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If $f: \mathbb R^2 \rightarrow \mathbb R$ is a continuous function such that $f(x)=0$ for only finitely many values of $x$, then prove/disprove the following :

$(a)$ Either $f(x) \geq 0 ~~\forall x\in \mathbb R^2 $ or $f(x) \leq 0 ~~\forall x\in \mathbb R^2 $

$(b)f $ is onto

$(c)$ $f$ is one-one

Attempt: I have an intuition that the continuous nature of function $f$ would play a part in this problem. however, unfortunately, I am not able to decipher how to move ahead.

Any directions on this problem shall be really great.

Thank you.

MathMan
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    $;f(x,y)=x^2+y^2;$ already disproves (2)-(3). – Timbuc Oct 24 '14 at 04:08
  • Thanks. Any hints on $(1)$ please.. – MathMan Oct 24 '14 at 04:13
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    Suppose $f(a)\gt0$ and $f(b)\lt0$. Now suppose you draw a continuous path from $a$ to $b$. Does $f(x)$ have to be $0$ for some $x$ on that path? Now, how many nonintersecting continuous paths can you draw from $a$ to $b$? – bof Oct 24 '14 at 04:14
  • @bof yeah, :-) $f(x)$ would need to be $0$ for some $x$ on that path if $f(a)>0,f(b)<0$. However, there can still be a single only continuous path from $a$ to $b$? – MathMan Oct 24 '14 at 04:21
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    The function $f$ is defined on the whole plane, right? There are lots of continuous paths from $a$ to $b$. Let $d$ be the distance between $a$ and $b$. For each $D\ge d$ you can draw a circle (two circles if $D\gt d$) of diameter $D$ which passes through $a$ and $b$, and those circles have no other point in common, so you get a continuum of nonintersecting continuous paths (circle arcs) from $a$ to $b$. – bof Oct 24 '14 at 04:26
  • (a) is always true, and so (b) is always false. (c) is false for any continuous function from $\mathbb{R}^2$ to $\mathbb{R}$, even without the other condition; it's false even when the domain is restricted to the boundary of the unit circle, ${ x\in\mathbb{R}^2 ;|; |x|=1 }$. – mjqxxxx Oct 24 '14 at 05:20
  • @mjqxxxx Could you please tell me why $(a)$ is always true? – MathMan Oct 24 '14 at 05:22
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    @VHP I thought I explained why (a) is always true in my last comment? You can easily draw an infinite number of nonintersecting continuous paths between those two points (where $f$ has opposite signs), and $f$ has at least one zero on each of those paths, so $f$ has an infinite number of zeros. – bof Oct 24 '14 at 05:34
  • @bof :-) I was just curious to know if mjqxxxx was also suggesting another approach as well. I understand that there are infinite continuous non intersecting paths between $a$ and $b$ . But, the function $f$ would be well defined and take only one path .. – MathMan Oct 24 '14 at 11:26

1 Answers1

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$f^{-1}\left(0\right)$ has a finite number of elements let's say $(x_1,y_1),(x_2,y_2),\ldots,(x_N,y_N)$. Let $$R=\frac{1}{3}\min_{1\leq i < j\leq N}\sqrt{\left(x_j-x_i\right)^2+\left(y_j-y_i\right)^2}$$ and $$B_k=\{(x,y):\sqrt{(x-x_k)^2+(y-y_k)^2}<R\}\quad\text{for}\;k=1,2, \ldots,N$$ It follows $f$ vanishes at $(x_k,y_k)$ and for no other point of $B_k$. Since $f$ is continuous $f$ have the same sign for all $(x,y)\in B_k\backslash(x_k,y_k)$.

Let's suppose there are two points $(x_a,y_a)\in B_a$ and $(x_c,y_c)\in B_c$ such that $f(x_a,y_a)$ and $f(x_c,y_c)$ has opposite sign, let's say $f(x_a,y_a)<0<f(x_c,y_c)$. Let $g:[0,1]\longrightarrow \mathbb{R}$ be the function defined by $$g(t)=f(tx_c+(1-t)x_a,ty_c+(1-t)y_a)$$ $g$ is a continuous function, Mean Value Theorem implies there is $t_0\in [0,1]$ with $g(t)<0$ for $t<t_0$ (in some neighborhood around $t_0$) and $g(t)>0$ for $t>t_0$ (in some neighborhood around $t_0$). Then $f$ vanishes at $(t_0x_c+(1-t_0)x_a,t_0y_c+(1-t_0)y_a)$ and this point is not in $\{(x_1,y_1),(x_2,y_2),\ldots,(x_N,y_N)\}$. Hence $f$ have the same sign for all $(x,y)\in B_k\backslash(x_k,y_k)$ for $k=1,2,\ldots,N$.

By continuity of $f$ we conclude $f$ has the same sign for all $(x,y)\in\mathbb{R}^2\backslash f^{-1}\left(0\right)$.

Ángel Mario Gallegos
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