Conditions for convergence of a derivative, given the function itself is convergent

Suppose $$\{f_n\}_{n \in \mathbb{N}}$$ is a family of bounded, differentiable, monotone increasing functions on $$[0,1]$$, which converge uniformly to a limit $$f$$. Also, suppose we know that $$f_n'$$ is $$\alpha$$-Lipschitz continuous for some constant $$\alpha$$ (not depending on $$n$$). I want to analyze the differentiability of $$f$$ and the convergence of $$f_n'$$, if this is even possible.

Of course, the monotone increasing property of $$f_n$$ implies that $$f$$ is also monotone increasing, so it is differentiable Lebesgue almost surely on $$[0,1]$$. Is it possible to say that $$f_n'$$ converges to $$f'$$ pointwise, wherever $$f'$$ exists (even if only on a subsequence)?

The answer is positive, and weaker assumptions suffice.

Claim: Suppose $$\{f_n\}_{n \in \mathbb{N}}$$ is a family of bounded, differentiable, functions on $$[0,1]$$, which converge pointwise to a limit $$f$$. Also, suppose we know that $$\{f_n'\}$$ are a uniformly equicontinuous$$^{(*)}$$ family of functions. Then $$f$$ is differentiable on $$[0,1]$$ and $$f_n' \to f$$ uniformly.

$$(*)$$ (Equicontinuity, defined in [1], certainly holds if $$f_n'$$ are $$\alpha$$-Lipschitz continuous for some constant $$\alpha$$ not depending on $$n$$. In fact, Holder continuity with uniform constants also suffices.).

Proof of claim: Since $$f_n$$ are uniformly bounded and $$\{f_n'\}$$ are uniformly equicontinuous, it follows$$^{\bf (**)}$$ that the derivatives $$f'_n$$ are uniformly bounded. The Ascoli-Arzela theorem [1] implies that $$f_n'$$ has a subsequence $$f'_{n(k)}$$ that converges uniformly to some $$g \in C[0,1]$$. Therefore, for all $$t$$ in $$[0,1]$$, $$f(t)-f(0)=\lim_k \int_0^t f'_{n(k)}(x) \,dx = \int_0^t g(x) \,dx \,,$$ so $$f'=g$$ in $$[0,1]$$ by the fundamental theorem of calculus.

If $$f'_n$$ do not converge uniformly to $$g=f'$$, then there exists some $$\epsilon>0$$ and another subsequence $$f'_{m(k)}$$ of $$f_n'$$, such that $$\|f'_{m(k)}-f'\|_\infty >\epsilon$$ for all $$k$$. But $$f'_{m(k)}$$ must also have a uniformly convergent subsequence by [1], and the argument above shows that the limit of this subsequence must be $$f'$$, a contradiction. Thus $$f'_n \to f'$$ uniformly in $$[0,1]$$.

$$(**)$$ Addendum: Since there exists $$M$$ such that $$|f_n| \le M$$ in $$[0,1]$$ for all $$n$$, the mean value theorem implies that for each $$n$$, there is some $$t_n \in (0,1)$$ such that $$|f_n'(t_n)|=|f_n(1)-f_n(0)| \le 2M$$. Since $$f_n'$$ are uniformly equicontinuous, there is some $$\delta_1<1$$ so that $$|x-y|<\delta_1$$ implies $$|f_n(x)-f_n(y)|<1$$ for all $$n$$. Thus for all $$n \ge 1$$ and all $$x \in [0,1]$$, we have $$|f_n'(x)-f_n'(t_n)| \le 1+1/\delta_1$$, so $$|f_n'(x)| \le 2M+1+1/\delta_1$$.

• Thank you Dr. Peres. I failed to see that these assumptions imply that $\{f_n'\}$ is uniformly bounded, from which Arzela-Ascoli can be immediately applied. Aug 24, 2022 at 16:21
• @qp212223 I added an explanation of this point. Aug 24, 2022 at 21:47

Yes. I will prove this via functional analysis; I do not know if there is a simpler method.

1. Note that $$f_n$$ are uniformly bounded, since they converge uniformly.
2. Suppose $$m\leq f_n\leq M$$ for any $$n$$. By the intermediate value theorem, there is some point $$a_n\in[0,1]$$ and $$b_n\in[0,M-m]$$ such that $$f_n'(a_n)=b_n$$.
3. Each $$f_n'$$ is uniformly bounded in $$L^{\infty}$$, since $$\{f_n'\}_n$$ are equi-Lipschitz on a compact domain, and attain the similar values of the $$b_n$$. The domain $$[0,1]$$ has finite measure; thus $$f_n'$$ are uniformly bounded in any $$L^p$$ for $$p\in[1,\infty]$$.
4. For any $$p\in(1,\infty)$$, the space $$L^p$$ is reflexive. Pick your favorite such $$p$$ (mine is $$p=2$$); then there is a subsequence (which I will also call $$f_n'$$) that converges weakly, by separable Banach-Alaoglu. Call the limit $$g$$, but note that $$g$$ is only well-defined up to a.e. equivalence.
5. Since $$[0,1]$$ has finite measure, if $$\phi\in L^\infty$$, then $$\phi\in L^2$$. Thus $$\int{\phi f_n'}\to\int{\phi g}$$ by weak convergence in $$L^2$$. So actually $$f_n'\rightharpoonup g$$ in $$L^1$$.
6. Fix $$\epsilon$$ and let $$U(\epsilon)_n=\{x:f_n'(x)-g(x)\geq\epsilon\}$$ and $$V(\epsilon)_n=\{x:g(x)-f_n'(x)\geq\epsilon\}$$. Then $$0\leq\epsilon|U(\epsilon)_n|\leq\int{1_{U(\epsilon)_n}(f_n'-g)}\to0$$ Likewise $$|V(\epsilon)_n|\to0$$, and so $$f_n'\to g$$ in measure as well as weakly.
7. Convergence in measure implies convergence a.e. up to a subsequence. Again, I will call that subsequence $$f_n'$$.
8. Let $$U$$ be a dense set on which $$f_n'\to g$$ pointwise. A pointwise limit of uniformly Lipschitz functions is also Lipschitz (exercise); thus any representative of the a.e.-equivalence class $$g$$ is uniformly continuous on $$U$$. A uniformly continuous function on a dense set has a unique continuous extension to that set's closure; thus $$g$$ has representatives that are continuous on $$[0,1]$$. Pick one such, and also call it $$g$$.
9. Pick $$x\in[0,1]$$. Then there exists $$\{x_t\}_{t=0}^\infty\in U^\omega$$ converging to $$x$$. For any $$t$$, \begin{align*} \limsup_{n\to\infty}{|g(x)-f_n'(x)|}&\leq\limsup_{n\to\infty}{|g(x)-g(x_t)|+|g(x_t)-f_n'(x_t)|+|f_n'(x_t)-f_n'(x)|} \\ &\leq|g(x)-g(x_t)|+0+\alpha|x_t-x| \\ &\to0 \end{align*} as $$t\to\infty$$. Thus $$f_n'\to g$$ pointwise everywhere.
10. For any set $$[0,x]$$, we have $$\int_0^x{g(x)\,dx}=\int{1_{[0,x]}g}=\lim_{n\to\infty}{\int{1_{[0,x]}f_n'}}=\lim_{n\to\infty}{(f_n(x)-f_n(0))}=f(x)-f(0)$$
11. Since $$g$$ is continuous (everywhere), $$f'=g$$ everywhere by Lebesgue's differentiation theorem.