A bounded convex function

Question

Suppose $f(x)$ is a convex function that is bounded from above by a constant.

I.e, $f(x)<C$. I want to show $f$ is a constant function.

Besides the definition of a covnex function I don't see what to use and how to prove this claim.

Can someone help me? thanks!

It's false if the domain is bounded. For unbounded domains, see https://math.stackexchange.com/questions/518091/show-bounded-and-convex-function-on-mathbb-r-is-constant?rq=1 — Tatin, Jul 28 '19 at 04:50

score 1 · Accepted Answer · answered Jul 28 '19 at 04:55

1

Let $0 \leq t \leq x$. Then $t=\frac t x x+(1-\frac t x) 0$ so $f(t) \leq \frac t x C+(1-\frac t x) f(0)$. Letting $ x \to \infty$ we get $f(t) \leq f(0)$. Similarly apply convexity to the points $-x, 0$ and $t$ and let $x \to \infty$ to get $f(0) \leq f(t)$. This proves thast $f(t)=f(0)$ for all $t \geq 0$. I will let you handle the case $t<0$.

answered Jul 28 '19 at 04:55

Kavi Rama Murthy

311,013

How do you show that $f(t) \ge f(0)$? – MathematicalPhysicist Jul 31 '19 at 08:31
$0=\alpha t +(1-\alpha) x$ where $\alpha =\frac {-x} {t-x}$ for $x >0$. So $f(0)\leq \alpha f(t)+(1-\alpha) f(x) <\alpha f(t)+(1-\alpha)C$. Let $x \to \infty$. – Kavi Rama Murthy Jul 31 '19 at 08:36

A bounded convex function

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