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

$\hat{f}(x):=\sum_{n=0}^\infty(-1)^n\frac{f^{(n)}(x)}{n!}x^n$ term by term we get $\frac{d}{dx}\hat{f}(x)=0$. Thus we may conclude that the series is convergent to the constant $f(0)$. But as it is well-known one can not be allowed to differentiate from a series term by term. Now the questions are:

i) Is the series convergent around origin?

ii) If it is convergent then is it convergent to the constant $f(0)$.

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.

c) 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

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If $f(x)=x$, then $\hat f(x)=-x\not=0=f(0)$. Is everything OK in the question? –  TaQ Jul 16 '13 at 9:30
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perhaps taking n = 0 in the lower limit would fix things? –  Aaron Hoffman Jul 16 '13 at 10:40
    
Sorry you are right. In fact we have $\sum_{n=0}^{\infty}(-1)^n\frac{f^{(n)}(x)}{n!}x^n$ –  E.Akrami Jul 17 '13 at 4:24

1 Answer 1

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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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! –  E.Akrami Jul 17 '13 at 7:30
    
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! –  E.Akrami Jul 17 '13 at 7:43
    
@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. –  Peter Michor Jul 17 '13 at 12:00

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