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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.

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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.

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  • $\begingroup$ That's why I use \top ;) The timing of the two questions was really lucky haha, thanks! $\endgroup$
    – Anvit
    Commented Feb 14 at 20:53

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