Let \(f:\mathbb{R}^2 \to \mathbb{R} \) be a continuous map such that \(f(x)=0 \) for only finitely many values of \(x\). Which of the following is true?

A. either \(f(x) \le 0 \)for all \(x\) or \(f(x) \ge 0 \) for all \(x\).

B. the map \(f\) is onto

C. the map \(f\) is one-to-one

D. none of the above


Let \(f\) be the map \(f(x_1,x_2)=x_1^2 +x_2^2 \). Then \(f\) is zero at only (0,0). \(f\) is continuous because \(x=(x_1,x_2) \to x_1 \to x_1^2 \) is continuous from \(\mathbb{R}^2 \to \mathbb{R} \to \mathbb{R} \). i.e, each arrow is continuous. The first arrow is the projection map, and such maps are always continuous, and the second arrow is just squaring, which is continuous. And composition of continuous functions are continuous, so \(x \to x_1^2 \) is continuous function from \(\mathbb{R}^2 \to \mathbb{R}\). Where here and henceforth \(x=(x_1,x_2)\in \mathbb{R}^2 \).

Similar reasoning will show that \(x \to x_2^2 \) is continuous function from \(\mathbb{R}^2 \to \mathbb{R}\).

Sum of continuous functions is continuous, so the map \(x \to x_1^2+ x_2^2 \) is continuous function from \(\mathbb{R}^2 \to \mathbb{R}\).

This function \(f\) is not one-one since \(f(1,0)=f(0,1)=1\) and it is not onto since it only takes values in \([0,\infty ) \).

So we now are sure that B,C are false options.

We will prove A now.

Let \(x=(x_1,x_2)\) and \(y=(y_1,y_2)\) be two points in \(\mathbb{R}^2\) such that \(f(x)>0\) and \(f(y)<0\).

We will prove that this will imply infinitely many zeros in between \(x\) and \(y\). But wait a second… what does between mean in this context? For that we consider the paths between \(x\) and \(y\). Note that there are infinitely many paths between any two points in \(\mathbb{R}^2\). Further, we can in fact have infinitely many paths completely disjoint except for the initial and final points. We show that corresponding to each path \(\alpha: [0,1] \to \mathbb{R}^2\) which connects \(x\) and \(y\) we have a zero in the path. Since there are infinitely many disjoint paths, we get infinitely many distinct zeros for \(f\).

Now, \(\alpha: [0,1] \to \mathbb{R}^2\) is a continuous function , \( \alpha(0)=x \), \(\alpha(1)=y\).

Consider the composition \(g=f o \alpha : [0,1] \to \mathbb{R}\). \(g\) is continuous. g(0)=f(x)>0 and g(1)=f(y)<0.

Therefore by the intermediate value theorem, \(g(c)=0\) for some \(c\in [0,1]\).

That means, \(f(\alpha(c))=0\). And using the discussion above we get a contradiction.

This proved the option A.