I am looking for an answer in the affirmative or the negative concerning the asymptotics of an entire function. The question, relatively simple to state, has proven very taxing to solve and requires an argument I am at a loss for. I have it stated in a more general case, and have reduced the general case into a single case.

Suppose $f(w)$ is an entire function, such that $a_n \to A$ as $n \to \infty$ and

$$f(w) = \sum_{n=0}^\infty a_n \frac{(-w)^n}{n!}$$

We are given the hypothesis that $f(w)w \to 0$ as $|w| \to \infty$ when $|\arg(w)|< \pi/2$.

Must it be that $|f(ix)| < C$ for all $x \in \mathbb{R}^+$?

The prototypical case is the exponential function $e^{-w}$ which satisfies this, and finite sums of exponentials $e^{-\lambda w}$ equally so for $0 < \lambda \le 1$. It proves to not work when we use $we^{-w}$ or $p(w)e^{-w}$ for some polynomial $p$, however in such cases the derivatives about zero are no longer bounded and do not tend toward a constant; despite the fact it satisfies the asymptotic condition.

The motivation as to why this result appears to be true can be observed by the fact

$$f^{(n)}(w) \to Ae^{-w}\,\,\,\text{as}\,\,\,n\to\infty$$ uniformly on compact subsets of $\mathbb{C}$. This would seem to imply that since $f^{(n)}(ix) \approx Ae^{ix}$ for large $n$ it should teeter off into a bounded function for $x \in \mathbb{R}^+$.

We are also given the stronger form of convergence, for $|\theta| < \pi/2$ and $\tau > 0$

$$\int_1^\infty |f^{(n)}(xe^{i\theta}) - A e^{-xe^{i\theta}}|x^{-\tau}\,dx \to 0\,\,\,\text{as}\,\,\,n\to\infty$$

Essentially the boundedness of $f$ would allow us to say for $\tau > 1$ that

$$\int_1^\infty |f(ix)|x^{-\tau}\,dx <\infty$$

and even stronger

$$\int_1^\infty |f^{(n)}(ix) - A e^{-ix}|x^{-\tau}\,dx \to 0\,\,\,\text{as}\,\,\,n\to\infty$$

which is more along the lines of the required result I am searching for. Any help or suggestions about how to show this would be greatly appreciated.

  • $\begingroup$ A simple counterexample is $f(z)=Ae^{-z}+ze^{-z/2}$ (I originally wrote a too complicated answer, now deleted, to find this). $\endgroup$ – Christian Remling Apr 7 '16 at 1:34
  • $\begingroup$ ahh you're right, I think I need to strengthen my conditions, thanks so much. $\endgroup$ – user78249 Apr 7 '16 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy