Constant sign

[Tsakanikas Nickos]

Prove that if \( \displaystyle f : S \subset \mathbb{R}^{3} \longrightarrow \mathbb{R} \) is a nonzero continuous function defined on a connected surface \( \displaystyle S \), then \( \displaystyle f \) does not change sign on \( \displaystyle S \).

[Papapetros Vaggelis]

Hello Nickos.

Suppose that \(\displaystyle{f(p)<0\,,f(q)>0}\) for some \(\displaystyle{p\,,q\in S}\). Obviously, \(\displaystyle{p\neq q}\) because if

\(\displaystyle{p=q}\), then \(\displaystyle{0<f(q)=f(p)<0}\), a contradiction. Since \(\displaystyle{p\,,q\in S\,,p\neq q}\) and the regular surface

\(\displaystyle{S}\) is connected, there exists a continuous cuve \(\displaystyle{c:\left[a,b\right]\longrightarrow S}\) on S, such that

\(\displaystyle{c(a)=p\,,c(b)=q}\) (or \(\displaystyle{c(a)=q\,,c(b)=p}\)) .

We define \(\displaystyle{g=f\circ c:\left[a,b\right]\longrightarrow \mathbb{R}\,,g(t)=\left(f\circ c\right)(t)=f(c(t))}\). This function is a well defined

real-valued function of one variable. Also, this function is continuous at \(\displaystyle{\left[a,b\right]}\) as a composition of continuous functions.

Also, \(\displaystyle{g(a)\cdot g(b)=f(c(a))\cdot f(c(b))=f(p)\cdot f(q)<0}\). According to \(\displaystyle{\rm{Bolzano's}}\) theorem, there exists

\(\displaystyle{t_0\in\left(a,b\right)}\) such that \(\displaystyle{g(t_0)=f(c(t_0))=0}\), where \(\displaystyle{c(t_0)\in S}\), a contradiction, since

\(\displaystyle{f(x)\neq 0\,,\forall\,x\in S}\) .

Therefore, the function \(\displaystyle{f}\) does not change sign on \(\displaystyle{S}\) .