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If $f:\mathbb{C}\rightarrow\mathbb{C}$ is a differentiable function and $f(x+2\pi)=f(x)$ for all $x\in \mathbb{R}$, would $f(z+2\pi)=f(z)$ for all $z\in \mathbb{C}$?

Is there any theorem/lemma concerning this? Are there any examples/counter examples for this?

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    $\begingroup$ If $f$ is holomorphic, then this works. But differentiability is too weak a condition (take for example $f(x+iy) = \sin((1-y)x)$). $\endgroup$
    – Joel Cohen
    Commented May 25, 2012 at 21:47
  • $\begingroup$ Do you have a proof? $\endgroup$
    – Potato
    Commented May 25, 2012 at 21:48
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    $\begingroup$ @Potato : If $f$ is holomorphic, then $g(z) = f(z+2\pi)$ also is. But since $g$ and $f$ are equal on $\mathbb{R}$, the identity theorem (en.wikipedia.org/wiki/Identity_theorem) tells us they are equal on $\mathbb{C}$. $\endgroup$
    – Joel Cohen
    Commented May 25, 2012 at 21:52
  • $\begingroup$ You should repost that below as an answer. $\endgroup$
    – Potato
    Commented May 25, 2012 at 21:56
  • $\begingroup$ @JoelCohen: This should really be an answer ... $\endgroup$
    – user26872
    Commented May 25, 2012 at 21:56

1 Answer 1

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If $f$ is holomorphic, then $g(z) = f(z+2\pi)$ also is. But since $f$ and $g$ are equal on $\mathbb{R}$, the identity theorem tells us they are equal on $\mathbb{C}$.

Now if you only assume differentiability as a function on $\mathbb{R}^2$, there are counter examples. Take for example $f(x+iy) = \sin(yx)$.

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    $\begingroup$ Arguably, if the OP says that the domain of the function is $\mathbb C$, then differentiable means holomorphic. $\endgroup$
    – Phira
    Commented May 25, 2012 at 22:13
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    $\begingroup$ @Phira : You're right, this is probably what was implied. I added the last part just to avoid confusion with being differentiable as a function on $\mathbb{R}^2$. $\endgroup$
    – Joel Cohen
    Commented May 25, 2012 at 22:28
  • $\begingroup$ Thank you, :) I didnt explicitly mention it, but I would guess that I did actually mean holomorphic. $\endgroup$
    – Derrick
    Commented May 25, 2012 at 23:46

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