Uniformly bounded derivative implies uniform convergence

Question

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.

Answer 1

I can conclude a subsequence of $f_n$ converges uniformly. But the existence of a uniformly convergent subsequence does not guarantee the original sequence converges uniformly.

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.

Answer 2

A slight alternative to the above great answer provided by Daniel Fischer is to first show that $f$ is continuous and then use the same technique, but directly to $f$, instead of showing the "Cauchy criterion".

To see why $f$ is continuous, fix $\epsilon>0$ and $x\in[a,b]$. Since $\forall n\in\mathbb{N}, \forall x\in[a,b]$ we have $|f'_n(x)| < M$ so using the mean value theorem, $$\exists\delta>0 : |y-x|<\delta\Rightarrow \forall n\in\mathbb{N}, |f_n(x) - f_n(y)|<\frac{\epsilon}{3}$$ For this same $\delta$, if $|y-x|<\delta$ and then we pick $n$ large, using pointwise convergence of $f_n$, we have

$$ |f(x)-f(y)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|< \frac{\epsilon}{3} +\frac{\epsilon}{3}+\frac{\epsilon}{3} = \epsilon$$

We found our $\delta$ so $f$ is continuous. The idea used here was to somehow shift the difference in $f$ to the difference in $f_n$, instead. The uniform bound in the derivatives of $f_n$ was necessary to ensure that our $\delta$ worked for any $n$. In any case, as $[a,b]$ is compact, $f$ is also uniformly continuous.

Now to show uniform convergence, again fix $\epsilon>0$. Now, for some large $N\in\mathbb{N}$, let us partition the interval $[a,b]$ into $N$ short enough pieces $[x_0,x_1), [x_1,x_2),...,[x_{N-1},x_N]$ such that $\forall n\in\mathbb{N}$, if $x,y$ are in the same piece then we have $|f_n(x) - f_n (y)|<\epsilon/3$ (possible since $f_n'(x)$ are bounded) and $|f(x)-f(y)|<\epsilon/3$ (possible since $f$ is uniformly continuous). This, again, used the uniform bound in the derivatives so that the length of the pieces work uniformly across all $f_n$'s. Finally, by picking $n$ large such that for $k=1, 2, \dots, N-1$, we have $|f_n(x_k)-f(x_k)|<\epsilon/3$, then $\forall x\in[a,b]$, $x$ is in the same piece as some $x_k$ and so

$$|f_n(x)-f(x)|\le |f_n(x) - f_n(x_k)|+|f_n(x_k)-f(x_k)|+|f(x_k)-f(x)| \le \frac{\epsilon}{3} +\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$$

Note that we picked $n$ before we picked $x$ so this shows uniform convergence.

Answer 3

A well known equivalence of uniform convergence on compact sets is that $f_n$ converges uniformly to $f$ on a compact set $K$ if and only if for every convergent sequence $x_n \to x$ in $K$ one has $f_n(x_n) \to f(x)$

So take any such sequence. Now,

$$|f_n(x_n) - f(x)| \le |f_n(x_n) - f_n(x)| + |f_n(x) - f(x)|$$

The term $|f_n(x) - f(x)|$ can be made arbitrarily small by choosing a big enough $n$, since $f_n \to f$

On the other hand by Lagrange's Mean Value Theorem,

$$|f_n(x_n) - f_n(x)| = |f'_n(c)| |x - x_n| \le M |x - x_n|$$

(With $c$ somewhere between $x$ and $x_n$) and the rightmost side can now be made arbitrarily small by again choosing a big enough $n$