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Let $D$ be a convex set in $\mathbb{R}^n$. Suppose there exists some $a>0, b>0$ and $a+b=1$ such that for any $x,y \in D$, the inequality $f(ax+by) \leq af(x)+bf(y)$ holds. Is it necessary that $f$ is convex on $D$? I know when $a=b$ it is just the definition of a convex function.

Any help will be appreciated.

Ѕᴀᴀᴅ
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gžd15
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    Take a look at https://math.stackexchange.com/questions/71019/example-of-a-function-such-that-varphi-left-fracxy2-right-leq-frac-va and https://math.stackexchange.com/questions/98826/example-of-function-satisfying-for-fixed-t-in-0-1-inequality-ftx1-ty?noredirect=1&lq=1 – Robert Z Sep 27 '18 at 05:14

1 Answers1

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In general the answer is "No". Take any $f\colon\mathbb{R}^n\to\mathbb{R}$ which is linear with respect to $\mathbb{Q}$ but non linear with respect to $\mathbb{R}$. Then there is some real $a'\in(0,1)$ and some $x_0\in\mathbb{R}^n$ such that $f(a' x_0)\not=a' f(x_0)$. We also may assume that $f(a' x_0)>a' f(x_0)$ by possibly using $-x_0$ instead of $x_0$. Then $f$ satisfies $f(ax+by) \leq af(x)+bf(y)$ with equality for $a$ (and also $b$) rational. But, using $b'=1-a'$, $f(a' x_0+b' 0)=f(a'x_0)>a'f(x_0)=a' f(x_0)+b'f(0)$.

Remark: The existence of a mapping $f$ being linear with respect to the field of rationals but not linear with respect to the reals may be proved by using a basis of the vector space $\mathbb{R}^n$ viewed as a rational vector space.

Jens Schwaiger
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