The definition I have of a convex function $f: \mathbb{R} \rightarrow \mathbb{R}$ is that for every $x, y \in \mathbb{R}$ and every $\lambda \in [0, 1]$, $$ f(\lambda x + (1-\lambda )y) \leq \lambda f(x) + (1- \lambda )f(y).$$ It is well known that convex function differentiable a.e. I want to show that dervitive of f is nondecreasing. That points that f is differentiable, difination of convex implies nondecreasing. So, I just need to show that that set of points that f is not differentiable, jumps up. Then, the $f^{'}$ is nondecreasing. Does one have any idea?

  • $\begingroup$ points are not differentiable, but functions are $\endgroup$
    – zhw.
    Aug 8, 2019 at 16:43
  • $\begingroup$ @zhw. : I meant that points that function.is not differentiable. $\endgroup$
    – Adam
    Aug 8, 2019 at 17:03

1 Answer 1


Let $x_{-} <x<x_{+}$. We can write $x=\frac{x_{+}-x}{x_{+}-x_{-}}x_{-}+\frac{x-x_{-}}{x_{+}-x_{1}}x_{+}.$

$f(x)=f(\frac{x_{+}-x}{x_{+}-x_{-}}x_{-}+\frac{x-x_{-}}{x_{+}-x_{1}}x_{+})=^{\textrm{convex property}} \leq \frac{x_{+}-x}{x_{+}-x_{-}} f(x_{-})+\frac{x-x_{-}}{x_{+}-x_{1}} f(x_{+})$. Then $$ \frac{f(x)-f(x_{-})}{x-x_{-}} \leq \frac{f(x_{+})-f(x)}{x_{+}-x}$$ Let $x \rightarrow x_{-}$ and $x \rightarrow x_{+}$. So, $f^{'}(x_{-}) <f^{'}(x_{+}).$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .