0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.