Prove that if a convex function defined on $(-\infty,+\infty)$ is non-monotonous, then there exists a point $c$ such that $f$ is (not strictly) monotonous decreasing on $(-\infty,c]$ and is (not strictly) monotonous increasing on $[c,+\infty)$.

This is trival if $f$ is differentiable on $(-\infty,+\infty)$, but how to prove it if $f$ is indifferentiable?

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  • Please don't make question titles just by taking the first words of an exercise and cutting them off at a completely random and meaningless point. – Henning Makholm Aug 10 at 14:12
  • 1
    Monotonic, not monotonous. – DanielWainfleet Aug 10 at 14:34
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It is well known that the right-hand derivative $g(x)=f'(x+)$ exists at every point and $$\int_a^{b} g(t)\, dt =f(b)-f(a) \cdots (\dagger)$$ whenever $a<b$. Also, $g$ is increasing. lf $g(t)\geq 0$ for all $t$ then $(\dagger)$ shows that $f$ is monotonically increasing on $\mathbb R$. Similarly if $g(t)\leq 0$ for all $t$ then $(\dagger)$ shows that $f$ is monotonically decreasing on $\mathbb R$. In the remaining case define $c=sup \{t: g(t)<0\}$. Then $g(t)\geq 0$ for all $t>c$ and $g(t)\leq 0$ for all $t<c$, so $f$ is decreasing on $(-\infty ,0)$ adn increasing on $(0,\infty)$. A good reference for properties of convex functions I have used is Real and Abstract Analysis by Hewitt and Stromberg.

We know that there are points $p < q < r$ such that $f(p) \ge f(q) \le f(r)$.

Consider $[p, r]$. The function is continuous, so by the extreme value theorem it attains a minimum $f(c)$ at $c \in [p,r]$.

Then it should be easy to show that the $c$ satisfies the condition.

  • How exactly does convexity come in? – Kavi Rama Murthy Aug 10 at 11:45
  • @KaviRamaMurthy each step uses convexity... – Kenny Lau Aug 10 at 11:46
  • To the proposer: A convex function is necessarily continuous. Also, if $f$ is convex and not monotonic then $p,q,r$ exist. – DanielWainfleet Aug 10 at 14:38

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