Suppose $0 \neq u \in \mathbb{R}^{N},\ \alpha \in \mathbb{R}$ and let $H^{-}(u,\alpha) = \left\{ x \in \mathbb{R}^{N} \mid \langle x,u\rangle \ \lt \alpha \right\}$
I need to prove that $H^{-}(u,\alpha)$ is an open set in $\mathbb{R}^{N}$. The hint suggests using the Cauchy-Schwarz inequality to check the definition is satisfied. I know what the inequality is but I'm not sure how to use it in this question or carry on after that.
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Andrés E. Caicedo
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Mathlete
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1You can use it to show that $x \mapsto \langle x,u\rangle$ is continuous. – Daniel Fischer Nov 16 '13 at 16:08
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How? And why would that help? – Mathlete Nov 16 '13 at 16:08
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It would help because $H^-(u,\alpha) = f^{-1}((-\infty,\alpha))$ is the preimage of an open set under $f \colon x \mapsto \langle x,u\rangle$. And a function is continuous if and only if the preimages of open sets are always open. – Daniel Fischer Nov 16 '13 at 16:11
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OK so how would I go about implementing the CS inequality? – Mathlete Nov 16 '13 at 16:21
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As pointed out in the comments, it suffices to show that $x \mapsto \langle x, u\rangle$ is continuous, and so it suffices to show that $|\langle y, u\rangle|$ is small whenever $y$ is close enough to $0$ - this follows from the linearity of the inner product in the first coordinate (if $z \approx x$, then $x - z \approx 0$). Now this can be done with the Cauchy-Schwarz inequality by noting that
$$|\langle y, u \rangle| \le \|y\| \|u\|$$
Hence if $\|y\| < \epsilon / \|u\|$, then $|\langle y, u\rangle| < \epsilon$, as desired. Since $u$ was fixed, this works.