$\theta (r\cos \theta,r\sin \theta)=\theta :\mathbb R^2\rightarrow \mathbb R$ is differentiable $?$

Question

$\theta :\mathbb R^2\backslash\{0\}\rightarrow \mathbb R$ is defined by $$\theta(x,y)=\Theta$$, the unique number from $(-\pi,\pi]$ such that $x=r\cos\theta,y=r\sin\theta$ where $r=\sqrt{x^2+y^2}.$

So basically this function $\theta$ assigns a point $(x,y)$ of $\mathbb R^2\backslash\{0\}$ to the Principal Value of its amplitude.

Now question is about the Boundedness,Continuity,Differentiability etc properties of the function.

Now its range is given to be $(-\pi,\pi]$ so it is bounded. That part is done.

Now before checking differentiability,we must see the continuity.If we take any poiny from the $x$ axis,that is the entire $x$-axis goes to $0$.Entire positive $y$-axis goes to $\pi$ and the entire of the negative $y$-axis goes to $-\pi$ but since that is not in range so we won't be taking any points from the $-y$-axis.

If we fix $x_0$ and vary the $y$ co-ordinate; then fix $y_0$ and vary the $x$ co-ordinate,i.e. find a sequence in one co-ordinate keeping the other fixed converging,then $\theta$ does converge,to $\theta(z_0)$ where $z_0$ is the limit of the sequence.So it is continuous.

But the answer says it is not continuous.So I am doing wrong or may be thinking in two dimension instead of three.Please just give me some hints,I hopefully can manage the rest.

Why is the negative $x$ axis mapped to $0$ and not $\pi$ and why is the positive $y$ axis mapped to $\pi$ rather than $\frac{\pi}{2}$? Similar question for the negative $y$ axis. — John Wayland Bales, Nov 21 '16 at 06:55
If we fix x_0 and vary the y co-ordinate; then fix y_0 and vary the x co-ordinate That doesn't prove continuity in $(x,y)$, just directional continuity along some lines parallel to the axes. Consider instead $(r \cos \theta, r \sin \theta)$ with $\theta \to \pi^+$ vs. $\theta \to \pi^-$. — dxiv, Nov 21 '16 at 07:11
@dxiv: But I thought a function from a product space to another was continuous iff both the co-ordinate functions were continuous. — user118494, Nov 21 '16 at 07:47
See for example Discontinuous functions that are continuous on every line in R^2. — dxiv, Nov 21 '16 at 07:55
This is basically argument function. There were some posts about this function, for example here and here. — Martin Sleziak, Nov 21 '16 at 09:16

score 1 · Answer 1 · answered Nov 21 '16 at 12:40

1

It is not continuous,and hence not differentiable, at any point in $\mathbb R^- \times \{0\},$ for if $r>0$ then $\lim_{\phi\to (\pi-)}\;\theta (re^{i\phi})=\pi$ while $\lim_{\phi \to (-\pi)+}\;\theta (re^{i\phi})=-\pi.$

answered Nov 21 '16 at 12:40

DanielWainfleet

57,985

$\theta (r\cos \theta,r\sin \theta)=\theta :\mathbb R^2\rightarrow \mathbb R$ is differentiable $?$

1 Answers1