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 $g(x)$.


To clarify the point raised by @DavidC.Ullrich.

Since ${f'_n}$ converges uniformly, there exists $N \in \mathbb N$ such that $\lVert f'_n - f'_m \rVert_\infty < \epsilon$ for all $n, m > N$, that is $$ |f'_n(x) - f'_m(x)| < \epsilon \qquad \forall m,n > N, \forall x\in I $$

So, by means of the mean value theorem, for each $m,n > N$ and for each $x \neq y\in I$ we can write

$$ \left\lvert \frac {f_n(y) - f_n(x)} {y - x} - \frac {f_m(y) - f_m(x)} {y - x} \right\rvert = \\ \left\lvert \frac {f_n(y) - f_m(y)} {y - x} - \frac {f_n(x) - f_m(x)} {y - x} \right\rvert = \\ \left\lvert \frac {(f_n - f_m)(y)- (f_n - f_m)(x)} {y - x}\right\rvert = \\ \lvert (f_n - f_m)'(\xi) \rvert = \\ \lvert f_n'(\xi) - f_m'(\xi)\rvert < \epsilon $$

  • $\begingroup$ 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 $\lim \limits_{x \rightarrow y} f_n(x) =f_n(y)$ ? $\endgroup$
    – user42761
    Oct 15 '12 at 13:10
  • $\begingroup$ Is it then sufficient to define $f(x) := \lim \limits_{n \rightarrow \infty} f_n(x)$ ? $\endgroup$
    – user42761
    Oct 15 '12 at 13:51
  • $\begingroup$ @André I added the details of the proof. $\endgroup$
    – AlbertH
    Oct 15 '12 at 20:35
  • $\begingroup$ For each $x,y$ with $x\ne y$ there exists $N$ so $\left\lvert \frac {f_n(y) - f_n(x)} {y - x} - \frac {f_m(y) - f_m(x)} {y - x} \right\rvert \leq \epsilon$ for all $n,m>N$. But how do you know $N$ can be taken independent of $x,y$??? $\endgroup$ Jan 5 '20 at 12:12
  • $\begingroup$ @DavidC.Ullrich Just a thought, could be wrong: since $\exists N \; \forall n,m>N\; : \; \lVert f_n'-f_m'\rVert_{\infty}\leq\epsilon \Rightarrow \forall w \in I, \; |(f_n'-f_m')(w)|\leq \epsilon$, then by the MVT, for arbitrary $x\neq y$ ($x<y$ WLOG), $\exists z\in (x,y)\subseteq I \; : \; |(f_n-f_m)(y)-(f_n-f_m)(x)|\leq (y-x) |(f_n'-f_m')(z)| \leq (y-x)\epsilon$, and rearrangement gives the relation. $N$ seems completely determined by $\epsilon$ and doesn't appear to play any role beyond setting the Cauchy relation for $\{f_k'\}_{k\geq 0}$. $\endgroup$ Jun 23 '20 at 12:45

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.

  • 4
    $\begingroup$ I think this assumes that the $f_n'$ are continuous. $\endgroup$ Apr 29 '18 at 15:08

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