# When is an analytic function in $L^2(\Bbb R)$?

Suppose $f:\Bbb R\to\Bbb C$ is real analytic. 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?

• Unless I am mistaken, your series for arctan does not converge for $x>1$. – Jonas Dahlbæk Sep 2 '14 at 16:36
• You're right. Thanks for that. – Cameron Williams Sep 2 '14 at 16:38
• Did you ever get to understand this question? – Boby Jan 4 '17 at 13:53
• I haven't really had any success with it. Seems to be a very difficult question. – Cameron Williams Jan 4 '17 at 15:36
• MathOverflow: When is an analytic function in $L^2(\Bbb R)$? – Martin Sleziak Jan 5 '17 at 1:57

## 1 Answer

Assume $f(x)=\sum _{n=0}^\infty a_n x^n$. Put $c_n=\sum_{m=0}^n a_m a_{n-m}$, so that $f(x)^2=\sum _{n=0}^\infty c_n x^n$. Then, invoking first the Monotone Convergence Theorem, then the Tonelli/Fubini Theorem and then removing vanishing terms, we see that $$\int f^2 dx=\lim_{k\rightarrow\infty} \int_{-k}^k f^2 dx=\lim_{k\rightarrow \infty} \sum_{n=0}^\infty \frac{c_n(k^{n+1}-(-k)^{n+1})}{n+1}=\lim_{k\rightarrow \infty} \sum_{n=0}^\infty \frac{2c_{2n}}{2n+1}k^{2n+1}.$$ Thus, $f\in L^2$ if and only if the increasing function $g:[0,\infty)\rightarrow [0,\infty)$ defined by $$g(k)=\int_{-k}^k f^2 dx=\sum_{n=0}^\infty \frac{2c_{2n}}{2n+1}k^{2n+1}$$ satisfies $$g(k)\leq M$$ for some $M<\infty$.

• Can we somehow convert the last expression into an asymptotics-based condition on $c_{2n}$? It's not immediately obvious to me as to how to do that. – Cameron Williams Sep 2 '14 at 15:33
• I doubt it. As you point out in the statement of the question, some sort of cancellation has to take place. Maybe someone more knowledgeable than me will come along if you let the question hover for a few days. Make sure to update if you make any progress on your own, I would be interested in knowing the result! – Jonas Dahlbæk Sep 2 '14 at 15:47
• Thanks a bunch. I considered a Fubini type argument as you did but I didn't see it coming together nicely so I didn't pursue it. At least it comes out to a nice expression. – Cameron Williams Sep 2 '14 at 15:52