I am reviewing for the complex analysis but I am struck by the following question,
Let $f(z)$ be a bounded analytic function on the right half plane. Suppose that $f(z)$ extends continuously to the imaginary axis and satisfies $|f(iy)|\leq M$ for all point $iy$ on the imaginary axis. Show that $|f(z)|\leq M$ for all $z$ in the right half-plane. Hint/ For $\epsilon > 0$ small, consider $(z+1)^{-\epsilon}f(z)$ on a large semidisk.
—— From Gamelin's Complex analysis section 3.5 question 5. I know this question is asking about the maximum principle in particular whenever $|f(z)|\leq M$ for all boundary points then this relation holds for all interior points too. However, for this question I don't know how to write a formal solution given the detail of the question and the hint. So I wonder if anyone can tell where I should begin with this problem, for example, how to write the solution in beginning? Thanks in advance.
Every point on the right half plane lies in a semi-disk with center on the imaginary axis passing through the origin. On the semicircular part we have $|\frac {f(z)} {(z+1)^{\epsilon}}| \leq \frac C {(R-1)^{\epsilon}} <M$ if $R$ is large enough. Here $R$ is the radius of the semicircle and $C$ is a bound for $|f(z)|$. Now apply Maximum Modulus Principle to the region bounded by the semicircle and a line segment on the imaginary axis. [Note that $(z+1)^{\epsilon}$ is defined in terms of the principal logarithm and is it analytic].