I'm working on old qualifying exam problems and this one came up.

Suppose $f$ is an entire function on $\mathbb{C}$, and that $|f(\frac{1}{n})| \leq n^{-n}$ for all $n\in \mathbb{Z}_{>0}$. Prove that $f$ is constant.

My approaches: One obvious thing to do is to try and show that $f$ is bounded and appeal to Liouville's theorem. However, our bounds are for small values of $z$, so this did not yield much success.

One can see that $f(0)=0$ from this condition. Also, $$ f'(0) = \lim_{n\to \infty} \frac{f(\frac{1}{n}) - f(0)}{\frac{1}{n}} = \lim_{n\to \infty} n f\left(\frac{1}{n}\right). $$ Note that $$ \left|nf\left(\frac{1}{n}\right)\right| \leq n^{-n+1} \to 0 \text{ as } n\to \infty. $$ Thus, $f'(0)=0$.

I would like to keep trying to show that $f^{(n)}(0)=0$ and use the fact that $f$ is entire to get that $f=0$ identically on $\mathbb{C}$. But I get stuck because I don't have an estimate for $f'(1/n)$.

I think what makes this problem harder is that we only have an estimate on $f$ at a countable set, instead of the usual problem where we know some kind of bound on $f$ everywhere.

Any hints are appreciated.

  • $\begingroup$ Try to use the maximum principle on the unit disk and then use the identity principle. $\endgroup$ – Nitrogen Sep 15 '15 at 18:58

If $f$ is not identically zero then $f$ has a zero of multiplicity $k \ge 0$ at $z=0$, i.e. $$ f(z) = z^k g(z) $$ where $g$ is an entire function with $g(0) \ne 0$.

Then $$ \bigl|g(\frac 1n)\bigr| = n^k \, \bigl|f(\frac 1n)\bigr| \le \frac{n^k}{n^n} $$ for all positive integers $n$. The right-hand side tends to zero for $n \to \infty$ and it follows that $g(0) = 0$, which is a contradiction.

  • $\begingroup$ I like this approach, but shouldn't it read $n^k/n^n $ in the last inequality? $\endgroup$ – Andrew Sep 15 '15 at 19:26
  • $\begingroup$ @Andrew: You are right of course, that was a stupid error. Thanks! $\endgroup$ – Martin R Sep 15 '15 at 19:28
  • $\begingroup$ @MartinR,How do you know that $f$ has a zero at z=0???Please explain it. $\endgroup$ – Suraj_Singh Sep 16 '15 at 14:20
  • $\begingroup$ @Kilimanjaro: It follows from the given estimate $|f(\frac{1}{n})| \leq n^{-n}$. But actually you don't need that, $k=0$ is an allowed value in the proof. $\endgroup$ – Martin R Sep 16 '15 at 14:21
  • $\begingroup$ @MartinR,but why?I mean we could have a zero at say,$z=1/2$ or so. $\endgroup$ – Suraj_Singh Sep 16 '15 at 14:23

Since $f$ is entire, we know $$f(z) = \sum_{j=0}^\infty a_j z^j \quad \forall z \in \mathbb{C}$$ Using the bound given, we see $$ \left |f\left (\frac{1}{n} \right) \right | = \left | \sum_{j=0}^\infty \frac{a_j}{n^j} \right | \leq \frac{1}{n^n}$$ You've already seen that $a_0=0$ by taking $n\to \infty$, now pull out a factor of $n$ to see $$\left |f\left (\frac{1}{n} \right) \right | = \frac{1}{n}\left | \sum_{j=1}^\infty \frac{a_j}{n^{j-1}} \right | \leq \frac{1}{n^n}\implies \left|n f\left ( \frac{1}{n} \right) \right | =\left | \sum_{j=1}^\infty \frac{a_j}{n^{j-1}} \right | \leq \frac{1}{n^{n-1}}$$ i.e. we have that $a_i =0$ for $i \in [0,n]$. Proceed with induction to conclude such a function is identically zero.

Remark to what you said: You don't need to have the estimate on $f'(1/n)$ since the function is entire and your representation holds for all $z \in \mathbb{C}$.

  • $\begingroup$ I don't see how you arrive at the conclusion that $a_i=0$ for $i\in [0,n)$. $\endgroup$ – uniquesolution Sep 15 '15 at 21:49
  • $\begingroup$ you iterate the bound and take the limit, i.e.$$|n^2 f(1/n) | \leq \frac{1}{n^{n-2}} \implies a_2 \leq 0 \ldots| n^{n-1}f( 1/ n) | \leq \frac{1}{n} \implies a_{n-1} = 0 $$ $\endgroup$ – Jeb Sep 16 '15 at 11:53

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