An analytic function $f$ bounded on the right half plane and $|f(z)|\leq 1$ on the imaginary axis Assume that $f$ is an analytic function that $|f(z)|\leq 1$ on the imaginary axis and that $f$ is bounded in the right half plane. Prove that in fact $|f(z)|\leq 1$ in the right half plane. Hint: Consider the function $f_{\epsilon}(z)=f(z) e^{-\epsilon \sqrt{z}\,\,\,}$ for small positive $\epsilon$.
Even though there is a hint I couldn't figure out how to approach to this question. If the function is conformal the statement is true but what we have otherwise? any help would be great.
 A: We take the branch of $\sqrt{z}$ with $\sqrt{1} = 1$ on the right half-plane. Then for $z = re^{i\varphi}$ in the right half plane - with $-\frac{\pi}{2} < \varphi < \frac{\pi}{2}$ - we have $\sqrt{z} = \sqrt{r}\cdot e^{i\varphi/2}$, so $\operatorname{Re} \sqrt{z} = \sqrt{\lvert z\rvert}\cos \frac{\varphi}{2} \geqslant \sqrt{\lvert z\rvert}\cos \frac{\pi}{4} = \sqrt{\frac{\lvert z\rvert}{2}}$.
Thus, if we know $\lvert f(z)\rvert \leqslant M$ in the right half-plane, we have
$$\lvert f_\epsilon(z)\rvert \leqslant M e^{-\epsilon\sqrt{\lvert z\rvert/2}} \leqslant 1\tag{$\ast$}$$
for $\lvert z\rvert \geqslant 2\bigl(\frac{\log M}{\epsilon}\bigr)^2$. But the maximum modulus principle then tells us that also
$$\lvert f_\epsilon(z)\rvert \leqslant 1$$
in the half-disk $\Bigl\{ z : \operatorname{Re} z > 0, \lvert z\rvert < 2\bigl(\frac{\log M}{\epsilon}\bigr)^2\Bigr\}$, since we know $\lvert f_\epsilon(z)\rvert \leqslant 1$ on the boundary of that half-disk. On the semicircle $\lvert z\rvert = 2\bigl(\frac{\log M}{\epsilon}\bigr)^2$, that follows from $(\ast)$, and on the part of the imaginary axis, it follows from the assumption that $\lvert f(z)\rvert \leqslant 1$ there and the fact that $\lvert e^{-\epsilon\sqrt{z}}\rvert \leqslant 1$ on the closed right half-plane.
So we have $\lvert f_\epsilon(z)\rvert \leqslant 1$ on the whole right half-plane for every $\epsilon > 0$. But then
$$\lvert f(z)\rvert = \lim_{\epsilon \searrow 0} \lvert f_\epsilon(z)\rvert \leqslant 1$$
for every $z$ in the right half-plane.
