Can a continuous function send bounded domain to unbouded domain?

Let $$f:D=\overline{D(a,r)}\subset \mathbb{C} \to \mathbb{C}$$ be a continuous function whose image of $$D'=\{z \ | \ |z-a|=r\}$$ is homeomorphic to circle $$S^1$$.

Then I guess the image of $$f$$ should be in the interior of $$f(D')$$. But I couldn't prove it.

I tried to find a counterexample. A almost counterexample is $$f:\overline{D(0,1)}-\{0\}\to \mathbb{C}$$ defined by $$z \to \frac{1}{z}$$. It sends a punctured disc $$D(0,1)-\{0\}$$ to unbounded region. But the domain is not disc.

My guess is true? If it is wrong, please suggest any counter example.

Thank you very much!

• Let $f$ map the point with polar coordinates $(r,\phi)$ to $(2-r,\phi)$ if $r\geq1/2$ and to $(3r,\phi)$ if $r\leq1/2$. – Andreas Blass Aug 1 '19 at 21:52
• I'm confused... are you asking if a continuous function can send a bounded domain to an unbounded domain, or are you concerned about the image of the boundary of a closed disk? What, precisely, is your question? – Xander Henderson Aug 1 '19 at 21:53
• Dear Blass, thank you. Your example is what I am expecting to. – user29422 Aug 2 '19 at 4:57
• Just play around with the example $\tfrac{1}{z}$ by shifting around the $z$ variable and scaling appropriately so it blows up along the boundary. – Brevan Ellefsen Aug 3 '19 at 4:17

2 Answers

Firstly, $$\overline{D(a,r)}$$ is compact, so the image of it by $$f$$ is compact, and consequently bounded.

Secondly, $$f(\overline{D(a,r)})$$ needs not to be bounded by $$f(D')$$ (nevertheless it is possible to show that $$f(\overline{D(a,r)})$$ must contain the bounded component of $$C \backslash f(D')$$). Intuitively, the image of $$f$$ might "go out and go in".

For instance it is possible to construct $$f : \overline{D(0, 2)} \rightarrow C$$ such that $$f$$ sends $$\overline{D(0,1)}$$ to $$\overline{D(0,2})$$ by multiplication by $$2$$, $$\overline{D(0,2)} \backslash D(0,1)$$ to $$\overline{D(0,2)}\backslash D(0,1)$$ by the natural inversion (i.e $$z\rightarrow 2\frac{z}{|z|^2}$$). Thus $$f(D')$$ is the circle of radius $$1$$, and $$f(\overline{D(0, 2)})$$ goes a little out of this circle (it goes to the circle of radius 2).

• I think $f(D’)$ is the circle of radius of 2 in your instance. – user29422 Aug 2 '19 at 4:48

Consider $$h(x, t) =\begin{cases} x & x < 0\\ \min(xt, 1) & x \ge 0 \end{cases}$$ for $$t = 1$$, this is the identity. For $$t = 2$$, it sends all of $$[1/2, 1]$$ to $$1$$. It sends the interval $$[-1, 1]$$ to $$[-1, 1]$$, no matter what value of $$t\ge 1$$ you pick.

Now define $$f(x, y) = (h(x, 2 - |y| )\sqrt{1 - y^2}, y)$$ That sends the unit disk to the unit disk. But the entire segment $$[0.5, 1] \times \{0\}$$ gets sent to $$(1, 0)$$, so the interior of the disk does not map to the interior of the disk.

Yeah, this is needlessly complicated, but it was the first thing I thought up.

Here's the second thing I thought up:

In polar coordinates, send $$(r, \theta)$$ to $$(\min(2r, 1), \theta)$$.

That maps the unit disk to itself continuously, but the annulus between $$r = 0.5$$ and $$r = 1$$ is all sent to the unit circle in the codomain, hence NOT in the interior of the image.