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In Terry Tao's notes https://terrytao.wordpress.com/2009/03/30/245c-notes-1-interpolation-of-lp-spaces/#lind, the interpolation $B_0^{1-\theta} B_1^\theta$ appearing in the Hadamard 3-lines theorem comes from the fact that $h(s):= B_0^{1-s} B_1^s$ is an entire nowhere-vanishing function s.t. $|h|=B_0$ on the imaginary line $i\mathbb R$ and $|h|=B_1$ on the line $1+i\mathbb R$. Is this the only such function (up to constant multiple)?

If I had another such function $f$ satisfying just the modulus conditions, Entire function such that $|f(z)| = 1$ on the real line I think tells me that $f$ is automatically nowhere-vanishing. Since $h$ is nowhere-vanishing, $\frac fh$ is entire, and has modulus $1$ on $i\mathbb R$ and $1+i\mathbb R$, and nowhere-vanishing $\implies \frac fh = e^g$ for some entire $g$. Then on $i\mathbb R$ and $1+i\mathbb R$, $$1=|e^g| = e^{\Re(g)} \implies \Re(g) \in 2\pi i\mathbb Z \cap \mathbb R\implies \Re(g)=0.$$ There seem to be some results about situations like this, e.g. The real part of a holomorphic function and its link to the function being constant, but I don't really know where to go from here.

EDIT: I suppose from the very same theorem there is the function $h(s):= \exp(-i\exp(i\pi s))$, where for $s$ with $\Re s= 0,1$, we get $$\begin{aligned} |h(s)| &= \exp(\Re(-i \exp(i\pi s)) = \exp(\Im (\exp(i\pi s)) = \exp(\Im(\exp(i\pi \Re s) \exp(-\pi \Im s))\\ &= \exp(\Im(\pm \exp(-\pi \Im s)) = \exp(0) = 1 \end{aligned}$$ so $h(s) B_0^{1-s} B_1^s$ is also an entire nowhere-vanishing function s.t. $|h|=B_0$ on $i\mathbb R$ and $=B_1$ on $1+i\mathbb R$. Are these the unique functions with these properties? It seems like the facts I already found above via other MSE posts should severely restrict the class of such functions.

D.R.
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  • Your assumptions do not imply that such a function $f$ needs to be non-vanishing. E.g., take a conformal map from the strip onto the unit disk (which can be written down explicitly using exponential function and fractional linear transformations), or such a map composed with any Blaschke product, which makes it possible to generate lots of zeros. – Lukas Geyer Oct 07 '22 at 22:48
  • @LukasGeyer hmm, I think your example is not entire. I did link another post about entire functions being modulus 1 on $\mathbb R$ means it's non-vanishing, and I think the same would hold if it were modulus 1 on $i\mathbb R$. – D.R. Oct 08 '22 at 03:46
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    Yes, it is not entire, but for the proof of the theorem it is really only required to be holomorphic on the strip between the two lines, so I misunderstood your question. If you are looking for entire functions, you are correct that they can not vanish. – Lukas Geyer Oct 10 '22 at 00:17
  • @LukasGeyer ok, I see. I think at the very least I want the function to be non-zero on the strip, since the proof (of Lindelof linked in my post) requires that dividing by the function produces a function that grows less than double-exponentially on the strip (and a pole with definitely break that condition). – D.R. Oct 10 '22 at 02:13

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