I asked this question on Math Stack Exchange some time ago and a similar question recently appeared regarding $L^1$ instead see here This has prompted me to bring it to this community in the hopes of getting an answer, partial answer, or even perhaps literature that addresses this question. Text from initial post below. Original post.

Suppose $f:\Bbb R\to\Bbb R$ is (real?) entire. In order for $f$ to be in $L^2(\Bbb R)$, clearly all terms in the power series cannot be positive since $f$ would diverge at $\pm\infty$. Likewise, the distribution of negative terms cannot go to zero so we see that the power series for $f$ must be alternating (in some fashion). However this does not tell us much.

Taking $f(x) = \sum\limits_{n=0}^{\infty} \dfrac{(-1)^n}{n!}x^{2n}$ gives $f(x) = \exp(-x^2)$ and is in $L^2(\Bbb R)$. In this case, the coefficients have factorial decay but it is not immediately obvious what kind of decay the coefficients can have while still giving rise to an $L^2$ function.

Are there sufficient conditions on the power series coefficients that will ensure that the function is in $L^2(\Bbb R)$? For instance, are there asymptotic bounds on the coefficients that will ensure that the function is in $L^2(\Bbb R)$ or is this an impossible task?

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    $\begingroup$ A distantly related problem is to characterize holomorphic functions on $\mathbb{C}$ which are square-integrable with respect to Gauss measure, in terms of their Taylor coefficients. This is the Segal-Bargmann transform. $\endgroup$ Jan 4, 2017 at 20:06
  • $\begingroup$ From $0$ to $\infty$ this is well studied. Look at Mellin's inversion theorem and Ramanujan's master theorem. If $F(x) = \sum_{n=0}^\infty \phi(n+1)\frac{w^n}{n!}$, for $\phi$ holomorphic such that $|\phi(z)|< Ce^{\tau|y| + \rho|x|}$ for $z = x + i y$ $x > 0$, $0 < \tau < \pi/2$ then $F(-x)x \to 0$ as $x\to \infty$. In fact you get something stronger, that $F(-w)w \to 0$ when $|\arg(w)| < \pi/2 - \tau$. $\endgroup$
    – user78249
    Jan 4, 2017 at 20:06
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    $\begingroup$ In light of @booksee's comment that a power series need not converge on the whole real line unless the function is actually entire... do you mean to ask about entire functions, one way or the other? Even then, I think this sort of issue is hard to understand... perhaps partly because polynomials themselves are never in $L^2(\mathbb R)$, but/and the question would ask which infinite sums/limits of polynomials are in $L^2(\mathbb R)$. $\endgroup$ Jan 4, 2017 at 21:12
  • $\begingroup$ @paulgarrett I definitely meant entire functions. I mistakenly assumed that was understood. My apologies. Yeah it is a very difficult problem I gather because typical arguments do not work. I considered a Laplace transform approach (i.e. a regularization) and Fourier transform approaches but it was only made more opaque. $\endgroup$ Jan 5, 2017 at 1:37
  • $\begingroup$ Compare this question mathoverflow.net/questions/257647/… Both questions are hopeless. $\endgroup$ Jan 5, 2017 at 6:09

1 Answer 1

  1. Since you ask about "real analytic functions", should it be $f:\mathbb{R}\to \mathbb{R}$ instead of $f:\mathbb{R}\to \mathbb{C}$?
  2. I think analyticity is a "local" concept according to its definition.
  3. Because of 2, $f$ may not have a global power series expansion. For example, $1/(1+x^2)$ is real analytic over $\mathbb{R}$, but the radius of convergence for its power series expansion at $0$ is $1$.
  4. Regardless of 1-3, even assuming you were talking about real analytic functions with a global power series expansion, we have $$\cos x = \sum_0^\infty (-1)^n\dfrac{x^{2n}}{(2n)!},$$ where the absolute values of coefficients decay much faster than the example you gave. However, $\cos x \notin L^1(\mathbb{R})\cup L^2(\mathbb{R})$.
  5. Actually, intuitively, the faster it decays, the less possible it will be in $L^2(\mathbb{R})$. Consider $f(x)=x$, where all coefficients vanish for terms $x^n (n > 1)$. Of course, $f$ is real analytic with a globally convergent power series expansion but $f\notin L^p(\mathbb{R})$ for any $0<p\leq \infty$.
  • $\begingroup$ Hi booksee. I definitely meant for this to be geared towards entire functions. I mistakenly assumed that was clear. My apologies. $\endgroup$ Jan 5, 2017 at 1:39
  • $\begingroup$ Stumbled back into this after doing some thinking on the Bargmann transform (or a generalization thereof in my own research). It's as close to an accepted answer as I think I'll end up getting. $\endgroup$ Jul 23, 2019 at 20:48

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