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I have to show that a bounded differentiable convex function $f: \Bbb R \rightarrow \Bbb R $ is constant.

When a function $f$ is differentiable and convex, then I have a Theorem in my book that says, that:

$ f(y) \ge f(x) + f'(x)(y-x) \quad \forall x,y \; \in \Bbb R $

I'm stuck from here.

Hisaab1
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  • it would be enough to get $f(y)\ge f(x) + m(y-x)$ for some $m\neq 0$ – Calvin Khor Sep 28 '20 at 12:03
  • $x=\frac x n (n)+(1-\frac x n )(0)$ gives $f(x) \leq \frac x n M+(1-\frac x n )(f(0))$where $M$ is an upper bound for $f$. Let $n \to \infty$ to get $f(x) \leq f(0)$ for all $x \geq 0$. Reverse inequality is similar. – Kavi Rama Murthy Sep 28 '20 at 12:07
  • Any bounded convex function $:\mathbb R \to \mathbb R$ is a constant. No need for differentiability. – Kavi Rama Murthy Sep 28 '20 at 12:17

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