4

After looking at this question, I am now wondering if the theorem proven in the first answer below can be generalized to a Banach space. See here for my attempt. But before doing that, I have the following problem:

NOTE: I use the notations in the first answer below that question.

I don't know how to justify why the series $\sum a_kf_k$ can be differentiated term by term, i.e. why $\partial^\alpha (\sum a_kf_k)=\sum a_k(\partial ^\alpha f_k) $, and why the convergence of $\sum a_k\partial ^\alpha f_k$ implies the existence of $\partial^\alpha (\sum a_kf_k)$

In Banach space, the multi-index notation does not mean anything, so I should write $\sum a_kD^nf_k$.

Ma Joad
  • 7,696
  • what do you mean by "term by term"? Do you want know why each $f_k$ is (infinitely) differentiable? – supinf Jan 08 '20 at 12:38
  • @supinf See my edits. – Ma Joad Jan 08 '20 at 12:39
  • Over $\mathbb{C}$ it seems to be false, since differentiable functions over $\mathbb{C}$ that are $0$ on a set with a limit point are $0$ everywhere. Although I must admit, I don't exactly know what Frechet differentiability is. – Lukas Rollier Jan 08 '20 at 12:50
  • You seem to suggest that we can upgrade smoothness to analyticity. This is in general not true. The identity theorem is true for any analytic function in any Banach space. That means, if $A$ has nonempty interior and $f$ is analytic, then $A=X$. I honestly don't know what kind of topological assumption on $A$ would save you. – Severin Schraven Jan 08 '20 at 12:55
  • @SeverinSchraven No I am not doing that. – Ma Joad Jan 08 '20 at 12:58
  • @SeverinSchraven Please read the first answer to that linked question. $\sum a_kf_k$ is NOT a power series. – Ma Joad Jan 08 '20 at 13:02
  • Note that the notation $\partial^\alpha$ does not work anymore in the Banach space $X$, because if there are no coordinates there cannot be partial derivatives. – supinf Jan 08 '20 at 13:07
  • @LukasRollier Analytic functions don't count, according to the conditions given:) – Ma Joad Jan 08 '20 at 13:31
  • @supinf Edited. – Ma Joad Jan 08 '20 at 13:32
  • @Jethro maybe you are interested in this question – supinf Jan 13 '20 at 18:12

1 Answers1

1

(Note: This is just a part of the answer, I am not even sure if the whole proof can be generalized to separable Banach spaces.)

I don't know how to justify why the series $\sum a_kf_k$ can be differentiated term by term, i.e. why $\partial^\alpha (\sum a_kf_k)=\sum a_k(\partial ^\alpha f_k) $, and why the convergence of $\sum a_k\partial ^\alpha f_k$ implies the existence of $\partial^\alpha (\sum a_kf_k)$

A possible way to solve this is by verifying that $\sum a_k D^nf_k$ is a Fréchet derivative of $\sum a_k D^{n-1} f_k$ (and then use induction). In order to show this the estimate $$ \| D^{n-1}f_k(x+h)- D^{n-1} f_k(x) - D^n f_k(x)h\| \leq \sup_{\|y\|\leq \|h\|} \frac12 \| D^{n+1} f_k(x+y)\| \|h\|^2 $$ can be useful (this estimate follows by using a Taylor series and estimating the remainder).

supinf
  • 13,593