2
$\begingroup$

Thinking about the (probably) well-known fallacy about approaching a unit square diagonal with staircase functions and thus concluding the diagonal length be $2$ instead of $\sqrt 2$ led me to an interesting question:

Given a sequence $(f_k)_{k\in\mathbb N}$ of differentiable functions converging towards a differentiable limit function $f$, when does the limit of derivatives coincide with the derivative of the limit function, that is, when do we have $f'(x)=\lim_{k\to\infty}f_k'(x)$ for all $x$ in the function's domain? And what about second or $n$-th derivatives, supposing all the functions $f_k$ as well as $f$ are twice or $n$ times differentiable?

No need to tell me staircase functions aren't differentiable - this is supposed to be a more general question about necessary and sufficient conditions for the limit of $n$-th derivatives to coincide with the $n$-th derivative of the limit.

$\endgroup$
2
$\begingroup$

Suppose all functions are defined on an open set.

If $f_n \to f$ pointwise and $f_n' \to g$ uniformly, then $f$ is differentiable and $f' = g$.

For higher derivatives you should have $f_n^{(m)}$ converging uniformly.

$\endgroup$
1
$\begingroup$

You can have weaker hypotheses. Namely:

Suppose $(f_n)$ is a sequence of real or complex-valued differentiable functions on an interval $I\subset\mathbf R$. If

  • The sequence $(f'_n)$ converges uniformly on every bounded closed interval contained in $I$ to a function $g$,
  • $\bigl(f_n(x_0)\bigr)$ converges for at least one point $x_0\in I$,

then the sequence $(f_n)$ converges uniformly on every bounded closed interval contained in $I$ to a function $f$, differentiable on $I$, and $f'=g$.

The same result is true if ‘differentiable’ is replaced by $C^1$.

$\endgroup$

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.