Analytic function on annulus bounded by $\log 1/|z|$ is zero

Question

Let $f(z)$ be an analytic function on $A(0,1)=\{z\in\mathbb{C}\mid0<|z|<1\}$ such that $$\forall z\in A(0,1)\quad|f(z)|\le\log\bigg(\frac 1 {|z|}\bigg).$$ Prove $f\equiv 0.$

Define $g(z)=e^{f(z)}$ and note that $$\forall z\in A(0,1),\quad |g(z)|=e^{\Re f(z)}\le e^{|f(z)|}\le e^{\log |z|^{-1}}=\frac{1}{|z|}.$$ Now I don't know how to prove $g\equiv c$. Suppose I did that, $f=\ln c$ but since $f(1)=0$ we get the result.

As said, I'm struggling with proving that g is constant. I thought doing it By applying Cauchy integral formula but I only succeeded bounding the derivative.

How can I prove $g$ is constant ?

Answer 1

The given bound for $|f(z)|$ implies $$ \lim_{z \to 0} z \, f(z) = 0 $$ and therefore (Riemann's theorem) that $f$ has a removable singularity at $z=0$, i.e. it can be continued to a holomorphic function in the unit disk $\Bbb D$.

Now you can apply the maximum modulus principle and conclude that for all $z \in \Bbb D$ and $|z| < r < 1$, $$ |f(z)| \le \max \{ |f(\zeta)| : |\zeta| = r \} \le \log \frac 1r $$ and with $r \to 1$ it follows that $f(z) = 0$.

Answer 2

The bound implies that the function $zf(z)$ has a removable singularity at $0$, hence an analytic function $g(z)$ on the open disc. Observe that Cauchy integrals running very close to the unit circle are as small as we want. More explicitly we have $$g^{(n)}(0)=\frac{n!}{2\pi i}\oint_\gamma\frac{g(z)}{z^{n+1}}\,\mathrm dz $$ where $\gamma(t)=re^{it}$, $0\le t\le2\pi$ is a circular curve of radius $r\approx 1$ so that $$\left|g^{(n)}(0)\right|\le \frac{n!}{2\pi r^n}\int_{t=0}^{2\pi}\left|g(z)\right|\,\mathrm dt \le \frac{n!}{r^{n-1}}\log\frac1r.$$ As the right hand side tends to $0$ as $r\to 1$, we conclude that $g(0)$ and all $g^{(n)}(0)$ are zero. Hence $g\equiv 0$ and also $f\equiv 0$.