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Suppose that f is an analytic function on the open unit disk and $\lvert f(z) \rvert \geq \sqrt{\lvert z \rvert}$ for all z on the unit disk. Show that $\lvert f(z) \rvert \geq 1$ on the unit disk.

We can write it down as

$\frac{\vert z \rvert}{(\lvert f(z) \vert)^2} \leq 1$,

provided that $f(0) \not= 0$. We know that f cannot be zero for other values since otherwise it would contradict the inequality in the supposition. The above function is holomorphic and is equal to zero at zero, again provided that f is not zero at zero. Thus, we can apply the Schwarz lemma and the result follows. Now, if f is zero is at zero then this argument does not work and I do not know how to handle that case separately, if there is any way. Thanks for your help.

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3 Answers 3

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It becomes a bit simpler if you consider the function $h(z) = z/f(z)$ instead of $z/f(z)^2$. $h$ is holomorphic in the punctured unit disk $\Bbb D \setminus \{ 0 \}$ with $$ |h(z)| \le \sqrt{|z|} \le 1 \, . $$ It follows that $h$ has a removable singularity at the origin, and the extended function $h:\Bbb D \to \overline {\Bbb D}$ satisfies $h(0) = 0$.

So by the Schwarz Lemma, $|h(z)| \le |z|$, and that implies $|f(z)| \ge 1$.

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If $f(0)=0$ then $f(z)=zg(z)$ with $g$ analytic at $0$. But then $|z||g(z)|\geq \sqrt {|z|}$ which gives $\sqrt {|z|}|g(z)|\geq 1$. Letting $z \to 0$ we get a contradiction. Hence, $f(0)\neq 0$.

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The square root is suggestive of using Parseval. For any $r\in (0,1)$
$ r = \frac{1}{2\pi}\int_0^{2\pi}r d\theta\leq\frac{1}{2\pi}\int_0^{2\pi}\big \vert f(r e^{i\theta})\big \vert^2 d\theta=\sum_{k=0}^\infty \vert a_k\vert r^{2k}$.
Supposing $a_0 =0$: $\implies 1 \leq\sum_{k=0}^\infty \vert a_k\vert r^{2k-1}=\sum_{k=1}^\infty \vert a_k\vert r^{2k-1} \leq\sum_{k=1}^\infty \vert a_k\vert r^{k}$
but the RHS may be made arbitrarily small, which is impossible.


For a different finish, observe that for any $r\in (0,1)$ then $f$ is non-zero on the closed disc $\overline B(0,r)$, so the minimum modulus theorem tells us that the minimum occurs on the boundary i.e. for some $w$ where $\vert w\vert =r$ and $\vert f(w)\vert \geq \sqrt{\vert w \vert}=\sqrt{r}$ and the lower bound may be made arbitrarily close to $1$, hence $\vert f(z)\vert \geq 1$ for all $z\in B(0,1)$.

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