Assume that $F(z)$ is analytic on the open strip $0<Re z < 1$ and bounded and continuous on the closed strip $0 \leqslant Re z \leqslant 1$. If $$ |F(it)| \leqslant M_0, \;\;\; |F(1+it)| \leqslant M_1, \;\;\; -\infty < t < \infty, $$ then we have $$ |F( \theta + it) | \leqslant M_{0}^{1-\theta} M_1^\theta , \;\;\; -\infty < t < \infty $$ for $\theta \in [0,1]$
Proof.
Let $\epsilon >0$ and $\lambda \in \mathbb R$. Put $$ F_{\epsilon} (z) = e^{\epsilon z^2 + \lambda z} F(z)$$ so that $$ \lim_{Im z \to \pm \infty} F_{\epsilon} (z) = 0, \;\;\; \left | F_{\epsilon} (it) \right | \leqslant M_0 , \;\;\; \left | F_{\epsilon}(1+it) \right| \leqslant M_1 e^{\epsilon + \lambda} . $$ Then by the Phragmen-Lindelof principle(or Maximum Modulus Principle), we have $$ | F_{\epsilon} (z) | \leqslant \max ( M_0, M_1 e^{\epsilon + \lambda})$$ which means that $$ |F(\theta + it) | \leqslant e^{ – \epsilon ( \theta^2 – t^2 )} \max ( M_0 e^{ – \theta \lambda} , M_1 e^{(1- \theta) \lambda + \epsilon} ). $$ Let $ \rho = e^{\lambda}$ and $\epsilon \to 0$, then $$ |F( \theta + it) | \leqslant \max ( M_0 \rho^{-\theta} , M_1 \rho^{1- \theta} ). $$ The right hand side is as small as possible when $ M_0 \rho^{-\theta} =M_1 \rho^{1- \theta}$, i.e. $\rho = M_0 / M_1$. So with this choice of $\rho$, we are done.
Note that this theorem is very useful to derive Riesz-Thorin Interpolation Theorem.
References.
[1] J. Bergh. Interpolation Spaces: An Introduction.
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