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Let $f$ be an entire function and let $A >0$, $B >0$ be constants such that $|f(z)| \leq A \exp(|z|^B )$ $\forall z \in \mathbb{C}$.

If $f(\log n)=0$ $\forall n>2$, then show that $f$ is constant.

I could prove this result if $B<1$ using the order of growth theorem and the fact that the harmonic series is divergent, but for the other cases of $B$, I am not able to give a proof. Thanks for any help.

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Assume to get a contradiction that $f$ is not the zero function. Then the set of zeros of $f$ is at most denumerable. By hypothesis, the set of zero of $f$ is infinite, so actually it can be ordered in a sequence, say $(z_n)_{n\in\mathbb{N}}$, such that $|z_1|\le|z_2|\le|z_3|\le...$ and, by the identity theorem, it also follows that $|z_n|\rightarrow+\infty, n\rightarrow \infty$. Define $$N:[0,+\infty)\rightarrow\mathbb{N},r\mapsto\ \sum_{n=0}^{\infty} \chi_{[0,r]}(|z_n|).$$ WLOG we can assume $f(0)=1$ (otherwise we can just divide $f$ by $f^{(n)}(0)z^n$ where $n\in\mathbb{N}$ is such that $$f(0)=0, f^{(1)}(0)=0,...,f^{(n-1)}(0)=0,f^{(n)}(0)\neq0,$$ and then notice that the function $g$ so obtained satisfies the same hypothesis, perhaps with a different $A$, and it is such that $g(0)=1$).

Fix $\delta>0$. By Jensen's formula: $$\forall r>0, \exp\left(\int_{-\pi}^\pi\log\left|f\left((1+\delta)re^{it}\right)\right|\frac{\operatorname{dt}}{2\pi}\right)\\=\prod_{n=1}^{N\left((1+\delta)r\right)} \frac{(1+\delta)r}{|z_n|}\ge\prod_{n=1}^{N\left(r\right)} \frac{(1+\delta)r}{|z_n|}\ge\prod_{n=1}^{N\left(r\right)} (1+\delta)=(1+\delta)^{N(r)},$$ so: $$\forall r>0, N(r)\le \frac{1}{\log(1+\delta)}\int_{-\pi}^\pi\log\left|f\left((1+\delta)re^{it}\right)\right|\frac{\operatorname{dt}}{2\pi}\le \frac{\log(A)+(1+\delta)^Br^B}{\log(1+\delta)}.$$ Then, if $R\ge2$ and $M_R\in\mathbb{N}$ is the largest integer such that $M_R\le R$, we have: $$R-2\le M_R-1\le N(\log(M_R))\le N(\log(R))\le \frac{\log(A)+(1+\delta)^B\left(\log(R)\right)^B}{\log(1+\delta)},$$ but for $R$ large enough this inequality cannot hold, so we have got a contradiction.

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