A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is called $\mathbb{R}$-analytic iff for every $x_0 \in \mathbb{R} $ there exist $R>0$ and power series $\sum_{n=0}^\infty a_n (x-x_0)^n$ convergent for $|x-x_0|<R$ and such that $f(x)=\sum_{n=0}^\infty a_n (x-x_0)^n$ for $|x-x_0|<R$.

For some strange continuous functions on $\mathbb{R}$, for example for Weierstrass continuous nondifferentiable function (i.e. $f(x)=\sum_{n=0}^\infty a^n \cos(b^n \pi x)$ for $x\in \mathbb{R}$, where $0<a<1$, $b$ is positive odd integer such that $ab> 1+\frac{3}{2}\pi$), there exist a sequence of $\mathbb{R}$-analytic functions (even entire functions) which converges to $f$ uniformly (in the case of Weierstrass function it is sufficient to take the sequence of partial sum of series defining this function).

Is it maybe true that for every continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ there exists a sequence of $\mathbb{R}$ analytic functions which converges uniformly to $f$ ?

  • $\begingroup$ Did you try for functions like $f(x)=\begin{cases}e^{-\frac 1{x^2}}&,x\neq 0\\\ 0&,x=0\end{cases}$? $\endgroup$ – Davide Giraudo Feb 11 '12 at 14:58
  • $\begingroup$ I believe this is true, yes. $\endgroup$ – Henno Brandsma Feb 11 '12 at 15:02
  • $\begingroup$ @Davide: How about $f_n(x)=\exp(-\frac{1}{x^2+1/n})$? $\endgroup$ – hmakholm left over Monica Feb 11 '12 at 15:45
  • $\begingroup$ I'm not certain but I think a positive answer to the question is implied by Lemma 6 of "Analytic extensions of differentiable functions defined in closed sets", H Whitney, Trans. Amer. Math. Soc, 1934. $\endgroup$ – Colin McQuillan Feb 11 '12 at 15:49
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    $\begingroup$ @Davide: you are mixing "approximable by polynomials" with "approximable by Taylor polynomials". As I learned not that many years ago, it is often possible to find polynomials that approximate your function better than the Taylor polynomials. $\endgroup$ – Martin Argerami Feb 11 '12 at 18:24

The proof of the assertion seems to be contained in the paper

Carleman, T., Sur un théorème de Weierstrass, Ark. Mat., Ast. Fysik 20B (1927), 1-5

a PDF of which is available (in French) from this page. In fact, he proves a bit more. A plain explanation can be found in other papers (e.g., here; I would post more links if I had more reputation). What I gather that he proved is, given a continuous function $f$ and a continuous notion of acceptable "error" (with error $\epsilon$ coming from choosing the constant function $\epsilon$), one can find an entire function approximating to the given precision.

A nice proof is reported to be contained in Lectures on Complex Approximation Theory by D. Gaier, Birkhaussen 1987 on page 49.

  • $\begingroup$ Dear Clinton: Thanks for this great answer!!! I was wondering if you could add to your answer the links you would have given if you had more reputation. I suppose you can give them in plain text mode, can't you? (I upvoted you answer.) $\endgroup$ – Pierre-Yves Gaillard Feb 11 '12 at 19:13
  • $\begingroup$ Unfortunately I closed the search. However, simply searching the web for the title turns up many results, including articles in English which feel forced to summarize the content for their English-speaking readers :) However, I will point out that there is a review in German linked from the same page that contains the PDF, and Google Translate seems to do well on it. Also, the review is in German, but the PDF is in French; much more readable to me at least! $\endgroup$ – Clinton Curry Feb 11 '12 at 20:52
  • $\begingroup$ Skimming the Carleman paper, I don't immediately see how to get the "better than uniform" approximation. Carleman only seems to discuss uniform approximation. I don't see how Boghossian extracts the claimed statement. $\endgroup$ – Nate Eldredge Dec 11 '17 at 15:07

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