# Set of points where a convex function is not differentiable

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?

• points are not differentiable, but functions are
– zhw.
Aug 8, 2019 at 16:43
• @zhw. : I meant that points that function.is not differentiable.
Let $$x_{-} . We can write $$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_{-})