# Uniform convergence of derivatives, Tao 14.2.7.

This is ex. 14.2.7. from Terence Tao's Analysis II book.

Let $I:=[a,b]$ be an interval and $f_n:I \rightarrow \mathbb R$ differentiable functions with $f_n'$ converges uniform to a function $g:I \rightarrow \mathbb R$. Suppose $\exists x_0 \in I: \lim \limits_{n \rightarrow \infty} f_n(x_0) = L \in \mathbb R$. Then the $f_n$ converge uniformly to a differentiable function $f:I \rightarrow \mathbb R$ with $f' = g$.

We are not given that the $f_n'$ are continuous but he gives the hint that $$d_{\infty}(f_n',f_m') \leq \epsilon \Rightarrow |(f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0))| \leq \epsilon |x-x_0|$$ This can be shown by the mean value theorem. My question is : How does this help me to prove the theorem ?

Since $\{f_n(x_0)\}$ converges, for each $\epsilon > 0$ and $n, m$ large enough we have \begin{align} \lvert f_n(x) - f_m(x) \rvert &\leq \left\lvert (f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0)) \right\rvert + \left\lvert f_n(x_0) - f_m(x_0) \right\rvert \\ &\leq \epsilon \left\lvert x - x_0 \right\rvert + \epsilon \\ &\leq \epsilon (b - a) + \epsilon \end{align} Hence $f_n$ converges uniformly on $I$ to a function $f$, moreover for each $\epsilon > 0$ and $m, n$ large enough, the inequality $$\left\lvert \frac {f_n(y) - f_n(x)} {y - x} - \frac {f_m(y) - f_m(x)} {y - x} \right\rvert \leq \epsilon$$ holds for each $x\neq y\in I$. (It is the same inequality of the hint but now we can assume it holds for generic $y\in I$, because we showed $f_n(y)$ converges for all $y \in I$)
The above relation implies that $\frac {f_n(y) - f_n(x)} {y - x}$ converges uniformly to $\frac {f(y) - f(x)} {y - x}$.

Now we can write $$\left\lvert\frac {f(y) - f(x)} {y - x} - g(x) \right\rvert \leq \\ \left\lvert\frac {f(y) - f(x)} {y - x} - \frac {f_n(y) - f_n(x)} {y - x} \right\rvert + \left\lvert \frac {f_n(y) - f_n(x)} {y - x} - f_n'(x)\right\rvert + \left\lvert f_n'(x) - g(x) \right\rvert$$ For each $\epsilon > 0$ and $n$ large enough we get $$\left\lvert\frac {f(y) - f(x)} {y - x} - g(x) \right\rvert \leq 2\frac \epsilon 3 + \left\lvert \frac {f_n(y) - f_n(x)} {y - x} - f_n'(x)\right\rvert$$ and for $y$ close enough to $x$ $$\left\lvert\frac {f(y) - f(x)} {y - x} - g(x) \right\rvert \leq \epsilon$$ So $f'(x)$ exists and is equal to $g(x)$.

• so $(f_n(x))_{n=0}^{\infty}$ is voor every $x \in I$ a Cauchy-sequence and since $\mathbb R$ is complete we know that $\lim \limits_{x \rightarrow y} f_n(x) =f_n(y)$ ? – user42761 Oct 15 '12 at 13:10
• Is it then sufficient to define $f(x) := \lim \limits_{n \rightarrow \infty} f_n(x)$ ? – user42761 Oct 15 '12 at 13:51
• @André I added the details of the proof. – AlbertH Oct 15 '12 at 20:35
• Great thanks for your time ! – user42761 Oct 15 '12 at 21:02

Due to the uniform convergence of the $f'_n$ you can find an $N$ for every $\epsilon$ such that $(f'_n(x) - g(x)) \leq \epsilon$ for all $n \geq N$, which is equivalent to $d_\infty(f'_n,g) \leq \epsilon$. Thus, $d_\infty(f_n',g) \to0$ as $n\to\infty$.

Now, $\int_{a}^x f_n'(y) - g(y) dy \leq d_\infty(f_n',g)|x-a| \leq d_\infty(f_n',g)|b-a|$. Since $d_\infty(f_n',g) \to0$ as $n\to\infty$ you get that $\int_a^x f_n'(y)dy$ converges uniformly to $\int_a^x g(y)dy$.

In general, that won't transfer to $f_n(x) = c_n + \int_a^x f_n'(y)dy$ because the $c_n$ could be chosen maliociously. But if $\lim_{n\to\infty}f_n(x) = \lim_{n\to\infty} c_n + \int_a^x f_n'(y)dy$ converges for one $x$, then $\lim_{n\to\infty}c_n$ must converge, since the second term converges too (every uniformly!).

Which in turn means the limit must actually converge for all $x$, ecause $\lim_{n\to\infty}c_n$ doesn't actually depend on $x$. And for the same reason (and because the other term converges uniformly), the convergence is even uniform.

• I think this assumes that the $f_n'$ are continuous. – punctured dusk Apr 29 '18 at 15:08