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Consider the analytic function $f : (0,\infty) \to (0,\infty)$ given by $$ f(x) = \bigg( \sum_{i=1}^n a_i b_i^{1/x} \bigg)^x $$ where $n\in\mathbb N$, $a_i>0$ and $0<b_1<\ldots<b_n<1$. I am interested in analytic continuations of $f$ to the right half plane $\mathbb H_+ = \{z\in\mathbb C : \mathrm{Re}(z)>0\}$.

My guess is that that such an analytic continuation to all of $\mathbb H_+$ exists only in the trivial case $n=1$ (where the continuation is $f(z) = b\cdot a^{z}$). Even the $n=2$ case seems to be quite hard.

Any suggestions and ideas would be much appreciated!

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Partial answer:

$g(z)=\sum_{k=1}^n a_kb_k^{1/z}$ is analytic on the punctured plane and clearly $f(z)=e^{z\log g(z)}$ defines $f$ analytic outside the zeroes of $g$ - as usual locally and then we can stitch together a global $f$ with appropriate cuts.

Now $g >0$ on the real axis outside the origin by definition, so it comes down to figuring out if $g$ has zeroes only in the left-hand plane.

Since $1/z$ preserves the right hand and left hand planes, we need to look at the zeroes of $\sum a_kb_k^w=\sum a_ke^{c_k w}, c_k=\log b_k$ which are known as exponential or Ritt polynomials and there is a large literature about.

Edit later - actually since in the case $n=2$ the zeroes of $h(w)=a_1e^{c_1w}+a_2e^{c_2w}, c_k=\log b_k, c_1<c_2<0$ are clearly periodic with period $\frac{2k \pi i}{c_2- c_1}$, so all have the same real part, the function $f$ will be analytic in the right half plane precisely when one, hence all, are in the left half plane as then the zeroes of $g(z)=h(1/z)$ are like that too as noted.

This happens when with $a=\frac{a_1}{a_2}$ so one such zero is $w_0=\frac{\log (-a)}{c_2-c_1}$ we have $\Re w_0 \le 0$ and that happens precisely when $|a| \le 1$ or $a_1 \le a_2$, so we can actually completely solve the case $n=2$

Edit later per comments: Since $f(x)$ seems to be defined the usual way for $x>0$ where we have a unique positive $1/x$ root for the positive number $g(x)$, $\log g(x)$ is defined uniquely and continuously (real analytically too) by its usual real value and we can extend it (uniquely) to an analytic $\log g$ on a small (simply connected) neighborhood of the positive real axis in $\Re z>0$ - note that it may not be a strip if zeroes of $g$ accumulate towards the real axis. So already $f$ has unique analytic continuation to a complex open set in $\Re z >0$ and the only question is how much more we can go and still be analytic in $\Re z >0$ and that is precisely until we hit zeroes of $g$ where we have to start making cuts. If there are no zeroes, then $\log g$ and hence $f$ have analytic continuation to the whole half plane.

If $n=1$ we have $g(x)=a^xb, x>0$ so $\log g(x)=x \log a +b, x>0$ and that clearly extends to $\Re z>0$ so $f(z)=b a^z$ is the analytic continuation to $\Re z >0$ and actually to all $z$ finite as the singularity at $0$ is removable as we see.

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  • $\begingroup$ Hi @Conrad, thanks for you answer! I am already confused by the first statement though. Can you explain how $g$ not having zeroes in the right-hand plane implies that $f(z)$ is analytic on the right-hand plane? Already for $n=1$ it does not suffice to pick one branch cut for all $z$. So we need to adapt the cut to $z$, right? But how does that give us a well-defined value of $f(z)$ for all $z$ (with positive real part)? $\endgroup$
    – Lau
    Commented Feb 10, 2023 at 17:23
  • $\begingroup$ edited with a more complete explanation of why $f$ always has an analytic continuation to a neighborhood of $x>0$ in $\Re z >0$ etc; not sure what you claim for $n=1$ since $ba^z$ is perfectly (and uniquely by specifying that $ba^1=ba$) well defined for all $\Re z >0$ (and actually all $z$ for that matter) when $a>0$ so there is no need for branch cuts etc $\endgroup$
    – Conrad
    Commented Feb 10, 2023 at 18:28
  • $\begingroup$ Thanks for the added explanation! I am still a bit confused though. Are you making use of the simple connectedness of the right-hand plane somewhere? You say that zeroes of $g$ are the only objections to an analytic continuation of $\log g$. I don't understand why we can't apply your logic to continuate, say, $\log g$ with $g(z)=z$ on the punctured plane. Sorry I am a bit rusty with complex analysis. $\endgroup$
    – Lau
    Commented Feb 11, 2023 at 15:08
  • $\begingroup$ Yes one uses that the right hand plane is simple connected for the no zero case since one can always put together local continuations in a global single valued functions on such; for a non simply connected domain one needs cuts to get a single valued function like in your example here with $\log z$; if there are zeroes in the right hand plane one still can extend $f$ but you need cuts - for example if the zeroes are vertical like in $n=2, a_1>a_2$ good cuts are the half lines joining them to infinity to the left since you can define $f$ in the whole plane minus those and a cut at zero $\endgroup$
    – Conrad
    Commented Feb 11, 2023 at 16:14
  • $\begingroup$ Ah I see. This can be seen by explicitly defining $(\log g)(z)$ for $\Re z>0$ as $\log(g(1))$ plus the line integral of the analytic function $g'/g$ over any curve joining $1$ and $z$ in the right-hand plane, right? Since $g'/g$ is analytic it follows from simple connectivity that this is independent of the chosen curve. And locally one can find an appropriate logarithm so that $(\log g)$ is actually equal to the logarithm of $g(z)$. All of this depens only on simple connectivity and $g$ not having zeroes, right?. $\endgroup$
    – Lau
    Commented Feb 12, 2023 at 12:00

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