TITLE: Improving Cauchy estimates?
QUESTION [7 upvotes]: Consider an entire function $f:\mathbb C \to \mathbb C$ that is real on the real line and even.
This function has a Taylor series of the form
$$f(z) = \sum_{i=0}^{\infty} a_i z^{2i} \text{ with } a_i \in \mathbb R.$$
I have an estimate of the form
$$\vert e^{z^2 \mu} f(z) \vert \le e^{\vert z \vert^4 \nu}$$
for all $z \in \mathbb C$ and $\mu, \nu >0.$
I could now bound the Taylor coefficients of this power series using Cauchy estimates, see for instance here, by using that
$$\vert f(z) \vert \le e^{\vert z \vert^4\nu + \vert z \vert^2 \mu}.$$
However, I feel that I am wasting something here, as the initial estimate already implies that on the real line $f$ grows at most like $e^{z^4\nu}$ when multiplied by a Gaussian.
So the bound I derived for my Cauchy estimate is very conservative on the real line.
Of course I cannot immediately improve upon this estimate since when $z$ is purely imaginary, that bound may be optimal and Cauchy estimates are supposed to hold on a circle in the complex domain. However, since the function is real on the real line, the growth should really be determined by my initial estimate somehow.
Therefore my question is: Can I do anything better here to bound the Taylor coefficients than applying Cauchy estimates to the estimate
$$\vert f(z) \vert \le e^{\vert z \vert^4\nu + \vert z \vert^2 \mu}.$$

REPLY [8 votes]: Let $z=re^{i\theta}$. Then your original inequality implies
$$|f(re^{i\theta})|\leq \exp(\nu r^4-\mu r^2\cos2\theta).$$
Then instead of Cauchy estimate you can use the exact formula for the coefficient:
$$|a_{2n}|=\left|\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z^{2n+1}}dz\right|\leq\frac{1}{2\pi}
\int_0^{2\pi}|f(re^{i\theta})|r^{-2n}d\theta$$
$$\leq r^{-2n}\exp(\nu r^4)\frac{1}{2\pi}\int_0^{2\pi}e^{-\mu r^2\cos2\theta}d\theta,$$
and now take the minimum,
with respect to $r$. (The last integral can be expressed in terms
of the modified Bessel function $I_0$, if needed.)
However, this can give you an advantage only for small $n$, since for large $n$, the term with $r^2$ in the first estimate is negligible.