Stein-Shakarchi 'Complex Analysis' Chapter 3 Exercise 15 b.

[caffeinemachine]

I am stuck on a problem given in Stein and Shakarchi's Complex Analysis.

(Chapter 3, Exercise 15b) Use the maximum modulus principle or Cauchy inequalities to solve the following:
Let $f$ be a bounded holomorphic function on the open unit disc, and suppose that $f$ converges uniformly to $0$ in the sector $\theta< \text{arg}(z)< \varphi$ as $|z|\to 1$. Then show that $f$ is $0$.

In other words, we have a bounded holomorphic function on the open unit disc with the following property:
For all $\varepsilon>0$, there is $r$ with $0< r< 1$ such that whenever $1>|z|>r$ with $\theta< \text{arg}(z)< \varphi$, we have $|f(z)|<\varepsilon$.

We want to show that $f$ is identically $0$.

I have been unable to make any progress on this problem. Can somebody help. Thanks.

 

[jvicens]

I can offer some guidance on how to approach this problem. First, let's recall the maximum modulus principle, which states that for a holomorphic function $f$ on a bounded open set $\Omega$, the maximum value of $|f|$ on $\Omega$ must occur on the boundary of $\Omega$. This can also be extended to the maximum modulus principle for harmonic functions, which states that the maximum value of a harmonic function on a bounded open set $\Omega$ must also occur on the boundary of $\Omega$.

Now, let's consider the function $g(z) = \frac{1}{z}$, which is holomorphic on the open unit disc. We know that $|g(z)| = \frac{1}{|z|}$, so as $|z| \to 1$, $|g(z)| \to 1$. This means that for any $\varepsilon > 0$, we can find $r$ such that whenever $1 > |z| > r$, we have $|g(z)| < \varepsilon$.

Next, we can use Cauchy's inequalities, which state that for a holomorphic function $f$ on a disc of radius $R$, we have $|f^{(n)}(0)| \leq \frac{n!}{R^n}\sup_{|z|=R}|f(z)|$. Applying this to our function $f$, we have $|f(z)| \leq \frac{n!}{R^n}$ for all $z$ with $|z| = R$.

Now, using the fact that $f$ converges uniformly to $0$ in the sector $\theta < \text{arg}(z) < \varphi$, we can find $r$ such that for all $z$ with $\theta < \text{arg}(z) < \varphi$ and $r < |z| < 1$, we have $|f(z)| < \varepsilon$. This means that $|f^{(n)}(0)| \leq \frac{n!}{R^n}\varepsilon$ for all $n$, and since $\varepsilon$ can be made arbitrarily small, we have $f^{(n)}(0) = 0$ for all $n$. This implies that $f$ is identically $0$, since its Taylor series at $