# Lemma 10.8 Rudin functional analysis, proof.

There's the following theorem I'm trying to understand (Lemma 10.8 Rudin, Functional Analysis).

Let $f$ be an entire function on $\mathbb{C}$ such that $f(0) = 1, f'(0) = 0, 0 < \lvert f(\lambda) \rvert \leq e^{\lvert \lambda \rvert}$ for all $\lambda \in \mathbb{C}$. Then $f(\lambda) = 1$ for all $\lambda \in \mathbb{C}$.

The maximum modulus theorem is used (Theorem 10.24 Rudin, Real and Complex analysis).

Suppose $\Omega$ is a region, $f \in H(\Omega)$, and $\overline{D}(a;r) \subset \Omega$. Then $$\lvert f(a) \rvert \leq \max_{\theta} \lvert f(a + re^{i\theta}) \rvert$$

For the lemma 10.8 the proof goes as follows.

Because $f$ doesn't have zeroes then there's a function $g$ such that $f = \exp\left\{ g \right\}$, $g$ entire in $\mathbb{C}$ and $Re[g(\lambda)] \leq |\lambda|$ with $\lambda \in \mathbb{C}$, $g(0) = g'(0) = 0$. Let $r \geq |\lambda|$, we can verify that

$$\lvert g(\lambda) \rvert \leq \lvert 2r - g(\lambda) \rvert$$

$$h_{r}(\lambda) = \frac{r^2g(\lambda)}{\lambda^2(2r - g(\lambda))}$$

such a function is holomorphic in $\left\{\lambda : |\lambda| < 2r \right\}$. Clearly $\lvert h_r(\lambda) \rvert \leq 1$ if $\vert \lambda \rvert = r$. By the maximum modulus theorem we have for $\lvert \lambda \rvert \leq r$ $$\lvert h_r(\lambda) \rvert \leq 1.$$

If we fix $\lambda$ and let $r \to \infty$, we have $g(\lambda) = 0$.

My main question is how the maximum modulus theorem is actually used in this context?

My function $h_r$ is holomorphic in $\Omega = D(0;2r)$ and also I have $\overline{D}(0,r) \subset D(0;2r)$, this should imply

$$\lvert h_r(0) \rvert \leq \max_{\lambda \in \partial \overline{D}(0,r)}\lvert h_r(\lambda) \rvert \leq 1$$

where with $\partial A$ I denote the boundary set of a set $A$. Why does this imply $\lvert h_r(\lambda) \rvert \leq 1$ if $| \lambda | \leq r$?. Also why if fix $\lambda$ and let $r$ diverging implies $g(\lambda) = 0$?

• Suppose $|h_r(\lambda)|=M>1$ for some $|\lambda|<r$, and suppose this is a maximum value for $h_r$ [which exists because the closed ball is compact]. Then in a small enough neighborhood of radius $\epsilon$ of $\lambda$ [so that the neighborhood is contained in the closed ball of radius $r$], the maximum modulus principle implies there's another point $\lambda_\epsilon$ (not equal to $\lambda$) where $|h_r(\lambda_\epsilon)|=M$. Thus there are points arbitrarily close to $\lambda$ where $|h_r|=M$, and thus $|h_r|$ is constant at value $M$ in the interior of the ball. See the contradiction? – Steve D Aug 11 '17 at 18:53

There is an alternative formulation of the maximum modulus theorem, which apears to be more general: https://en.wikipedia.org/wiki/Maximum_modulus_principle.

Roughly, it says that if $|f|$ has a local maximum in the interior of its domain, then it is constant.

This can be applied for your case as follows: Let $\lambda_0 \in \overline D(0,r)$ be such that it maximizes $|h_r(\lambda_0)|$. If $|\lambda_0|=r$, then $|h_r(\lambda)|\leq 1$ follows. In the other case $|\lambda_0|<r$ we are in the interior of the domain, so $h_r$ is constant.

Edit: i forgot to adress "$r\to \infty \Rightarrow g(\lambda)=0$".

From $|h_r(\lambda)|\leq 1$ it folows that $$|g(\lambda)| \leq \frac{|\lambda^2(2r-g(\lambda))|}{r^2}$$ Taking the limit $r\to\infty$ yields $g(\lambda)=0$.

• Is this alternative formulation equivalent to the one I've mentioned? – user8469759 Aug 11 '17 at 13:06
• I dont know if there is a simple equivalence. But i think the author was refering to a statement similar to the wikipedia version. – supinf Aug 11 '17 at 13:16
• I believe it was referring to the one I wrote actually, given the author is the same. But I'll read through to check whether or not such formulation is proven to be equivalent. – user8469759 Aug 11 '17 at 13:19
• @supinv Sorry, I'm still reading reading your argument. Basically you proved that for each $r$ I have $h_r$ is a constant, right? But how do I infer that $g(\lambda) = 0$ when $r$ diverges? – user8469759 Aug 11 '17 at 15:13
• @user8469759 i edited to adress that part of your question – supinf Aug 12 '17 at 11:44