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I would like to know an example of function $f: \mathbb{R} \rightarrow \mathbb{R}$ which is not convex but satisfies for fixed $t\in (0,1)$ the following inequality: $f(tx+(1-t)y) \leq t f(x)+(1-t)f(y) \textrm{ for all } x,y \in \mathbb{R}.$

I know only that such functions have to be discontinuous everywhere.

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    Examples based on a Hamel basis of $\mathbb R$ over $\mathbb Q$ will satisfy your inequality for all rational $t$.2012-01-13

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Zoltán Daróczy, Zsolt Páles: Convexity with given infinite weight sequences. Stochastica: revista de matemática pura y aplicada, ISSN 0210-7821, Vol. 11, No. 1, 1987, 5-12, link to pdf.

Lemma 1. Let $D\subseteq\mathbb R^n$ be a convex compact set. Let $f:D\to\mathbb R$ be an $\alpha$-convex function for some fixed $\alpha\in(0,1)$, i.e. assume that $f(\alpha x+(1-\alpha)y)\le \alpha f(x)+(1-\alpha)f(y) \qquad (\ast)$ holds for all $x,y\in D$. Then $f$ is also Jensen convex, i.e. $(\ast)$ is satisfied with $\alpha=\frac12$.

According to the autors, the same result was obtained earlier by N. Kuhn in the paper A note on t-convex functions, but I was not able to find this paper online.


Based on the above result, it suffices to consider $t=\frac12$. It was explained in this answer that some form of AC is needed to construct non-convex Jensen convex function. (Since every measurable Jensen convex function is convex.)

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    Thanks. It is very help$f$ul.2012-01-13