# Determine if a vector valued function has a root based on the behaviour on the boundary

Say $$f: \mathbb R^n \to \mathbb R^n$$ is a vector valued continuous function. Say we further know that $$t^\top f(t) \ge 0$$ $$\,\forall \,\,||t||=1$$. Does this imply that $$f$$ has a root in the unit ball.

This is pretty trivial to prove when $$n=1$$ as you have that $$f(1)\ge0$$ and $$f(-1)\le 0$$. I can further show that each individual component of $$f$$ will have some root but I cannot show that a single $$t$$ can be the root for each of the component at the same time.

First of all, let me change notation: we are given a function $$f\colon \mathbb R^n\to \mathbb R^n$$ such that $$x^Tf(x)\ge 0$$ for $$\lvert x \rvert=1$$. (I can't write $$t^T$$, that seems too terrible to me. :-) ).
The answer is affirmative. Indeed, define $$g(x):=-f(x)$$. Now we have a continuous function such that $$x^Tg(x)\le 0$$ for $$\lvert x \rvert=1$$. We can thus invoke this recent question from MathOverflow: https://mathoverflow.net/q/463957/13042, showing that $$g$$ must vanish somewhere for $$\lvert x \rvert\le 1$$. Therefore, $$f$$ must vanish also.