The convex function $f$ is differentiable at $a$ if and only if for every $n$ there exists $a_n>0$ such that $|f(a+a_n)+f(a−a_n)−2f(a)|<\frac{a_n}{n}$

Question

Let $f:\mathbb{R}→\mathbb{R}$ a convex function.

Show that f is differentiable in $a\in \mathbb{R}$ ⇔ for every $n∈\mathbb{N}$ there exists $a_n>0$ such that $|f(a+a_n)+f(a−a_n)−2f(a)|<\frac{a_n}{n}$.

''$\Leftarrow$'' I know that $a-a_n<a<a+a_n$ anf $f$ is convex, so

$\frac{f(a)-f(a-a_n)}{a_n}<\frac{f(a+a_n)-f(a)}{a_n}\Rightarrow\frac{f(a)-f(a-a_n)}{a_n}<\frac{f(a+a_n)-f(a)}{a_n}<\frac{f(a)-f(a-a_n)}{a_n}+\frac{1}{n} $

Because $f$ is convex the derivates to left and right in $a$ exists. But I am stuck.

Answer 1

Using the convexity, it is easy to show that we have for any $x_1 < x_2 < x_3$ $$\frac{f(x_2)-f(x_1)}{x_2-x_1} \leq \frac{f(x_3)-f(x_1)}{x_3-x_1} \leq \frac{f(x_3)-f(x_2)}{x_3-x_2}.$$ So we see that for any $0<h < a_n$ $$\tag{1}\frac{f(a)-f(a-a_n)}{a_n}\leq \frac{f(a)-f(a-h)}{h} \leq \frac{f(a+h)-f(a)}{h} \leq \frac{f(a+a_n)-f(a)}{a_n}.$$

Thus, by the 'sandwich theorem', it is enough to show the convergence of the difference quotient for one sequence only. Moreover, we see that the left- and right derivate exist.

"$\Leftarrow$": First rewrite the conditon by $$\tag{2}\left| \frac{f(a+a_n)-f(a)}{a_n} + \frac{f(a-a_n)-f(a)}{a_n} \right| < \frac{1}{n}.$$ Letting $n \rightarrow \infty$, this gives $D_+f = D_-f$.

Edit: At this step we need that there exists a subsequence with $a_{n_k} \rightarrow 0$.

It is simple to see that (1) implies that the left-term in (2) is non-negative. Thus, we may drop the absolute value. Now, let us assume that we have $a_n \geq \varepsilon$ for some $\varepsilon>0$ and all $n \geq N$. This implies for all $0<h< \varepsilon$ that $a-a_n < a -h < a < a+h < a+a_n$ and thus by (1) again $$\frac{f(a+h)-f(a)}{h} + \frac{f(a-h)-f(a)}{h} \leq \frac{f(a+a_n)-f(a)}{a_n} + \frac{f(a-a_n)-f(a)}{a_n} < \frac{1}{n}.$$ Since this is valid for any $0 <h < \varepsilon$, we can argue as in the first case.

"$\Rightarrow$": This step is easy: Just use the definition of the convergence of the difference quotient.