Find all $f$: (a) $f$ is analytic on $B(0,1)$, (b) $f(0)=1$ and (c) $|f(z)| \geq 1 $ for $|z|<1$

Question

Find all $f: \mathbb{C} \to \mathbb{C}$ such that

(a) $f$ is analytic on $B(0,1)$,

(b) $f(0)=1$ and

(c) $|f(z)| \geq 1 $ for $|z|<1$.

Attempt:

Looks like minimum modulus theorem to me, which would imply that $f$ has to be the constant function $f(z)=1$. But $f$ need not be analytic on $\bar B(0,1)$, that is it can fail to be analytic on the boundary cirlce $\{z \in \mathbb{C}:|z|=1\}$, so the MMT wouldn't apply. So how does this work?

Many thanks!

Answer 1

One way to prove this is to consider the point $f(0)=1$ and show that it cannot be a local minimum. To this end, consider the Taylor series for $f(z)$ centered at $z=0$. We know that this series exists and is convergent within some radius $r$ as $f(z)$ is analytic at the origin. We have

$$f(z)=f(0)+f^{'}(0)z+\frac{f^{''}(z)z^2}{2}+\cdots=1+\sum_{n=1}^\infty\frac{f^{(n)}(0)z^n}{n!}$$

Now, unless $f(z)$ is a constant function, we know there exists a minimum $n$ such that $f^{(n)}(0)\neq 0$. Call this $n=N$. Then $f(z)$ is

$$f(z)=1+\frac{f^{(N)}(0)z^N}{N!}+\sum_{n=N+1}^\infty\frac{f^{(n)}(0)z^n}{n!}$$

Now, consider the point $z=\epsilon e^{i\theta/N}$ where $0<\epsilon<r$ (that is, this point is within the radius of convergence). Then we have

$$f(\epsilon e^{i\theta/N})=1+\frac{f^{(N)}(0)\epsilon^N e^{i\theta}}{N!}+\sum_{n=N+1}^\infty\frac{f^{(n)}(0)(\epsilon e^{i\theta/N})^n}{n!}$$

Denote the coefficient in front of $\epsilon^N e^{i\theta}$ as $x+i y$ (and note that it is non-zero) and rearrange terms to get

$$f(\epsilon e^{i\theta/N})=1+(x+iy)\epsilon^N e^{i\theta}+\left(\sum_{n=N+1}^\infty\frac{f^{(n)}(0)\epsilon^{n} e^{in\theta}}{n!}\right)$$

Taking the modulus, we get

$$|f(\epsilon e^{i\theta/N})|=\left|1+(x+iy)\epsilon^N e^{i\theta}+\left(\sum_{n=N+1}^\infty\frac{f^{(n)}(0)\epsilon^{n} e^{in\theta}}{n!}\right)\right|$$

$$\leq \left|1+(x+iy)\epsilon^N e^{i\theta}\right|+\left|\sum_{n=N+1}^\infty\frac{f^{(n)}(0)\epsilon^{n} e^{in\theta}}{n!}\right|$$

$$\leq \left|1+(x+iy)\epsilon^N e^{i\theta}\right|+\sum_{n=N+1}^\infty\left|\frac{f^{(n)}(0)\epsilon^{n} e^{in\theta}}{n!}\right|$$

$$\leq \left|1+(x+iy)\epsilon^N e^{i\theta}\right|+\sum_{n=N+1}^\infty \epsilon^{n}\left|\frac{f^{(n)}(0)}{n!}\right|$$

Since the series converges, we are assured that

$$\lim_{n\to\infty}\left|\frac{f^{(n)}(0)}{n!}\right|=0$$

Define

$$M=\max\left\{\left|\frac{f^{(n)}(0)}{n!}\right|:n\geq N+1\right\}$$

Then

$$\sum_{n=N+1}^\infty \epsilon^{n}\left|\frac{f^{(n)}(0)}{n!}\right|\leq \sum_{n=N+1}^\infty \epsilon^{n}M=M\frac{\epsilon^{N+1}}{1-\epsilon}$$

We may also analyse the other absolute value

$$\left|1+(x+iy)\epsilon^N e^{i\theta}\right|=\sqrt{\epsilon ^{2 N} \left(x^2+y^2\right)+2 \epsilon ^N (x \cos (\theta )-y \sin (\theta ))+1}$$

However, note that $x \cos (\theta )-y \sin (\theta )$ has a minimum value of $-\sqrt{x^2+y^2}$ at some complicated $\theta$ (this is easily shown using derivatives). Then

$$\sqrt{\epsilon ^{2 N} \left(x^2+y^2\right)+2 \epsilon ^N (x \cos (\theta )-y \sin (\theta ))+1}$$

$$= \sqrt{\epsilon ^{2 N} \left(x^2+y^2\right)-2 \epsilon ^N \sqrt{x^2+y^2}+1}$$

$$=\sqrt{\left(1-\epsilon ^{N} \sqrt{x^2+y^2}\right)^2}=1-\epsilon ^{N} \sqrt{x^2+y^2}$$

Putting everything together, we get that at a certain $\theta$

$$|f(\epsilon e^{i\theta})|\leq 1-\epsilon ^{N} \sqrt{x^2+y^2}+M\frac{\epsilon^{N+1}}{1-\epsilon}=1+\epsilon^N\left(\frac{\epsilon M}{1-\epsilon}-\sqrt{x^2+y^2}\right)$$

Since

$$\lim_{\epsilon\to 0}\frac{\epsilon}{1-\epsilon}=0$$

choose $\epsilon$ such that

$$\epsilon=\min\left\{\frac{1}{2},\frac{r}{2},\frac{\sqrt{x^2+y^2}}{2(M+\sqrt{x^2+y^2})}\right\}$$

Note that this is non-zero as $\sqrt{x^2+y^2}\neq 0$.This implies

$$\frac{\epsilon M}{1-\epsilon}-\sqrt{x^2+y^2}<0$$

and therefore

$$|f(\epsilon e^{i\theta})|\leq 1-\epsilon^N\left|\frac{\epsilon M}{1-\epsilon}-\sqrt{x^2+y^2}\right|<1$$

We conclude that if $f(z)$ satisfies all the conditions above, then $f(z)=1$ and we are done.

Answer 2

$f$ has no zeros in $B(0, 1)$, so that $$ g: B(0, 1) \to \Bbb C, \, g(z) = 1/f(z) $$ is holomorphic in the unit disk. Also $$ \forall z \in B(0,1): |g(z)| \le 1 = |g(0)| $$ so that $|g|$ has a maximum at $z=0$. According to the maximum modulus principle, $g$ (and consequently $f$) is constant in $B(0, 1)$.

If $f$ is holomorphic in $\Bbb C$ then it follows that $f(z) = 1$ everywhere, according to the identity theorem. Otherwise nothing can be said about $f(z)$ for $|z| \ge 1$.

Answer 3

If $f$ is not constant, then $f(B(0,1))$ is open in $\mathbb C$ by the open mapping theorem. Thus, since $1\in f(B(0,1)),$ $f(B(0,1))$ contains $B(1,r)$ for some $r>0.$ But $B(1,r)$ contains points of modulus $<1,$ contradicting (c). Therefore $f$ is constant, and since $f(0)=1,$ we must have $f\equiv 1.$