Show $f$ is constant when $\forall h>0$, $\forall y \in \mathbb{R}$, $f(y+h)-f(y) \leq f(y)-f(y-h)$.

Asked Nov 22 '22 at 04:16

Active Nov 22 '22 at 05:14

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For $f:\mathbb{R} \to \mathbb{R}$, $f''(x) \leq 0$ for all $x \in \mathbb{R}$. Let $\forall y \in \mathbb{R}$, $\forall h>0$, $f(y+h)-f(y) \leq f(y)-f(y-h)$.

Show "$f$ is a bounded below, then $f$ is constant function."

I'm trying to prove $f$ is a constant. But I don't have any idea how to approach those. Would you please give me any ideas or solutions?

asked Nov 22 '22 at 04:16

kechang lee

@geetha290krm If $f''\leq 0$, the inequality holds. But the question is showing f is constant If $ f$ does hold property either that inequality or $f''\leq 0$. – kechang lee Nov 22 '22 at 05:08
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Does this answer your question? Function bounded below, with negative 2nd derivative. Show function is constant. – geetha290krm Nov 22 '22 at 05:21
A solution can be found in the link on the right pane of this page. Please do check for duplicates before posting a question. – geetha290krm Nov 22 '22 at 05:22

Show $f$ is constant when $\forall h>0$, $\forall y \in \mathbb{R}$, $f(y+h)-f(y) \leq f(y)-f(y-h)$.

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