3
$\begingroup$

Let $f_{n}$ be some sequence of functions.

If $f_{n}$ uniformly converges to $f$ then is it true that $f^{\prime}_{n}\rightarrow f^{\prime}$?

Is there an example that proves/disproves this?

$\endgroup$
  • $\begingroup$ No I think you need another condition on $f'_n$. $\endgroup$ – Gregory Grant Jan 16 '16 at 12:35
  • 1
    $\begingroup$ Welcome to our site! $\endgroup$ – kjetil b halvorsen Jan 16 '16 at 12:41
  • $\begingroup$ @kjetilbhalvorsen, Thank you. $\endgroup$ – gbd Jan 16 '16 at 12:44
7
$\begingroup$

Counterexample: $f_n = \frac1n\sin(n^2 x)$.

$\endgroup$
  • $\begingroup$ What does $f_{n}$ converge to? $\endgroup$ – gbd Jan 16 '16 at 12:40
  • 1
    $\begingroup$ @gbd, $f_n\to 0$ because $|f_n|\le 1/n$. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 16 '16 at 12:41
  • $\begingroup$ Does it converge to zero? $\endgroup$ – gbd Jan 16 '16 at 12:41
  • $\begingroup$ What is the other condition need that @GregoryGrant mentions? $\endgroup$ – gbd Jan 16 '16 at 12:43
  • $\begingroup$ @gbd, see the other answer. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 16 '16 at 12:44
5
$\begingroup$

This is false in quite a strong sense. The Weierstrass function is a uniform limit of differentiable functions, but is nowhere differentiable. By the Stone—Weierstrass Theorem, any continuous function on a closed bounded interval is a uniform limit of polynomials.

$\endgroup$
2
$\begingroup$

The additional condition required is that a function $g$ exists to which the $f'_n$ uniformly converge (in which case of course $g=f'$). In particular, this PDF proves on p. 9 that this condition suffices, but on p. 8 it shows that $f_n=\frac{x}{1+nx^2}$ is a uniformly convergent sequence whose derivatives form a convergent but not uniformly convergent counterexample.

$\endgroup$
2
$\begingroup$

As mentioned above this not true in general. But one might still aks what a sufficient condition for "$f_n' \to f'$ uniformly" might be. If you already know that $f_n$ converges uniformly to $f$ and $f_n'$ converges uniformly to some function say $g$, then by the fundamental theorem of calculus one indeed obtains: $g=f'$ hence $\space$$f_n' \to f'$ uniformly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.