I'm looking at an entire function of the form $$ f(\lambda):=p(\lambda)e^{-\lambda}+q(\lambda)\;, $$ where $p$ and $q$ are polynomials and $\lambda\in\mathbb{C}$. I need to establish that $f$ is an entire function of finite exponential type, and to determine its order. This would be helpful in what I'm planning on doing down the road. Any ideas are welcome.

This is what I have done so far. Begin by noting that $$ e^{-\lambda}=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n\lambda^n}{n!}\;. $$ Thus, $$ \begin{array}{lcl} |f(\lambda)| &\leq& \displaystyle\sum_{n=0}^{\infty}\left|\frac{\lambda^n p(\lambda)}{n!}\right|+|q(\lambda)|\\ &\leq& e^{|\lambda|+\ln|p(\lambda)|}+e^{\ln|q(\lambda)|}\;. \end{array} $$ We say that $f$ is of order $\rho$ if $|f(z)|\leq ce^{{|z|}^{\rho}}$ for $|z|$ large enough, so that $|f(z)|e^{-{|z|}^{\rho}}$ is bounded.

  • $\begingroup$ To find the order, try to prove that for each $\rho > 1$, $|f(\lambda)| e^{-|\lambda|^\rho} \to 0$ as $|\lambda| \to \infty$. Then show that $|f(\lambda)|e^{-|\lambda|}$ does not necessarily tend to $0$ as $|\lambda| \to \infty$. $\endgroup$ – Antonio Vargas Jun 23 '15 at 1:25
  • $\begingroup$ Suppose on the contrary that $\rho\leq 1$. Then, in the limit as $|\lambda|\rightarrow\infty$, we have that $|f(\lambda)|\rightarrow\infty$ significantly faster than $e^{{|\lambda|}^{\rho}}$. This means that $|f(\lambda)|e^{-{|\lambda|}^{\rho}}\rightarrow\infty$, which is a contradiction as it is supposed to be bounded above by $c$. Hence, $\rho>1$. Ok, but does this establish that the order is finite? $\endgroup$ – cantor_paradise Jun 23 '15 at 4:54

The contribution of $q$ is negligible: $|q(z)|\exp(-|z|^\rho)$ tends to zero as $|z|\to\infty$, for every $\rho>0$.

If $p$ happens to be a constant polynomial, then $f$ has order $1$, since $|e^{-z}|e^{-|z|}=e^{-\operatorname{Re}z -|z|}\le 1$ for all $z$.

Otherwise, with your definition $f$ is not of order $1$, but it is of every order $1+\epsilon$. Just combine the two facts above to get $$ |p(z)|e^{-\epsilon |z|}\;|e^{-z}|e^{-|z|} \to 0,\qquad |z|\to\infty$$

However, I observe that the definition of order that you gave disagrees with Wikipedia (and other sources familiar to me), which define $$ \rho = \limsup_{r\to\infty} \frac{\ln\ln M(r)}{\ln r},\qquad M(r) = \max_{|z|\le r} |f(z)| $$ In your case $\ln M(r) =r+ O(\ln r) $, hence $\frac{\ln\ln M(r)}{\ln r} \to 1$ as $r\to\infty$. This means $f$ is of order $1$.

Same estimate for $\ln M(r)$ also gives exponential type $\sigma = 1$.

  • $\begingroup$ You are correct about the definition of order and type. Indeed, Levin defines the order $\rho$ of an entire function the way you have it above. However, he defines your function $M$ as $$\displaystyle M_f(r)=\displaystyle{\mbox{max}}_{|z|=r}|f(z)|\;.$$ Is there a reason why you have $|z|\leq r$ instead? My usage of the notion of order above was based on some comments made by AntonioVargas elsewhere on these boards. But now I'm not so sure after having a look at the work of B. Ya. Levin. Could it be that these two formulations of order are equivalent? $\endgroup$ – cantor_paradise Jun 27 '15 at 0:11
  • $\begingroup$ By the maximum principle, it makes no difference whether to write $|z|\le r$ or $|z|=r$. No, two notions of order are not equivalent which is what my post says: $ze^{z}$ has order $1$ by one definition but not by the other. You may have misunderstood the comment. $\endgroup$ – user147263 Jun 27 '15 at 0:13
  • $\begingroup$ You may find the comment here math.stackexchange.com/questions/274704/… $\endgroup$ – cantor_paradise Jun 27 '15 at 0:20
  • $\begingroup$ I left a comment for Antonio there. Keep in mind that comments are like quick notes on a napkin, not meant to be used for references. In particular, one cannot edit a comment even if one realizes a slight mistake later. $\endgroup$ – user147263 Jun 27 '15 at 0:28

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