Limit of derivatives of convex functions

Question

Let $(f_n)_ {n\in\mathbb{N}}$ be a sequence of convex differentiable functions on $\mathbb{R}$.

Suppose that $f_n(x)\xrightarrow[n\to\infty]{}f(x)$ for all $x\in\mathbb{R}$.

Let $D:=\{x\in\mathbb{R}\,|\,f\text{ is differentiable in }x\}$. I read that $f_n'(x)\xrightarrow[n\to\infty]{}f'(x)$ for all $x\in D$.

Why is it true? How can I prove it?

Furthermore is it true that $f$ is a convex function? As consequence $\mathbb{R}\smallsetminus D$ would be at most countable.

Edit after did's comment: clearly $f$ is convex.

Answer 1

Since $f$ is convex, there exist the left and right derivative $f'_{-},f'_{+}$ in every point.

For any $x\in\mathbb{R}$ and for any $\epsilon>0$ is possible to prove that there exists $N\in\mathbb{N}$ such that $$f'_{-}(x)-\epsilon < f_n'(x) < f'_{+}(x)+\epsilon$$ for all $n\geq N$. As a consequence if $x\in D$ then $f_n'(x)\xrightarrow[n\to\infty]{}f'(x)$.

How to prove it? First by definition of right derivative there exists $h>0$ such that $$\frac{f(x+h)-f(x)}{h} < f'_{+}(x) + \epsilon$$ Then since $f_n$ converges to $f$ there exists $N\in\mathbb{N}$ such that $$\frac{f_n(x+h)-f_n(x)}{h} < f'_{+}(x) + \epsilon$$ Now use the convexity and differentiability of $f_n$ to observe that $$f_n'(x)\leq\frac{f_n(x+h)-f_n(x)}{h}$$ Conclude $f_n'(x)<f'_{+}(x) + \epsilon$. A similar reasoning holds to prove the other inequality.

Answer 2

We can go further using the particular case already proved: if $(x_{n})$ is a sequence in $\mathbb{R}$ converging to a point $x\in D$, then

$\lim\limits_{n\rightarrow\infty}f_{n}'(x_{n})=f'(x)$

In order to prove this, we will use that $D$ is dense on $\mathbb{R}$, and the derivative $f'$ es continuous relative to $D$ (this theorem is in the classical book Convex Analysis by Rockafellar)

Lets take sequences of positive numbers $(\varepsilon_{n})_{n}$ and $(\widetilde{\varepsilon}_{n})$ converging to $0$ such that $x-\varepsilon_{n}, x+\widetilde{\varepsilon}_{n}\in D$ for all $n$.

Fix $n$, and lets take $m_{0}$ such that for all $m\geq m_{0}$,

$-\varepsilon_{n}<x_{m}-x<\widetilde{\varepsilon}_{n}$, equivalently $x-\varepsilon_{n}<x_{m}<x+\widetilde{\varepsilon}_{n}$

Using that $f_{m}'$ is increasing, we have

$f_{m}'(x-\varepsilon_{n})\leq f_{m}'(x_{m})\leq f_{m}'(x+\widetilde{\varepsilon}_{n})$ for all $m\geq m_{0}$

So, if $L$ is a limit point of the sequence $(f'_{m}(x_{m}))$, using your proposition we obtain

$f'(x-\varepsilon_{n})\leq L\leq f'(x+\widetilde{\varepsilon}_{n})$

Making $n\rightarrow\infty$ and using the continuity of $f'$, we get that $L=f'(x)$. Since $L$ is an arbitrary limit point of $(f_{m}'(x_{m}))$, we conclude that

$\lim\limits_{m\rightarrow\infty}f_{m}'(x_{m})=f'(x)$