Is it necessary that either $f(x) \leq 0$ for all $x$, or, $f(x) \geq 0$ for all $x$

Question

Let $f : \mathbb{R}^2 \to \mathbb{R}$ be a continuous map such that $f(x) = 0$ for only finitely many values of $x$. Is it necessary that either $f(x) \leq 0$ for all $x$, or, $f(x) \geq 0$ for all $x$

I don't think its necessary because if $f(x,y)=x$ then $f$ is continous and it takes both positive and negative values.

how many pairs $(x,y)$ make your proposed function equal to $0$? — Mark S., Dec 22 '18 at 03:02
Your example $f(x,y)=x$ has infinitely many zeros -- $f(0,k)=0$ for any real $k.$ — coffeemath, Dec 22 '18 at 03:02
What does $f(x)$ mean if $f$ has domain $\mathbb{R}^2$? Is $x$ a point in the plane? — Randall, Dec 22 '18 at 03:15

score 4 · Accepted Answer · answered Dec 22 '18 at 03:37

Yes, this statement is true. And one idea could be to prove it by contradiction.

So, assume that $f(\mathbf{x})=0$ only for finitely many $\mathbf{x}\in\mathbb{R}^2$, and yet it's not true that either $f(\mathbf{x})\le0$ for all $\mathbf{x}\in\mathbb{R}^2$ or $f(\mathbf{x})\ge0$ for all $\mathbf{x}\in\mathbb{R}^2$. The last part means that there exist at least one $\mathbf{x}_1\in\mathbb{R}^2$ such that $f(\mathbf{x}_1)<0$ and at least one $\mathbf{x}_2\in\mathbb{R}^2$ such that $f(\mathbf{x}_2)>0$. Now the key idea is that there are infinitely many paths connecting the points $\mathbf{x}_1$ and $\mathbf{x}_2$ in the plane, and since $f$ is continuous, there exist points at which $f$ is zero on each such path.

For example, let's say $f(-1,0)<0$ and $f(1,0)>0$. Consider $f$ restricted to the arcs of parabolas $y=a(x^2-1)$, $-1\le x\le1$, for all $a>0$. All these arcs connects these two points, and at the same time they don't have any common points other than the endpoints. Since $f$ is continuous along each arc and goes from being negative to being positive, $f$ is equal to zero at least once on each of them.

do you mean that considering over those distinct paths we can apply intermediate value theorem of real variable to get infinitely many paths? I can imagine it how it works but how do I write it properly? — , Dec 22 '18 at 04:49

score 3 · Answer 2 · answered Dec 22 '18 at 03:37

Given two points $A$ and $B$ in the plane, consider the family of all circular arcs including both points. Each such arc is the locus of $X$ such that $\angle AXB = \theta$ for some fixed $\theta$ (as a signed angle), so they don't overlap at all.

There are infinitely many such arcs. If $f$ is continuous and $f(A)$ and $f(B)$ have opposite signs, then each arc must contain a point $X$ at which $f(X)=0$. As such, there are infinitely many points at which $f$ is zero.

There it is - the contrapositive of your statement. If a continuous function from $\mathbb{R}^2$ to $\mathbb{R}$ takes both positive and negative values, then it has infinitely many zeros. Done.

Is it necessary that either $f(x) \leq 0$ for all $x$, or, $f(x) \geq 0$ for all $x$

2 Answers2