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I have a continuous function $f:\mathbb{R}^n\to\mathbb{R}$ such that $\forall x,y\in\mathbb{R}^n$ I have that $f(x)\ge\cfrac{1}{2}f(x-y)+\cfrac{1}{2}f(x+y)$.

How can I prove that $\forall x,y\in\mathbb{R}^n$ and $\forall t\in[0,1]$ I have that $f(tx+(1-t)y)\ge tf(x)+(1-t)f(y)$ ?

r2d2
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1 Answers1

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Firstly, $f(x)\ge\frac{1}{2}f(x-y)+\frac{1}{2}f(x+y)$ is no different from $f(\frac{x+y}{2})\ge\frac{1}{2}f(x)+\frac{1}{2}f(y)$.

Secondly, Midpoint-Convex and Continuous Implies Convex

Thirdly, A Lebesgue measurable function on an interval C is concave if and only if it is midpoint concave: https://en.wikipedia.org/wiki/Concave_function

Ian Zhang
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  • Sure, my function is midpoint-convex, but what is the most easy way to show that midpoint-convex and continuous implies convex? – r2d2 Feb 10 '21 at 21:54