3

Let $B = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2\le 1\}$ and $D = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 < 1\}$. Please help me to pick out the true statements.

(a) Given a continuous function $g : B \to \mathbb{R}$, there always exists a continuous function $f : \mathbb{R}^2 \to \mathbb{R}$ such that $f = g$ on $B.$

(b) Given a continuous function $g : D \to \mathbb{R}$, there always exists a continuous function $f : \mathbb{R}^2 \to \mathbb{R}$ such that $f = g$ on $D.$

(c) There exists a continous function $f : \mathbb{R}^2 \to \mathbb{R}$ such that $f = 1$ on the set $\{(x, y) \in \mathbb{R}^2 : x^2+y^2 = \frac{3}{2}\}$ and $f= 0$ on the set $B\cup\{(x, y) \in \mathbb{R}^2 : x^2+y^2 \ge 2\}$.

Do I need to use Pasting/Gluing Lemma?

Sriti Mallick
  • 6,137
  • 3
  • 30
  • 64

3 Answers3

2

Hint for (b): Consider the continuous function $\frac{1}{1-x^2-y^2}$ on $\{(x,y):x^2+y^2<1\}$.

Cameron Williams
  • 29,432
1

Since @CameronWilliams has answered (b) fully, I deal with the other two only.

Calling in a deep theorem to answer (a) and (c) seems to me to be overkill. For (a), define $f$ by extending it radially. That is, if $x^2+y^2\ge1$, let $f(x,y)=g(x/r,y/r)$ where $r=\sqrt{x^2+y^2}$. This makes $f$ constant on radii outside the unit disk.

For (c), the requirements also are radial, so that you can make the behavior the same on all the rays emanating from the origin. So, define $r(t)$ to be zero on $[0,1]$ as well as on $[2,\infty\rangle$, then $2x-2$ on $[1,3/2]$, and $4-2x$ on $[3/2,2]$. You see that $r$ is a piecewise linear continuous function on $[0,\infty\rangle$, onto $[0,1]$. Now let $f(x,y)=r(x^2+y^2)$.

Lubin
  • 62,818
0

Take a look at the Tietze Extension Theorem.

LASV
  • 3,108