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Let $\{f_n\}_n$ be a sequence of functions $\mathbb R \rightarrow \mathbb R$ which converges pointwise to $f$, ie:

$$f_n(x) \rightarrow f(x) \hspace{10pt}\hbox{for all $x$}.$$

What additional conditions are needed so that the derivatives at $0$ of $\{f_n\}_n$ converge to the derivatives at $0$ of $f$ ?

One condition that would make sense is "uniform convergence on all compact sets" though I can't seem to find references

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  • $\begingroup$ Two functions that are uniformly close may have rather different derivatives... $\endgroup$
    – Siminore
    Aug 26, 2015 at 15:18
  • $\begingroup$ there is no such type of theorem (that I am aware of) $\endgroup$
    – user251257
    Aug 26, 2015 at 15:23
  • $\begingroup$ Uniform convergence on compact sets would work if these were analytic functions on $\mathbb C$. $\endgroup$
    – Robert Israel
    Aug 26, 2015 at 15:32
  • $\begingroup$ Robert: Doesn't the $ sin(nx)/n $ example of siminore have uniform convergence on compacts ? $\endgroup$
    – Guillaume Dehaene
    Aug 27, 2015 at 9:13
  • $\begingroup$ See wikipedia-'Uniform convergence'. $\endgroup$
    – Dreamer
    Nov 6, 2018 at 17:14

1 Answer 1

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Unlike integration, differentiation is a very unstable operation. It is very hard to make assumptions on $\{f_n\}_n$ so that $\{f'_n\}_n$ converges. For instance, let $f_n(x)= \frac{\sin (nx)}{n}$: $\{f_n\}_n$ converges to zero uniformly, but the derivatives $f'_n$ are oscillating.

The only "elementary" theorem about differentiation of sequences of functions assumes that $\{f'_n\}_n$ converges uniformly; the conclusion is that pointwise convergence (even at a single point) of $\{f_n\}_n$ implies uniform convergence.

All this is strictly related to the fact that the differentiation operator is unbounded as a linear operator in essentially every useful function space.

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  • $\begingroup$ I see. So a condition that would work is if all the derivatives of the $p_n$ are bounded (with a different bound for each derivative), no ? That seems doable for the problem I'm tackling $\endgroup$
    – Guillaume Dehaene
    Aug 27, 2015 at 9:15

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