Analytic function on the right half-plane, continuous and bounded on the closed right-half plane Let $\mathbb{C}_+$ be the right half-plane defined by
\begin{align}
\mathbb{C}_+ \stackrel{\text{def}}{=} \{z \in \mathbb{C} \ \colon \Re(z) > 0\}.
\end{align}

I am trying to understand why for a function $f$ analytic on $\mathbb{C}_+$, and also continuous and bounded on the closure of $\mathbb{C}_+$, we have
  \begin{align*}
\sup_{z \in \mathbb{\mathbb{C}_+}} |f(z)| = \sup_{y \in \mathbb{R}} |f(iy)|.
\end{align*}

I rewrote $f(z)$, for $z \in \mathbb{C}_+$, as
\begin{align*}
f(z) = \int_{-i\infty}^{+i\infty} k_z(\zeta)f(\zeta)dy, \qquad \text{with } \  k_s(\zeta) \stackrel{\text{def}}{=} \frac{ \Re(z)}{i\pi (\zeta - z)(\zeta + \overline{z})}
\end{align*}
and then wanted to show that 
\begin{align*}
\sup_{z \in \mathbb{\mathbb{C}_+}} \|k_z\|_{L^1(i\mathbb{R})} < \infty,
\end{align*}
but I failed to do so. I was wondering if someone could help me with this last part, or explain another way to proceed.
 A: Well first, the statement you ask about in the body of the question cannot be literally true, because a priori there's no such thing as $f(iy)$. (In fact the hypotheses do imply that the boundary values $f(iy)$ exist for almost every $y$, but that's a little deeper than this.)
One wonders how you "rewrote" $f$ as claimed. Anyway, one can get a true statement by adding a hypothesis from the title:


If $f$ is continuous and bounded on $\overline{\Bbb C_+}$ and analytic in $\Bbb C_+$ then $$\sup_{z\in\Bbb C_+}|f(z)|=\sup_{y\in\Bbb R}|f(iy)|.$$


This is a simple Phragmen-Lindelof sort of argument. For $\epsilon>0$ define $$f_\epsilon(z)=\frac{f(z)}{1+\epsilon z}.$$Since $f$ is bounded it follows that $$\lim_{z\to\infty}f_\epsilon(z)=0.$$So $|f_\epsilon|$ achieves a maximum at some point of $\overline{\Bbb C_+}$. The Maximum Modulus Theorem says the maximum cannot happen at an interior point, so for every $z\in\Bbb C_+$ we have $$|f_\epsilon(z)|\le\sup_{y\in\Bbb R}|f_\epsilon(iy)|\le\sup_{y\in\Bbb R}|f(iy)|.$$Let $\epsilon\to0$.
