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It is stated here that if a sequence $f_n : [a,b] \to \mathbb{R}$ of convex continuous functions converges pointwise to a continuous function $f$, then the convergence is in fact uniform.

The answers given seem me to be correct proofs of the statement, however I was able to construct a counterexample pretty easily. Consider the functions

\begin{equation*} f_n = \begin{cases} -nx, & \text{if } 0 \leq x \leq \frac1n, \\ -2 + nx, &\text{if } \frac1n \leq x \leq \frac2n,\\ 0, &\text{otherwise}. \end{cases} \end{equation*} It's clear that $f_n$ is convex and continuous, and moreover converges pointwise to $f \equiv 0$. However, the convergence is not uniform, since $\sup |f_n| = 1$.

I'm not sure whether there's I overlooked something when constructing this counterexample or the original statement about uniform convergence of convex functions is incorrect.

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Try plotting your functions, are they actually convex? (Hint: you need to specify the domain of your $f_n$)

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  • $\begingroup$ Ah, I made a silly mistake! Thanks. $\endgroup$ – Reavered Aug 20 '19 at 21:24

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