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Cauchy-Schwarz inequality for pre-inner product

mozganutyj

New member
Jan 12, 2014
2
Dear all,

I've encountered some problems while looking through the book called "Operator Algebras" by Bruce Blackadar.
At the very beginning there is a definition of pre-inner product on the complex vector space: briefly, it's the same as the inner product, but the necessity of x=0 when [x,x]=0 holds is omitted.

The point is that this stuff is followed by the script that "Cauchy-Schwarz inequality" holds for pre-inner product and here is the place I'm stuck in:

I can't prove the trivial case: [y,y]=0 implies [x,y]=0 for every x (otherwise the CBS inequality won't hold as the right-hand side equals to zero in this case - therefore, the only chance for the CBS inequality to be satisfied is [x,y]=0).

I suspect that CBS holds when we've got the inner product, not only the pre-inner.

I'll appreciate every effort of assistance from your side!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,713
Dear all,

I've encountered some problems while looking through the book called "Operator Algebras" by Bruce Blackadar.
At the very beginning there is a definition of pre-inner product on the complex vector space: briefly, it's the same as the inner product, but the necessity of x=0 when [x,x]=0 holds is omitted.

The point is that this stuff is followed by the script that "Cauchy-Schwarz inequality" holds for pre-inner product and here is the place I'm stuck in:

I can't prove the trivial case: [y,y]=0 implies [x,y]=0 for every x (otherwise the CBS inequality won't hold as the right-hand side equals to zero in this case - therefore, the only chance for the CBS inequality to be satisfied is [x,y]=0).

I suspect that CBS holds when we've got the inner product, not only the pre-inner.

I'll appreciate every effort of assistance from your side!
Hi mozganutyj and welcome to MHB!

The proof that $[y,y]=0$ implies $[x,y]=0$ is the same as the standard proof of the CBS inequality. If $[y,y] = 0$ then for every scalar $\lambda$ $$0\leqslant [\lambda y+x, \lambda y+x] = 2\text{re}\lambda[y,x] + [x,x].$$ By taking $\lambda = -t\overline{[y,x]}$ where $t$ is large and positive, you get a contradiction.

Don't hesitate to come back here if you have queries about Bruce Balackadar's book. As you can see from my user name, Operator algebras are my speciality.
 

mozganutyj

New member
Jan 12, 2014
2
Thanks, Opalg, for warm greetings!

Yes, the proof is pretty straightforward indeed - I've stumbled over the moment when I've decided to prove it in different way, than the standard technique. Maybe, Blackadar has trapped me (Smile) when he had considered the [y,y]=0 case as trivial, thus making me think of even more simplified machinery to prove it (Happy).

Once again - I'm grateful to you, dear Opalg, for your help.
God bless you ... and save the Queen for sure (Happy)

Hi mozganutyj and welcome to MHB!

The proof that $[y,y]=0$ implies $[x,y]=0$ is the same as the standard proof of the CBS inequality. If $[y,y] = 0$ then for every scalar $\lambda$ $$0\leqslant [\lambda y+x, \lambda y+x] = 2\text{re}\lambda[y,x] + [x,x].$$ By taking $\lambda = -t\overline{[y,x]}$ where $t$ is large and positive, you get a contradiction.

Don't hesitate to come back here if you have queries about Bruce Balackadar's book. As you can see from my user name, Operator algebras are my speciality.