Prove that $f^{-1}((- \infty ,0 )) \subset B(0; \|a\|)$

Question

Given $f : \mathbb{R}^n \rightarrow \mathbb{R}$ with $n \geq 2$ and $$f(x) = \langle x, x \rangle ^2 - \langle a,x \rangle ^2, a \in \mathbb{R}^n \setminus \{0\}, $$ Prove that $f^{-1}((- \infty ,0 )) \subset B(0; \|a\|)$.

I know that the pull-back image $f^{-1}((- \infty ,0 ))$ gives us all $x$ with $\langle x, x \rangle ^2 < \langle a,x \rangle ^2$, but how does follow from this that $x \in B(0, \|a \|)$?

Answer 1

$f^{-1}((-\infty, 0)) \subseteq B(0; \|a\|)$ is equivalent to saying that "$\forall x \in \mathbb R^n$ if $ f(x) \le 0$ then $\|x\| \le \|a\|$", or the contrapositive claim "$\forall x \in \mathbb R^n$, if $\|x\| > \|a\|$ then $f(x) > 0$".

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So, let $x \in \mathbb R^n$ with $\|x\| > \|a\|$. Then $f(x) := \|x\|^2 - a^Tx \ge \|x\|^2 - \|a\|\|x\| \ge \|x\|^2 - \|a\|^2 > 0$, where

• the first inequality is by Cauchy-schwarz, namely $|a^Tx| \le \|a\|\|x\|$, and
• the second is by hypothesis, namely $\|x\| > \|a\|$.