TITLE: On the convergence of the function series $\sum_{n=0}^\infty(-1)^n\frac{f^{(n)}(x)}{n!}x^n$
QUESTION [8 upvotes]: 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.

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 $f(0).$?

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.?

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?

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.

REPLY [4 votes]: Check the definition of quasianalytic functions and Denjoy-Carleman ultradifferentiable functions - the results in this field should be helpful for you.