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](https://math.stackexchange.com/questions/114349/how-is-cauchys-estimate-derived/114363), 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  

$$\vert f(z) \vert \le e^{\vert z \vert^4\nu + \vert z \vert^2 \mu}.$$



Improving Cauchy estimates?
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.


