Can *YOU* understand this functional analysis proof?

[AxiomOfChoice]

My professor tried to show the following in lecture the other day: If T is a linear operator on a Hilbert space and (Tz,z) is real for every z in H, then T is bounded and self-adjoint.

Below, I use (*,*) to indicate the Hilbert space inner product.

He told us to use the identity (which I've worked out and verified)

$$(Tx,y) = (1/4)[(Tx+Ty,x+y) - (Tx-Ty,x-y) + i(Tx+iTy,x+iy) - i(Tx-iTy,x-iy)].$$

But this is also supposed to be equal to

$$(x,Ty) = \overline{(Ty,x)}.$$

Of course, this would complete the proof, since then we'd have (x,Ty) = (Tx,y).

I'm not sure how to show this. Our professor told us to "polarize" - i.e., to use the polarization identity - but I can't figure it out. The identity he gave us looks kind of like the polarization identity, but they're clearly different.

Also, we're supposed to use (Tx,x) and (Ty,y) are real for all x,y in H. But the identity above holds regardless of whether this is true! So I guess we're just supposed to use "(Tx,x) and (Ty,y) are real" when we show that identity also equals (x,Ty).

Does anyone see what to do? Thanks!

 

[AxiomOfChoice]

UPDATE: Looking through one of my textbooks (Stein-Shakarchi's Real Analysis), I see that the identity I listed (the one with 1/4's and i's and such) is actually referred to as a polarization identity.

I'm still clueless as to where to go from here though :(

 

[AlephZero]

If (Tz,z) is real for any z in H, the four inner products in the identity are all real.

Write down the values of (Tx,y) and (Ty,x) using the identity and compare them term by term.

 

[tiny-tim]

Hi AxiomOfChoice!

In that (Tx,y) formula, swap x with y to get (Ty,x), then take the conjugate.

 

[AxiomOfChoice]

Yep! That did it! Thanks, guys! Don't know why I didn't think of that...oh yeah! I'm an idiot!