0
$\begingroup$

Suppose M is any linear manifold in H. H is a hilbert space. Define the orthogonal complement of M to be $$M' =\{f \in H | \langle f,g\rangle= 0 ,\forall g\in M\}.$$

To see that M' is a closed linear manifold, suppose $\{u_n\}$ is a sequence in M', and that $\lim_{n\to \infty}u_n=u$ is in $H$. For every $g\in M$ $$\langle u,g\rangle =\langle\lim_{n\to \infty}u_n,g\rangle = \lim_{n\to\infty} \langle u,g\rangle =0.$$

The interchange of limits here is valid since (once again, by the Schwarz inequality) the linear operator $\langle u, g\rangle$ is a bounded linear operator for any fixed $g\in M$.

I dont understand the "interchange of limits" Where is the second limit?

$\endgroup$

1 Answer 1

1
$\begingroup$

It's just the English that is messing up with you. The plural is used like when you say "crossing streets without paying attention is dangerous"; where's the "second street"?

And interchange refers to exchanging the position of $\lim$ and $\langle$.

$\endgroup$
2
  • 1
    $\begingroup$ it says the interchange of limits? @Martin Argerami $\endgroup$ May 5, 2015 at 17:26
  • $\begingroup$ Please see the edit. $\endgroup$ May 5, 2015 at 17:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .