# Is it true that if $f_{n}\rightarrow f$ uniformly converges then $f^{\prime}_{n}\rightarrow f^{\prime}$?

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?

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

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

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

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.

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.

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.