23/06/2018
I'm looking for a proof of the following result. I don't know if it is true or not, but it seems to be true. I have tried it from different points of view, but I can not formalize them correctly. My ideas were to apply the intermediate value theorem for each radius or maybe in a topological manner considering the closed sets $Z_2 := \{z\in\mathbb{C}\;|\;g_2(z) = 0\}$ and $Z := \{z\in\mathbb{C}\;|\;g(z) = 0\}$.
Let $g_1,g_2:\overline{\mathbb{D}}\to [0,\infty)$ be continuous functions, where $\mathbb{D}:=\{z\in\mathbb{C}\;|\;|z| < 1\}$, and define $g:= g_1 - g_2$. Suppose that
- $g_2(0) = 0$;
- $g_2(z) > 0$ for all $z\in\mathbb{S}^1$;
- $g_1(z) > 0$ for all $z\in \overline{\mathbb{D}}$.
Then, there exists some simple and closed curve (homeomorphic to $\mathbb{S}^1$) $\gamma\subset \overline{\mathbb{D}}$ such that
- $0\in \text{int }\gamma$;
- $g_2(z) > 0$ for all $z\in \gamma$;
- $g(z) > 0$ for all $z\in \gamma\cup\text{int }\gamma$.
Solved. It is false and a counterexample given by a collegue at the University of Barcelona (Jordi) is the following: $g_1(z) = 1$ for all $z\in \overline{\mathbb{D}}$ and $g_2 := g_2' + g_2''$, where \begin{equation} g_2'(z) = \begin{cases} 0, & \text{if }|z| \leq b\\ |z| - b,& \text{if }|z| > b \end{cases} \end{equation} and \begin{equation} g_2''(z) = \begin{cases} 1 - \frac{2}{a}|z - a|, & \text{if }|z - a| \leq \frac{a}{2}\\ 0,& \text{if }|z - a| > \frac{a}{2} \end{cases} \end{equation} for some $0<b<1$ and $a < b/2$.
25/06/2018
I forgot an extra hypothesis, which is that the connected components of $Z_2$ are simply connected, so that this is the new statement:
Let $g_1,g_2:\overline{\mathbb{D}}\to [0,\infty)$ be continuous functions, where $\mathbb{D}:=\{z\in\mathbb{C}\;|\;|z| < 1\}$, and define $g:= g_1 - g_2$. Assume that the connected components of $Z_2:= \{z\in\mathbb{C}\;|\;g_2(z) = 0\}$ are simply connected and that
- $g_2(0) = 0$;
- $g_2(z) > 0$ for all $z\in\mathbb{S}^1$;
- $g_1(z) > 0$ for all $z\in \overline{\mathbb{D}}$.
Then, there exists some simple and closed curve (homeomorphic to $\mathbb{S}^1$) $\gamma\subset \overline{\mathbb{D}}$ such that
- $0\in \text{int }\gamma$;
- $g_2(z) > 0$ for all $z\in \gamma$;
- $g(z) > 0$ for all $z\in \gamma\cup\text{int }\gamma$.
It's still false.
Let $X$ be a copy of the Warsaw circle embedded into $D$ with $(0,0)$ on $X$. Note that the Warsaw circle is simply connected.
Define $g_1$ to be the constant $1$ function. Let $g_2 = C *dist(p,X)$ for some positive constant $C$. Pick $C$ large enough so that $g_2 > g_1$ for some point "inside" $X$. Since $X$ is compact, $Z_2 = X$. Also, $g_2(0) = 0$ since $0\in X$, and $g_2 > 0$ for any point not on $X$.
Now, let $\gamma$ be any curve satisfying all the relevant hypothesis. Since $0$ is in the interior of $\gamma$, $\gamma$ must contain some points on the "outside" of $X$. Because the image of $\gamma$ cannot intersect $X$, $\gamma$ must go around $X$. But then $\operatorname{int} \gamma$ contains points where $g<0$.