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For functions in a quasianalytic Denjoy-Carleman class we have the property that their Taylor expansions at a point (the origin) determines the function. For classes that don't only contain analytic functions these power series may not be convergent.

In principle, given the logarithmically convex sequence $\{M_n\}_{n=0}^{\infty}$ that determines the Denjoy-Carleman class and given the sequence of Taylor coefficients $\{f^{(n)(0)}\}_{n=0}^{\infty}$ it should be possible to determine the function in a neighborhood of the origin. This problem is not easy.

There are classical results that show, for example, that there exists a linear summation method that assigns a sum to the, possibly divergent, Taylor series at the origin which gives you the corresponding function. The same summation method working for all the functions in the given Denjoy-Carleman class. This means the existence of an infinite matrix $(w_{ij})$ such that for every sequence $\{f^{(n)}(0)\}_{n=0}^{\infty}$ of Taylor coefficients of a function $f\in C(\{M_n\}_{n=0}^{\infty})$, we have

$$f(x)=\lim_{i\rightarrow\infty}\sum_{j=0}^{i}w_{i,j}f^{(j)}(0)x^j.$$

The matrix $(w_{i,j})$ depends only on $\{M_n\}_{n=0}^{\infty}$ but it is not easy to compute from it.

There are also results (see this paper or the book by the same author on quasipower series and quasianalytic classes of functions) that show how to retrieve the function from its Taylor coefficients using these quasipower series.

Now, all these works, it seems to me, is done for functions of one variable.

Question: Are there results on representation of quasianalytic functions of several variables given their Taylor coefficients and the Denjoy-Carleman class? Are these results, for example the existence of a linear summation method that retriesves the function from its Taylor expansion, true in several variables?

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  • $\begingroup$ I'm not sure what exactly you are looking for here. If you restrict your function of several variables to the line connecting the origin with the point at which you want to know the value, you get a function of one variable in pretty much the same class and with all derivatives at the origin known, after which the usual recovery by an appropriate summation method can be made for that value. There is, of course, some trouble for non-convex natural domains but it is there even for the standard analytic continuation and is resolved the same way in both cases. $\endgroup$
    – fedja
    Jan 15, 2014 at 2:58

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