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Every convex function is continuous. It usually says "draw this and it will become obvious that the epigraph is not convex. However, when I draw the epigraph of $f: [0,3] \to \mathbb{R}, f(x) = x^2$ for $x \in [0,3)$, $f(3)=10$ it appears to be convex to me. How come this happens?

Also, I'd love some verification. I've found that for $f_{\beta} : [0,\infty) \to \mathbb{R}, f_{\beta}(x) \left\{ \begin{array}{l l} 3x^{2}-2x & \quad \text{if $0 \le x < 11$}\\ \beta & \quad \text{$x\ge 11$} \end{array} \right. $

, there does not exist a $\beta$ for which f is strictly convex. Is this correct?

Asaf Karagila
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JH-
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    A convex function $f\colon [a,b] \to\mathbb{R}$ is continuous in $(a,b)$, it need not be continuous in the endpoints of its domain. Regarding the second question, indeed $f_\beta$ isn't convex for any $\beta\in\mathbb{R}$. – Daniel Fischer Nov 05 '13 at 14:16
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    As W. Rudin states clearly in his book Real and complex analysis, a convex function on an open interval is continuous. You can't get continuity on a closed interval. – Siminore Nov 05 '13 at 14:17
  • @DanielFischer So that would mean that the first function is indeed convex, since it is continuous in (a,b)? – JH- Nov 05 '13 at 14:42
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    The function is convex, but not "since it is continuous in $(a,b)$" (that isn't either what you meant, I think). That is a necessary condition. It is continuous because it is convex (and hence continuous) on $(a,b)$, and the values in the endpoints of the interval are not smaller than the limits of $f$ as the endpoint is approached from within $(a,b)$. – Daniel Fischer Nov 05 '13 at 14:58
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    To be clear, your sentence "Every convex function is continuous" is false, unless you assume that its domain is an open set. In other words, convex functions are not necessarily continuous up to the boundary. – Siminore Nov 05 '13 at 15:18

1 Answers1

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Convex functions are not continuous necessarily: This is incorrect. You can't do analysis by drawing! Mathematics doesn't work like that. However sometimes convex functions are continuous on the interior of their domains. For instance:

Let $f:\mathbb{H}\to [-\infty,\infty]$ be proper and convex. Suppose $f$ is bounded above on some neighbourhood, or that $\mathbb{H}$ is finite dimensional, then $f$ is continuous on the interior of its domain. -- Corollary 8.30 Convex analysis and Monotone Operator Theory in Hilbert Spaces (Bauschke, Combettes)

Your other question: Yes this is not even convex (let alone strictly convex) for an $\beta$ except $\infty$. For $\infty$ $f$ is strictly convex. To be convex, the gradient, where defined, needs to be increasing. The gradient at $x=10$ is $58$, and the gradient at $12$ is $0$.

RobertHannah89
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