Continuous function on the open unit disc in $\mathbb R^2.$

Question

Suppose $U=(0,\frac{1}{2})\times(0,\frac{1}{2}),V=(-\frac{1}{2},0)\times (-\frac{1}{2},0)$ and $D$ be the open unit disc with centre at the origin of $\mathbb R^2.$ Let $f$ be a real valued continuous function on $D$ such that $f(U)=0.$ Then is follows that

1. $f(v)=0$ for every $v$ in $V$
2. $f(v)\neq 0$ for every $v$ in $V$
3. $f(v)=0$ for some $v$ in $V$
4. $f$ can assume any real value on $V$.

If we take $f(x,y)=\sqrt{x^2+y^2}$ for all $(x,y)\in V$ then $(1)$ and $(3)$ are not true.

If we take $f\equiv0$ on $\mathbb R^2$ then $(2)$ is not true.

So I think $(4)$ is true. Am I correct?

Any help is appreciated. Thank you.

Answer 1

Counterexamples 1. $f(x,y) = 0,x\ge 0,$ $f(x,y) = x,x\le 0.$ 2. $f\equiv 0.$ 3. See 1.

4. Is it possible for $f(V)=\mathbb R?$ No, because $V$ is contained in a compact subset of $D$ (namely $[-1/2,0]\times [-1/2,0].$) Hence $f(V)$ is bounded. Is it possible, given an arbitrary $a\in \mathbb R,$ to construct an $f$ satisfying the given hypotheses such that $f=a$ somewhere in $V?$ Sure: Let $f$ be as in 1. above. Then the function $-4af$ takes on the value $a$ at $(-1/4,-1/4)\in V.$