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Define g to be the function $g(z)=re^{\frac{i\theta}{2}}$ if $z=re^{i\theta}$ with $r>0$ and $-\pi<\theta\le\pi$, and $g(z)=0$ when $z=0$.

Is $g$ continuous from $\mathbb{C}\longrightarrow\mathbb{C}$?


Claim: $g$ is not continuous from $\mathbb{C}\longrightarrow\mathbb{C}$.

[It seems that all the function seems to be doing is halving $arg(z)$ for each $z\in\mathbb{C}$.]

If we consider points on the negative real line, i.e. $z=-a$ for $a\in\mathbb{R}$, then let $\epsilon=\frac{a}{2}$. If $g$ is continuous, then $\exists \delta>0$ such that $g(B_{\delta}(z))\subset B_{\epsilon}((g)z)$. Clearly $g(-a-(i\delta/2))\in g(B_{\delta}(-a))$ does not lie in $B_{a/2}(g(-a))$ for any $\delta>0$ as $-a$ is mapped to $ai$ and $-a-(i\delta/2)$ is mapped to a point in the lower half-plane.

We can conclude then that $g$ is not continuous from $\mathbb{C}\longrightarrow\mathbb{C}$.


Is this proof valid (if I play around to make it a little more formal), i.e. is the reasoning behind the claim correct?

Any verification/objection would be appreciated. Thanks, TJO.

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I think it is correct - alternatively you can consider a sequence $z_n = -1+(-1)^n\frac{\mathrm{i}}{n}$ which converges to $-1$ while $g(z_{2n})\to\mathrm{i}$ and $g(z_{2n+1})\to-\mathrm{i}$ – Ilya Dec 9 '11 at 12:45
Looks good to me, thanks. – Mathmo Dec 9 '11 at 12:51
up vote 3 down vote accepted

Your argument looks correct, but you can do it simpler. If $g$ is continuous then for any $z \in \mathbb{C}$ and $(z_n)_{n \in \mathbb{N}} \subset \mathbb{C}$ such that $z_n \to z$ it follows that $g(z_n)$ converges to $g(z)$.

In your case let $z=-1$ (or any other negative real number). Then let $z_n := e^{i \theta_n}$, where for example $\theta_n = -\pi + \tfrac{1}{n}$. Of course $z_n \to z$ since $\theta_n \searrow -\pi$. Then $$g(z_n) = g(e^{i(-\pi + 1/n)}) = e^{i(-\pi/2 + 1/2n)} \to e^{-i \pi/2} = -i \neq -1.$$

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Thanks; not sure why I went for a topological proof on reflection. – Mathmo Dec 9 '11 at 13:14

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