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Could anyone give an example of a sequence of differentiable (real) functions that uniformly converge to a differentiable function, but the derivatives of which don't converge to the derivative of the limit function?

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Why the downvote?.. – Pandora Jun 30 '15 at 20:54
up vote 14 down vote accepted

This is from the book, Counterexamples in Analysis by Gelbaum and Olmstead. (Google books link:

This is under:

A sequence of infinitely differentiable functions converging uniformly to zero, the sequence of whose derivatives diverges everywhere.

$$f_{n}(x) = \frac{\sin nx}{\sqrt{n}}$$

This is on page 76, Chapter 7.

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+1!$\,\,\,\,\,\,$ – AD. Mar 10 '12 at 21:44

Consider the function $f_{n}: [0,2\pi] \to \mathbb{R}$ defined by $f_{n}(x)=n^{-1/2}\sin{nx}$ and let $f:[0,2\pi] \to \mathbb{R}$ be the zero function, that is $f(x)=0$.

Ok, here i work out the details. Since $\sin{x}$ oscillates between -1 and 1, we have $d_{\infty}(f_{n},f) \leq n^{-1/2}$, where $d_{\infty}$ is the uniform metric defined as $d_{\infty}(f,g)= \sup_{x \in [0, 2\pi]} |f(x)-g(x)|$. Since $\frac{1}{\sqrt{n}} \to 0$, using squeeze test we see that $f_{n} \to f$ uniformly.

Whereas $f_{n}'(x)= \sqrt{n}\cos{nx}$ so we have $|f_{n}'(0-f'(0)|=\sqrt{n}$, which says that $f_{n}'$ does not converge pointwise to $f'$.

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