Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
what is $a$ exactly? –  akkkk Oct 3 '12 at 14:10
add comment

2 Answers

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

share|improve this answer
    
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
add comment

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 $\langle x_n,a \rangle - \langle x_m,a \rangle = \langle x_n-x_m,a \rangle$ ...

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.