Piecewise linear approximations of Convex function

Question

How to prove the following:
If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a convex function then there exists sequence of piecewise linear convex functions $\{f_n\}$ such that $f_n \rightarrow f.$

Answer 1

No : "enumeration of rational numbers" without precising how is done this enumeration will lead you nowhere.

I suggest you to take a subset of $\mathbb{Q}$ which is the set of values defined for all $k \in \mathbb{Z}$ and all $n \in \mathbb{N}$ by

$$x_{k,n}:=\frac{k}{2^n}.$$

(the so-called "dyadic numbers").

Then define $f_n$ as the continuous piecewise linear coinciding with $f$ with breakpoints $x_{k,n},f(x_{k,n})$ for all $k \in \mathbb{Z}$.

In this way, for any $n$ $f_{n+1}$ appears as a "refinement' of $f_n$, with convergence to $f$ that isn't difficult to establish.