Bound on $f_n'$ implies uniform convergence of $f_n$?

Question

Let $f_n$ be a sequence of functions that converge pointwise to a function $f$.

Suppose I know that $|f_n'(x)| \leq C(x)$ where the constant doesn't depend on $n$. How do I conclude that $f_n \to f$ uniformly on compact subsets?

Answer 1

For completeness, I give an example to show that the bound $|f_n'(x)| \leq C(x)$, with some finite-valued $C(x)$, is not sufficient to obtain uniform convergence. Example: let $$f_n(x) = \begin{cases} 1-\cos nx, \quad & 0\le x\le 2\pi/n \\ 0 \quad & \text{otherwise} \end{cases}$$ We have $f_n\to 0$ pointwise. Also, $$f_n'(x) = \begin{cases} n\sin nx, \quad & 0\le x\le 2\pi/n \\ 0 \quad & \text{otherwise} \end{cases}$$ satisfies $f_n'\to 0$ pointwise. Since every convergent sequence is bounded, it follows that $\sup_n |f_n'(x)|$ is finite at every point. Yet, the convergence $f_n\to 0$ is not uniform since $f_n(\pi/n)=2$.

If you assume that $C$ is bounded on every compact subset (which is the case when $C$ is continuous), then the Arzelà–Ascoli theorem applies, as David Mitra described:

you start with a given sequence $(f_n)$ and wish to show that it converges uniformly to $f$ on any compact set. So fix $K$ compact. By the Arzelà–Ascoli theorem, every subsequence of $(f_n)$ has a further subsequence that converges uniformly to $f$ on $K$. So $(f_n)$ converges uniformly to $f$ on $K$.