MHB How is this inequality obtained?

Boromir
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Introduction to Operator Theory and Invariant Subspaces - B. Beauzamy - Google BooksIn page 144 of this preview I don't know how they obtain the inequality in (1). It looked like cauchy schwarz but I don't think it is.

I also don't know how they connect the norm of the integral to the supremum of an inner product.

Finally, is it necessary to show the integral is bounded? Is that the motivation?
 
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Re: integral inequality

Boromir said:
Introduction to Operator Theory and Invariant Subspaces - B. Beauzamy - Google BooksIn page 144 of this preview I don't know how they obtain the inequality in (1). It looked like cauchy schwarz but I don't think it is.

I also don't know how they connect the norm of the integral to the supremum of an inner product.

Finally, is it necessary to show the integral is bounded? Is that the motivation?
On my computer, the Google Books preview of this text only goes up to p.139. Please state the problem and its context more explicitly.
 
Re: integral inequality

Opalg said:
On my computer, the Google Books preview of this text only goes up to p.139. Please state the problem and its context more explicitly.

How can it be different for different computers?
 
Re: integral inequality

Boromir said:
How can it be different for different computers?

On my computer pages 144-151 are not shown in the preview. It would really be best if you took the time to state the problem yourself.
 
Re: integral inequality

Boromir said:
Opalg said:
On my computer, the Google Books preview of this text only goes up to p.139. Please state the problem and its context more explicitly.

How can it be different for different computers?
It depends entirely on what Google's server chooses to send to different users.
 
Here is what the inequality says. It seems like a bunch of notation is required to be explained to make sense of this. But it seems like Opalg said just CS inequality.

$$
\left| \left< (\int |f| ~ dE)x,y\right> \right|^2 \leq \left< (\int |f| ~ dE)x,x \right> \left< \int (|f| ~ dE)y,y\right> $$
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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