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Let $f(t)$ be a convex function and let $g(t)$ denote its running average. Here $g(t)$ can be shown to be $$g(t) = \int_{0}^{1}f(ts)ds$$ Here, $g(t)$ is to be proved to be convex. A nice answer to this question is already posted here. My question is can this statement be proved by the following steps : (1) A convex function determines an epigraph (2) The epigraph is a convex set. Under the above integral operator which is linear , convexity is preserved (3) We hence get another epigraph which determines the convex function $g(t)$

Are all my steps correct and can convexity of $g(t)$ be proved in this way ?

varunmarda
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  • One immediate observation is linearity of an operator T implies that -T is also linear, so convexity does not follow from linearity. – Kavi Rama Murthy Nov 10 '17 at 06:19
  • If $T:U\rightarrow V$ is a linear function and $\mathcal{X} \subseteq U$ is convex, then indeed the following set is convex: $$T(\mathcal{X}) = {v \in V : v=T(u) \mbox{ for some $u \in U$}}$$ and so linear functions preserve convex sets. But in your above argument, you start with a convex set (being the epigraph) and then you introduce a "linear operator," but what is it linear on? The integral seems to be a linear transformation on functions, not points of a set. – Michael Nov 10 '17 at 07:02
  • Aaah got it. Thanks. – varunmarda Nov 10 '17 at 07:07

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