the derivative of a sequence of convex function coverge

Question

Continue of the discusstion in: Limit of derivatives of convex functions

It proves that:

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 $x\in D$.

What can be said about the converge rate of the derivative? i.e. assume that $|f_n(x)-f(x)| \leq \delta_n $ what can be said about $|f_n'(x)-f'(x)|$?

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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 $$\frac{f(x+h)-f(x)}{h} < f'_{+}(x) + \epsilon$$

Here I need somehow to bound $h$ as a function of $\epsilon$ or even better find a sequence $\epsilon_n$ and $h_n$ that converge to 0 and satisfy this.

Then since $f_n$ converges to $f$ there exists $N\in\mathbb{N}$ such that $$\frac{f_n(x+h)-f_n(x)}{h} < f'_{+}(x) + \epsilon$$

Here also I think that $h_n$ should converge small enaough so that we can move to $f_n$ instade of $f$

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 1

Given $x$ and $\epsilon > 0$, take $\delta> 0$ small enough that $$\eqalign{f'_+(x) - \epsilon &< \dfrac{f(x+y)-f(x)}{y} < f'_+(x) + \epsilon \ \text{for}\ 0 < y \le 2\delta \cr f'_-(x) - \epsilon &< \dfrac{f(x+y)-f(x)}{y} < f'_-(x) + \epsilon \ \text{for}\ -2\delta \le y < 0 }$$ Take $n$ so large that $|f_n(t) - f(t)| < \epsilon \delta$ at $t = x - 2\delta, x - \delta, x+\delta$ and $x + 2\delta$. Then $$ \eqalign{&\left| \dfrac{f_n(x + 2\delta)- f_n(x + \delta)}{\delta} - f'_+(x)\right|\cr \le & \left| \dfrac{f_n(x+2\delta) - f(x+2\delta)}{\delta}\right| + \left|\dfrac{f(x+\delta) - f_n(x+\delta)}{\delta}\right|\cr & + \left| \dfrac{f(x+2\delta)-f(x)}{\delta} - 2 f'_+(x) \right| + \left|\dfrac{f(x+\delta)-f(x)}{\delta} - f'_+(x) \right|\cr < & 5 \epsilon }$$

By the Mean Value Theorem, $$ f_n'(\xi) = \dfrac{f_n(x+2\delta) - f_n(x+\delta)}{\delta} < f'_+(x) + 5 \epsilon$$ for some $\xi \in (\delta,2\delta)$. Since $f_n$ is convex, $f'_n(t) \le f_n'(\xi) \le f'_+(x) + 5 \epsilon$ for all $t < x + \delta$. Similarly, on the other side we get $f'_n(t) \ge f'_-(x) - 5 \epsilon$ for all $t > x - \delta$. In particular, $f'_-(x) - 5 \epsilon < f'_n(x) < f'_+(x) + 5 \epsilon$.

Answer 2

Let $\lambda_n$ be such that $|f_n(x)-f(x)|\leq \lambda_n$.

Since $f_n(x)$ is convex we have: $$ f_n^+(x) \leq \frac{f_n(x+\delta_n)-f_n(x)}{\delta_n}$$ for a sequence $\delta_n$ to be detemined later.

By using the uniform converge of $f_n(x)$ we have: $$ f_n^+(x) \leq \frac{f(x+\delta_n)-f(x)+2\lambda_n}{\delta_n} $$ $$ \leq f^-(x+\delta_n)+2\frac{\lambda_n}{\delta_n}$$ $$ \leq f^+(x+\delta_n)+2\frac{\lambda_n}{\delta_n} \rightarrow f^+(x)$$

where we used the fact that $f^-(x) \leq f^+(x)$ and the right continuity of $f^+(x)$. The sequence $\delta_n$ should be chosen such that $\frac{\lambda_n}{\delta_n}$ converge to 0 and caluclating the converge rate depend of the sequences and the behaviour of $f$ at $x$. By similar calculations we can also have $$ f_n^-(x) \geq f^-(x-\delta_n)-2\frac{\lambda_n}{\delta_n}$$.

When $f^-(x) = f^+(x)$ we have converge of the derivative sequence.