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This is the question we were asked at the university by our professor for complex analysis. Not as an exam, but as a challenge. I don't think he knew the answer himself.

Find a nontrivial example of a function $f$ defined on a neighbourhood of $z\in \Bbb C$ and a path from $z$ to $z$ so that the analytic continuation of $f$ along the path is $f'$.

It's easy to find trivial example: $e^x$. What I want is $f\ne f'$.

A variation to this problem, but much easier, is to find $f$ whose analytic continuation is $f+a$, $af$, or $-f$ (for some $a\in \Bbb C$).

They are a logarithm, a power and a square root.

But the question in title I never knew how to tackle. Can anyone shed some light on it, please?

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    $\begingroup$ If the path is say $\gamma(t),$ does the requirement mean that at each $t$ we have $h(t)=f'(t),$ where $h$ is continuation of $f$ along $\gamma(t)$, or does this only need to hold at the endpoint of $\gamma$ (i.e. on the return to $z$)? $\endgroup$
    – coffeemath
    Dec 18, 2014 at 20:09
  • $\begingroup$ I'm asking just for the endpoint: $f$ and its continuation along the path are both defined in a neighbourhood of $z$; that's where the latter must be a derivative of the former. Of course, as a consequence, if you follow the path again, you end up having a function and its derivative on every point of the path (or in the neighbourhood of every point) and end up with $f''$ at $z$. $\endgroup$
    – Heimdall
    Dec 18, 2014 at 22:41
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    $\begingroup$ If as this comment starts you only want the continuation to match the derivative at the endpoint $z$ then consider $f(z)=z^2$ in a neighborhood of zero. Then the continuation of $f$ is itself, and its derivative at zero is $0$, matching $f'(0)=0.$ But in the rest of the comment you seem to say the opposite, that at each point of the path the continuation matches the derivative... So I don't (yet) see which you mean, that is, your last comment seems to start with one version and continue with the stronger condition along the whole path. $\endgroup$
    – coffeemath
    Dec 19, 2014 at 5:19
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    $\begingroup$ I said in a neighbourhood of $z$, not just matching the value at $z$ like $z^2$. Sorry I misunderstood your comment. $\endgroup$
    – Heimdall
    Dec 19, 2014 at 12:08
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    $\begingroup$ I've just noticed the answer has been deleted. Although it didn't actually answer the question, it certainly demonstrated an idea how it could be tackled and therefore I think it should be kept. (I had +1'd it.) $\endgroup$
    – Heimdall
    Jan 1, 2015 at 0:36

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Not a positive answer, but an observation about the sort of analytic continuation which would satisfy your challenge problem. If $f$ has the desired property, then it has that property all the way around the path $\gamma$. That is, at any point on $\gamma$, if we continue $f$ one full time around $\gamma$ we obtain the derivative. Following is justification (it was not immediately obvious to me, though it may be to you).

First some notation.

Let $\gamma$ denote the path in question, assumed to be parameterized with domain $[0,1]$. Since $\gamma$ is closed, for $s\in[0,1]$, we can define $\gamma_s$ to be the concatenation of the restriction $\gamma|_{[s,1]}$ followed by $\gamma|_{[0,s]}$. Using modular arithmetic (modulo $1$) we can think of $\gamma_s$ having domain $[s,s+1]$.

Let $f_s$ denote the analytic continuation of $f$ along the path $\gamma|_{[0,s]}$ (viewed as having for its domain a neighborhood of $\gamma(s)$). We are assuming that ${f_0}'=f_1$.

Let $A\subset[0,1]$ be the subset such that $r\in A\Leftrightarrow$ the analytic continuation of $f_r$ along $\gamma_r$ is ${f_r}'$. We are assuming that $A$ contains $0$ (and is thus non-empty). It is immediate from the definition of analytic functions that $A$ is open. It is also not hard to show that $A$ is closed (since if the points where two functions are equal have an accumulation point in their common domain, they are equal in their common domain). Thus $A=[0,1]$.

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  • $\begingroup$ Thank you. And from that follows that after 2 cycles you get $f''$. Yes, I have made that observation pretty much immediately after I first looked at that challenge. $\endgroup$
    – Heimdall
    Jan 8, 2015 at 23:13

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