Existence of a Zero of a Continuous Map

Question

Suppose that $f: \mathbb{R}^n \to \mathbb{R}^n$ is a continuous map, satisfying $$ \langle x,f(x) \rangle \geq 0, \forall x\in S^{n-1}=\{x\in \mathbb{R}^n: \Vert x\Vert =1 \} .$$ Prove that there exists $x\in B(0,1)=\{x\in\mathbb{R}^n: \Vert x\Vert\leq 1\}$, such that $f(x)=0$.

When $n=1$, this is just the intermediate value theorem. And the general case seems to be closely related to topology, like fixed point theorem. Since I haven't learned nonlinear functional analysis or algebraic topology, I have no idea how to get started. What theorem can I use here? And is there an elementary proof that uses only mathematical analysis?

Answer 1

Idea:

We will use induction, with the base case being $n=1$ proved by intermediate value theorem.

For $n=k+1$:

Divide $B(0,1)$ into slices parametrized with $t$ with $t \in [-1,1]$: $$B(0,1,t) = \{x \in B(0,1) \mid x_1 = t\}$$ where $x_1$ denotes the first coordinate of $x$.

So, for $t=\pm1$, $B(0,1,t)$ is a point, and for $t \in (-1,1)$, $B(0,1,t)$ is a ball one dimension lower.

For $t \in (-1,1)$, define $f_t : B_k(0,1) \to \Bbb R^k$ as follows, where $B_k$ denotes the ball of dimension $k$: $$f_t(x) = (\operatorname{sgn} t) f(t,\sqrt{1-t^2}x)_{2,\cdots,n}$$ where $\operatorname{sgn}$ dentoes signum.

Then, by the induction hypothesis, there is $x \in B_k(0,1)$ such that $f(x)=(g(t),0)$, where $g$ is a function $(-1,1) \to \Bbb R$.

In the first line I wrote "idea" instead of "proof", because one needs to justify that there is a choice of such $x$ for each $t$ such that $g$ is continuous and can be extended to $g':[-1,1] \to \Bbb R$ with $g'(-1) = f(-1,0,\cdots,0) \le 0$ and $g'(1) = f(1,0,\cdots,0) \ge 0$. (No, $g'$ is not about derivatives).

Then, using the base case, there is $t \in [-1,1]$ with $g(t)=0$, which becomes a zero of $f$.

Answer 2

[This proof uses degree theory.]

If $f(x) = 0$ for some $x\in S^n$ then we are done.

Assume now that $f(x)\neq 0$ for every $x\in S^n$.

Let $h\colon [0,1]\times B\to\mathbb{R}^n$ be the continuous map defined by $$ h(t,x) := (1-t)x + t f(x). $$ We have that $h(t,x)\neq 0$ for every $t\in [0,1]$ and for every $x\in S^n$. Namely, if $t=1$ then $h(1,x) = f(x) \neq 0$ for every $x\in S^n$, whereas for $t\in [0,1)$ one has $$ x\cdot h(t,x) = (1-t)|x|^2 + t x\cdot f(x) \geq 1-t > 0 \qquad \forall t\in [0,1),\ \forall x\in S^n. $$ Hence $\text{deg}(f, B, 0) = \text{deg}(id, B, 0) = 1$, so that $f$ has a zero in $B$.

Answer 3

I will elaborate my comment here. We use the fact that the identity map of $S^n$ is not homotopic to any constant map, and I will argue that there can be no simpler proof than this, since this fact (along with a lot of more elementary stuff) is equivalent to the statement to be proved.

First, assume for contradiction that $0 \notin f(B)$ ($B = B(0,1)$). Then define $g:S^n\rightarrow S^n$ by $g(x) = \frac{f(x)}{|f(x)|}$. We first show that this is homotopic to the identity. The homotopy is simply given by $F(t,x) = \frac{(1-t)x +tg(x)}{|(1-t)x +tg(x)|}$, which is defined and continuous if the denominator is nowhere zero. We compute \begin{equation} |(1-t)x +tg(x)|^2 = (1-t)^2|x|^2 + t^2 g(x)^2 + 2t(1-t)x\cdot g(x). \end{equation} By the condition on $f$ each of these terms is nonnegative, and by our "assumption-for-contradiction" one of the first terms is always positive. So $F$ is a continuous map. Clearly $F(0,x)=x$ and $F(1,x)= g(x)$.

Next we show that $g$ is also homotopic to a constant map. This homotopy is given by $G(t,x) = \frac{f(tx)}{|f(tx)|}$, which is a well-defined continuous map by our "assumption-for-contradiction". Clearly $G(1,x) = g(x)$ and $G(0,x) = x_0 = \frac{f(0)}{|f(0)|}$. Now since homotopy is an equivalence relation (in particular transitive), this means that the identity of $S^n$ is homotopic to a constant map, which is not true. Therefore the assumption $0\notin f(B)$ is false!

Now I'll attempt to show that you need to use the fact that the identity is not homotopic to a constant map, and by this I mean that the "zero-statement" you wish to prove in fact implies the "homotopy-statement". To show this, assume the converse of the "homotopy-statement", i.e. that there exists a homotopy $F:I\times S^n\rightarrow S^n$ from the identity of $S^n$ to a constant map. From this we construct a counterexample to the "zero-statement", which I call $f: \mathbb{R}^{n+1}\rightarrow \mathbb{R}^{n+1}$.

For $|x|\geq 1$, let $f(x) = x$. For $1\geq|x|> 0$ let $f(x) = F(|x|,\frac{x}{|x|})$, and let $f(0) = F(0,-) = x_0$. Here we take $F(1,-)$ to by the identity and $F(0,-)$ to be constant. By the pasting lemma, $f$ is continuous even where $|x| = 1$. It remains to check that it is continuous at $0$. To see this, pick any neighborhood $V$ of $f(0)=x_0$. Then $F^{-1}(V)\subset I\times S^n$ contains the strip $\{0\}\times S^n$, and by the tube lemma also some tube $[0,\varepsilon)\times S^n$. This proves that when $|x|<\varepsilon$, $\ f(x)\in V$, so the open ball $B(0,\varepsilon)$ is mapped into $V$, proving continuity at $0$.

The map $f$ satifies $x\cdot f(x) = x\cdot x =1\geq 0$ for $x\in S^n$, but by construction there is no point at all which is mapped to zero, so it contradicts the "zero-statement"!