Let $f_n$ be a sequence of differentiable functions on $[a, b] \subset \mathbb{R}$.
Suppose
- $\lim_{n \rightarrow \infty} f_n(x) = f(x)$ exists for all $x \in [a, b]$, and
- the derivatives $|f_n'(x)| < M$ are uniformly bounded over $n$ and $x$
Prove that $f_n$ converges to $f$ uniformly.
The closest I can get is to try to adapt Arzela-Ascoli somehow, but it is not working.
I can conclude the existence of a subsequence that converges uniformly, but this doesn't guarantee the original sequence converges uniformly.
Best Answer
Generally, that indeed does not imply the uniform convergence of the full sequence.
However, here we are in a special situation, since we know that the full sequence converges pointwise, and an equicontinuous sequence that converges pointwise converges uniformly on compact sets.
Since the sequence in question is not only equicontinuous but equilipschitz, the proof is easier:
Let $\varepsilon > 0$ be given. Since $\lvert f_n'(x)\rvert \leqslant M$ for all $n$ and $x\in [a,b]$, we have
$$\lvert f_n(x) - f_n(y)\rvert \leqslant \frac{\varepsilon}{4}$$
for all $x,y\in [a,b]$ with $\lvert x-y\rvert \leqslant \frac{\varepsilon}{4M}$.
Choose $N$ large enough that $\frac{b-a}{N} < \frac{\varepsilon}{4M}$, and let $x_k = a + k\frac{b-a}{N}$ for $0 \leqslant k \leqslant N$. For each $k$, there is an $n_k \in \mathbb{N}$ such that $\lvert f_n(x_k) - f(x_k)\rvert < \frac{\varepsilon}{4}$ for all $n \geqslant n_k$. Let $n(\varepsilon) = \max \{ n_k : 0 \leqslant k \leqslant N\}$.
Then, for every $x \in [a,b]$ and $n, m \geqslant n(\varepsilon)$ we have
$$\begin{align} \lvert f_n(x) - f_m(x)\rvert &\leqslant \lvert f_n(x) - f_n(x_k)\rvert + \lvert f_n(x_k) - f(x_k)\rvert + \lvert f(x_k) - f_m(x_k)\rvert + \lvert f_m(x_k) - f_m(x)\rvert\\ &\leqslant \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} = \varepsilon, \end{align}$$
for $k = \left\lfloor N\frac{x-a}{b-a}\right\rfloor$, so $\lvert x-x_k\rvert \leqslant \frac{\varepsilon}{4M}$.
Thus, for every $\varepsilon > 0$, we found an $n(\varepsilon)\in\mathbb{N}$ with
$$\sup \{ \lvert f_n(x) - f_m(x)\rvert : x \in [a,b], \, n,m \geqslant n(\varepsilon)\} \leqslant \varepsilon,$$
i.e. the sequence converges uniformly.