# Truncated Dirichlet series take their supremum on the imaginary axis

Hi there, I am struggling with a theorem about truncated Dirichlet series. I am trying to prove the following theorem:

Let $(a_n)_n \subset \mathbb{C}$ and $N \in \mathbb{N}$. Then $\sup_{t \in \mathbb{R}} \vert \sum_{n=1}^N a_n n^{-it} \vert = \sup_{\Re s \ge 0} \vert \sum_{n=1}^N a_n n^{-s} \vert$.

I think the trick is to use the maximum modulus principle for holomorphic functions (a holomorphic function, continuous up to the boundary attains its maximum on the boundary). Using this for the domain $-a < \Im s < a, 0 < \Re s < b$ could lead to the answer ($a,b \to \infty$), but how do I show that such a truncated Dirichlet series attains its maximum on the left border?

I would appreciate any hint on the proof or reference to any literature.

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Observe that $\vert \sum_{n=1}^N a_n n^{-s} \vert$ is bounded in the half-plane $\Re(s)\geq 0$, and it is very small for $\Re(s)$ large. As a result, there is a strip $0\leq\Re(s)\leq b$ such that the supremum here is the same as in the half-plane, but the supremum on the right edge $\Re(s)=b$ is much smaller. Now your statement follows from the Phragmén-Lindelöf theorem applied to this strip.