Suppose that $\{f_n\}$ is a sequence of functions that are differentiable on interval $[a,b]$ such that the sequence $\{f_n(x_0)\}$ converges at some point $x=x_0$ in $[a,b]$. If the sequence $\{f'_n\}$ converges uniformly to a function $g$ on $[a,b]$, then {$f_n$} must converge uniformly to $f$ on $[a,b]$ where $f'(x)\equiv g(x)$.
So I've tried writing out what this really means. I think it effectively comes down to proving that
$$\lim_{n \to \infty} \left[\lim_{h \to 0}\frac{f_n(x+h)-f_n(x)} h\right] = \lim_{h \to 0} \left[ \lim_{n \to \infty}\frac{f_n(x+h)-f_n(x)} h \right]$$
$$g(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)} h$$
And for any given $\varepsilon>0$ there exists a $\delta$ so that we have $$0<|h|<\delta \implies \left|\frac{f(x+h)-f(x)} h – g(x)\right|<\varepsilon$$
How do I continue from here, and if I shouldn't, which other path should I take?
Best Answer
Fix $\epsilon > 0$, and choose $N$ sufficiently large that $|f_n(x_0)-f_m(x_0)| < \epsilon$ and $|f_n'(t)-f_m'(t)| < \frac{\epsilon}{(b-a)}$ for $n,m\ge N$. Now, use mean value theorem on the function $f_n-f_m$ for two arbitrary points $x_1,x_2\in [a,b]$ to get $|f_n(x_1) - f_m(x_1) + f_m(x_2) - f_n(x_2)| = (f_n-f_m)'(c)|x_1-x_2|$. By our choice of $N$ previously, this is bounded by $\epsilon$. Then $\|f_n-f_m\|_{\textrm{sup}} < 2\epsilon$, so the sequence converges uniformly.
Now, for some $x\in [a,b]$, let $g_n(t) = \frac{f_n(t)-f_n(x)}{t-x}$. Then $|g_n(t)-g_m(t)| = \frac{1}{|t-x|}|f_n(t)-f_m(t)+f_m(x)-f_n(x)| < \frac{\epsilon}{b-a}$, so that the $g_n$ converge uniformly to $\frac{f(t)-f(x)}{t-x}$. The, by interchange of limits, $f'(x) = \lim_{t\to x} \lim_{n\to \infty} g_n(t) = \lim_{n\to \infty} \lim_{t\to x} g_n(t) = g(x)$