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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.

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    $\begingroup$ If $f(x)=x$, then $\hat f(x)=-x\not=0=f(0)$. Is everything OK in the question? $\endgroup$
    – TaQ
    Jul 16, 2013 at 9:30
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    $\begingroup$ perhaps taking n = 0 in the lower limit would fix things? $\endgroup$ Jul 16, 2013 at 10:40
  • $\begingroup$ Sorry you are right. In fact we have $\sum_{n=0}^{\infty}(-1)^n\frac{f^{(n)}(x)}{n!}x^n$ $\endgroup$
    – E.Akrami
    Jul 17, 2013 at 4:24
  • $\begingroup$ This is series is the value of the (Taylor series of $f(t)$ in point $t_0=x$) at point $t=0$. So you may use the bounds for the remainder for Taylor series to estimate the differences between $f(0)$ and the partial sum of your series. $\endgroup$ Jun 21, 2020 at 8:52
  • $\begingroup$ @Fedor Taylor series of $f$ may converge not to $f$ itself but to a function depending to $x$. Thus $\hat{f}(x)$ may converge to a sum depending to $x$. However, I do not know a function whose hat converges to a non-constant function! $\endgroup$
    – E.Akrami
    Jun 21, 2020 at 10:47

1 Answer 1

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

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  • $\begingroup$ Thank you Peter. I checked quasianalytic functions and Denjoy-Carleman ultradifferentiable functions but unfortunately I could not find anything new in relation with my question! $\endgroup$
    – E.Akrami
    Jul 17, 2013 at 7:30
  • $\begingroup$ The only thing which might be useful is that if $f$ satisfies $|f^{(n)}(x)|\le K^nM_n$ where $M_n$ is a sequence of positive numbers and $K$ is positive and if $L:=K\limsup_n\sqrt[n]{\frac{M_n}{n!}}<\infty$ then for $|x|<L^{-1}$ we have $\hat{f}(x)$ is convergent, see arXiv:1105.2611v2 [math.GM] 5 Jun 2011, Theorem 6. But the condition $L<\infty$ implies $M_n<A^nn!$ for some constant $A$ and then this implies that $f$ is analytic! $\endgroup$
    – E.Akrami
    Jul 17, 2013 at 7:43
  • $\begingroup$ @E.Akrami: I think that any function such that your series converges on an interval absolutely, has to be analytic. The reason is the description of quasianalytic functions. $\endgroup$ Jul 17, 2013 at 12:00

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