MHB Cauchy-Schwarz inequality for pre-inner product

[mozganutyj]

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]

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]

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.