On the convergence of the function series $\sum_{n=0}^\infty(-1)^n\frac{f^{(n)}(x)}{n!}x^n$

Question

Let $f$ be a smooth real function defined around origin. If we differentiate term by term the series

$\hat{f}(x):=\sum_{n=0}^\infty(-1)^n\frac{f^{(n)}(x)}{n!}x^n$, we get $\frac{d}{dx}\hat{f}(x)=0$. \begin{eqnarray}\frac{d}{dt}\hat{f}(t)&=& \sum_{n=0}^{\infty}(-1)^n\frac{f^{(n+1)}(t)}{n!}t^n+ \sum_{n=1}^{\infty}(-1)^n\frac{f^{(n)}(t)}{(n-1)!}t^{n-1}\nonumber\\&=& \sum_{n=0}^{\infty}(-1)^n\frac{f^{(n+1)}(t)}{n!}t^n-\sum_{n=0}^{\infty}(-1)^n\frac{f^{(n+1)}(t)}{n!}t^n \nonumber\\&=&0\nonumber \end{eqnarray} Thus $\hat{f}(t)$ should be constant. But in fact we are not allowed to differentiate term by term from a series.

Next, suppose that $f$ is a smooth periodic function which by the Fourier analysis we know that it has Fourier expansion. That is we suppose $f(t)=\sum_{m=-\infty}^\infty c_me^{im\omega t}.$ Then it is well known that we can differentiate to get $f^{(n)}(t)=\sum(im\omega)^n c_me^{im\omega t}.$ Thus \begin{eqnarray}\hat{f}(t)&=&\sum_n\sum_m\frac{(-1)^n(im\omega)^{n}}{n!}c_me^{im\omega t}t^n\nonumber\\&=& \sum_m\sum_n(\frac{(-1)^n(im\omega)^{n}}{n!}t^n)c_me^{im\omega t}\nonumber\\&=& \sum_me^{-im\omega t}c_me^{im\omega t}\nonumber\\&=&\sum_{m=-\infty}^{\infty}c_m\nonumber\\&=&f(0).\nonumber \end{eqnarray} Thus again it seems that the series should convergence to a constant. But in the above we have exchanged the order of two infinite sums which are not allowed.

The function $$f(t):=\sum_{m=0}^{\infty}\frac{1}{m!}e^{i2^mt}$$is smooth nowhere analytic, in the sense that convergence radius of the Taylor's series of $f$ at each point is zero and therefore $\hat{f}(t)$ diverges for all $t\ne0$.

The function $$f(t):=\sum_{m=1}^{\infty}\frac{1}{m!}e^{i2^{-m}t}$$is analytic at $t=0$ whose convergence radius is infinity. Thus $\hat{f}(t)$ converges for all $t$ to $f(0)$.

In fact one can show that

a) If $f$ is analytic at origin then the series $\hat{f}$ is convergent uniformly to the constant $f(0)$.

b)If $f$ is nowhere analytic in the sense that the radius of convergence of the Taylor's series is zero then of course the series is divergent. But if $f$ is nowhere analytic in the sense that the radius of convergence of the Taylor's series is positive but the Taylor's series does not converge to the function $f$ then the series $\hat{f}$ may converge.

c) About the function $f(x):=e^{\frac{-1}{x^2}},f(0)=0$ one can show that if the series is convergent then its sum is constant.

d) There are nowhere analytic functions such that the series is convergent in a dense subset to the constant $f(0)$ and there are nowhere analytic functions such that the series is divergent everywhere.

Now the main questions are.

1. Is there a smooth function $f$ which is not analytic at origin and the series $\hat{f}$ is convergent in an interval around origin and the sum is the constant $f(0).$?

2. Is there a smooth function $f$ which is not analytic at origin and the series $\hat{f}$ is convergent in an interval around origin and the sum is not constant.?

3. If we define a linear differential operator of infinite order $f\mapsto \hat{f}-f(0)$. Then in above we said that analytic functions at origin are contained in the space of eigenfunctions of the zero eigenvalue of this operator. Now the question arises that: are there nonzero eigenvalues for this operator?

4. For the function $f(x):=e^{\frac{-1}{x^2}},f(0)=0$, is the series $\hat{f}$ convergent? Please see the preprint arXiv:1105.2611v2 [math.GM] 5 Jun 2011 and the paper: Journal of Applied Analysis, Volume 25, Issue 2, Pages 131–139, DOI: https://doi.org/10.1515/jaa-2019-0014.

Answer 1

Check the definition of quasianalytic functions and Denjoy-Carleman ultradifferentiable functions - the results in this field should be helpful for you.