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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 ?

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2 Answers 2

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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 $g(x)$.

Edit

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

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

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  • $\begingroup$ 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, 2012 at 13:10
  • $\begingroup$ Is it then sufficient to define $f(x) := \lim \limits_{n \rightarrow \infty} f_n(x)$ ? $\endgroup$
    – user42761
    Oct 15, 2012 at 13:51
  • $\begingroup$ @André I added the details of the proof. $\endgroup$
    – AlbertH
    Oct 15, 2012 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, 2020 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, 2020 at 12:45
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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.

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    $\begingroup$ I think this assumes that the $f_n'$ are continuous. $\endgroup$ Apr 29, 2018 at 15:08
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    $\begingroup$ Just for those perusing old answers! This "proof" assumes that the functions $f_n'$ are integrable in some sense. But derivatives are not necessarily continuous, are not necessarily Riemann integrable, nor even Lebesgue integrable. The hint in Tao's book tries to force the issue by giving a hint that relies only on the mean-value theorem, not on any extra hypothesis about integrability. $\endgroup$ Apr 19 at 22:53

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