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Let $\mathcal H$ a Hilbert space over $\mathbb R$ and $A = \{x\in \mathcal H : \langle x, a \rangle \geq 1 \}$. I'm trying to prove that $A$ is closed.

Let $(x_n) \subset A$ be a Cauchy-sequence. Since $\mathcal H$ is a Hilbert space, the sequence converges to an $x \in \mathcal H$.

Suppose the sequence $(\langle x_n, a \rangle)\subset \mathbb R$ is also Cauchy. Since $[1,\infty)$ is closed, its limit, $\langle x, a \rangle$, is also in $[1,\infty)$. So $x \in A$, and $A$ is closed.

However, why is the sequence $(\langle x_n, a \rangle)$ Cauchy whenever $(x_n)$ is?

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what is $a$ exactly? –  akkkk Oct 3 '12 at 14:10

2 Answers 2

By Cauchy-Schwartz $\def\skp#1{\left\langle#1\right\rangle}\def\abs#1{\left|#1\right|}$\[\abs{\skp{x_n, a} - \skp{x_m, a}} = \abs{\skp{x_n - x_m, a}}\le \|x_n - x_m\|\cdot \|a\| \to 0, \quad n,m \to \infty. \]

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Thank you, I was trying really weird stuff with the triangle inequality and the parallelogram rule, but totally forgot about Cauchy-Schwarz. –  Garogolun Oct 3 '12 at 14:20

The map $\phi \colon x \mapsto \langle x,a \rangle$ is continuous. Hence $A=\phi^{-1}([1,+\infty))$.

It is a consequence of the Cauchy-Schwartz inequality: for every $x$ and $y$ in $\mathcal{H}$, $$ \left| \langle x,y \rangle \right| \leq \|x\| \|y\|. $$ If you remark that $\langle x_n,a \rangle - \langle x_m,a \rangle = \langle x_n-x_m,a \rangle$ ...

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