you please give a few references concerning sufficient conditions that allow the implication from pointwise convergence to uniform convergence? I know of Dini's Theorem and Arzelà-Ascoli Theorem.

Thank you!

Edit 1 to add context: on page 390 of (Monfort and Gourieroux (1995, Vol II)), it is stated:

a family $Q_n(\omega,\cdot)$ of concave functions converging pointwise to a function $Q_\infty(\cdot)$, which is necessarily concave, converges uniformly to that function over any compact subset

Here, $\omega$ can be taken as given and $\cdot$ is a place holder for $\theta\in\Theta$ with $\Theta\subset\mathbb{R}^p$ is compact and convex. Hence, the question above.

  • $\begingroup$ Can you quote the definition of "concave" being used in that book? Is it possible there are other standing assumptions you have overlooked? $\endgroup$ – Nate Eldredge Apr 14 '15 at 16:26
  • $\begingroup$ @NateEldredge You are right and, after some thought, I agree with your answer below. But to be honest, I'm stumped because I don't see which details I am missing. The authors don't define concavity explicitly so I'm assuming they are using the standard definition. I'll give it some more thought and I'll get back to this thread later. The result above is part of Property 24.5 on page 390. $\endgroup$ – Kim Jong Un Apr 14 '15 at 17:53

This is not true. Take $X = [0,1] \subset \mathbb{R}^1$ and $f_n(x) = 1-x^n$.

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    $\begingroup$ If $f(x)=1$ for $x\in[0,1)$ and $f(1)=0$, it's not obvious to me that the convergence $f_n\to f$ is not uniform. Can you please elaborate? $\endgroup$ – Kim Jong Un Apr 14 '15 at 14:41
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    $\begingroup$ If $f_n$ is continuous on a (closed?) interval $f$ for all $n$ and $f_n \rightarrow f$ uniformly then $f$ is continuous too. $\endgroup$ – Christopher Apr 14 '15 at 15:24
  • $\begingroup$ @KimJongUn: Given $\epsilon > 0$ and $n \ge 1$, you can find an $x$ very close to 1 such that $|f_n(x) - f(x)| > \epsilon$. To get a sense of what is happening, try graphing $|f_n - f|$ for large values of $n$; note that what you see is not uniformly close to 0. $\endgroup$ – Nate Eldredge Apr 14 '15 at 16:22
  • $\begingroup$ @NateEldredge that helps! $\endgroup$ – Kim Jong Un Apr 14 '15 at 18:03
  • $\begingroup$ @user73985 that is indeed another to see that $f_n\to f$ cannot be uniform for $f$ is not continuous at $x$ whereas each $f_n$ is. Thanks. $\endgroup$ – Kim Jong Un Apr 14 '15 at 18:04

Probably, the authors of the book you refer to were meant to make use of Theorem 10.8 of "Convex Analysis" (second printing 1972) by Rockafellar; hereinafter Theorem 10.8. If this is the case, the domain should be convex and relatively open.

Observe that for the sequence provided in the counterexample above, the maximal set for which the Theorem 10.8 holds is (-1,1). Otherwise, the limit function will not be finite and this would violate another condition of Theorem 10.8.


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