Constant sign

Let $f:\mathbb{S} \subset \mathbb{R}^3 \rightarrow \mathbb{R}$ be a non zero continuous function on a connected surface $\mathbb{S}$. Prove that $f$ does not change sign on $\mathbb{S}$.

Solution



Suppose that $f(p)<0, \; f(q)>0$  for some $p, q \; \in \mathbb{S}$. Obviously, \(\displaystyle{p\neq q}\) because if $p=q$, then $0<f(q)=f(p)<0$ and we have a contradiction. Since $p\,,q\in \mathbb{S}$ and $p \neq q$ and the regular surface $\mathbb{S}$  is connected, there exists a continuous curve $\gamma:[a, b] \rightarrow \mathbb{S}$ on $\mathbb{S}$, such that

$$\gamma(a)= p, \;\;\; \gamma(b)=q$$

We define $g=f \circ \gamma:[a, b] \rightarrow \mathbb{R}$ as $g(t)=(f\circ \gamma)(t)= f(\gamma(t))$ . this function is obviously continuous on $[a, b]$.

Also  $g(a)\cdot g(b)=f(\gamma(a))\cdot f(\gamma(b))=f(p)\cdot f(q)<0$. So by Bolzano's theorem, there exists $t_0\in(a, b)$  such that $g(t_0)=f(\gamma(t_0))=0$, where $\gamma(t_0)\in \mathbb{S}$, a contradiction, since $f(x)\neq 0, \;\forall x\in \mathbb{S}$

 Therefore the function $f$ does not change sign of $\mathbb{S}$.

This exercise can also be found in mathimatikoi.org