DavisT

@leslietownes I get your point, thanks... This is all far more involved than I anticipated, wow

good point, so $f^{1/2}$ suffers from the same problems as $\sqrt{f}$ (at least in general -- there are exceptions, like $f(z)=e^z$ we discussed)

put differently: is it meaningful to regard this "holomorphic square root" as the proper analytic continuation of $\sqrt{f}$ on the reals?

I have a quick follow-up question regarding what Thorgott wrote: any nowhere vanishing holomorphic function $f:\mathbb{C}\rightarrow\mathbb{C}\setminus\{0\}$ admits a holomorphic square root. I understand that this "square root" is (in general) not literally $\sqrt{f}$, but a function whose square gives the original function. But is it unambiguous to write $f^{1/2}$ for that function, in particular when trying to define an analytic continuation of $\sqrt{f}$ from the reals?

@robjohn @leslietownes @Thorgott thanks for all of your help, I'll have to think about everything, but it's much clearer now

at least that fits to the indicated cut at $\pm\pi i$ in my example

and Wolfram Alpha generally assumes the principal branch for sqrt, I suppose?

@robjohn right, bad example...

for example, $\sqrt{\sinh(x)}$ still suffers from the same problems due to the zeros of $\sinh$, right?

concerning analytic continuation from the reals

Is it generally so that taking the power 1/2 is better than sqrt

it potentially does, thanks

oh joy

at the moment I was writing this comment, I was wondering the same

yes

I think I have to deal with $\sqrt{e^z}$ as I want to consider an analytic continuation of $\sqrt{e^x}$

(at least not everywhere)

@robjohn ok, but $e^{z/2}$ is NOT the principal square root of $e^z$, right?

but $e^{z/2}$ is continuous everywhere

@robjohn but if we take the principal branch, is it still true that $\sqrt{e^z}$ becomes discontinuous at $\pi i$? How does that fit together with what you said before?

ah, so it's a 2nd root in the sense that its square gives the original function e^z, but it's not THE square root in the sense of $\sqrt{e^z}$

@Thorgott : but $z\mapsto e^z$ is a nowhere vanishing holomorphic function $\mathbb{C}\rightarrow\mathbb{C}\setminus\{0\}$, so why doesn't it admit a holomorphic square root?

@robjohn thanks, I'll do that

I'm doing research, but my analysis knowledge is really bad and too long ago... I'm not an analyst at all, but sometimes there's no way around, haha

thanks, that clears it up a lot...

@robjohn I'm not, sorry

okay, so the misconception on my side was that I thought it suffices that 0 is not a branch point to conclude that the function is entire, while in reality one still has to consider the cut along the negative real axis

that comes from simplifying my notes when asking a question, haha

I was thinking e^(pi*z), that's why I wrote $i$

oh yes of course, sorry

and did I put it right with what I've written above: " so the problem here is that while 0 is not a branch point of sqrt(e^z), we still have a branch cut on the negative real axis in the image of e^z (that is, at z=i) where sqrt(e^z) becomes discontinuous"

ok, but \sqrt(e^z) is continuous (in fact, analytic) inside an open strip around the real axis with imaginary part between (-i,i), right?

@leslietownes is that correct? so the problem here is that while 0 is not a branch point of sqrt(e^z), we still have a branch cut on the negative real axis in the image of e^z (that is, at z=i) where sqrt(e^z) becomes discontinuous

it's fine in a strip around the real axis, but not where the exp takes negative values, which is at $i$

oh wait a moment

I thought that since sqrt has a branch point at 0, but e^z never vanishes, sqrt(e^z) has no branch point and hence everything is fine

I'm confused

@leslietownes why not, if we choose the principal square root function?

@robjohn exactly, I never claimed something else

n=2 for the square root

where's the mistake here?

@robjohn I don't think I'm mixing that up... z\mapsto e^z is a holomorphic function on all of C, which takes values in C\{0}. Hence this function has a holomorphic logarithm, and thus holomorphic n-th roots for all n.

I suppose wolframalpha doesn't understand that there exists an analytic continuation, and thus messes up

maybe wolframalpha just messes something up... but from my understanding sqrt(e^z) should be analytic everywhere

can someone explain this to me please?

if \sqrt is the principal square root function on C, then sqrt(e^z) should be an entire function, because e^z does not vanish anywhere, and so we have no branch points. However, in the function plot on wolframalpha there appears to be a cut at z=x+iy for y=pi; see the contour plots here: wolframalpha.com/input/?i=plot+sqrt%28exp%28z%29%29

thanks, I'll have a look

yes please, that would be very helpful

@leslietownes thanks, I'll try that. If you happen to recall anything later, please let me know

do you have any suggestions where I could look for those generalizations of classical hardy classes you were mentioning?