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Let $T\colon H\to H$ be a selfadjoint continuous operator on a complex Hilbert space. Show:

$$ \lVert (T\pm i\mbox{Id})x\rVert^2=\lVert Tx\rVert^2+\lVert x\rVert^2~\forall~x\in H. $$


How can I show that?

I started with $$ \lVert (T\pm i\mbox{Id})x\rVert^2=\langle Tx\pm ix,Tx\pm ix\rangle. $$

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Hint: Try using the bilinearity (technically, sesquilinearity) of the inner product. – Nate Eldredge Feb 8 '13 at 1:25
Don't forget that pulling constants things out of the second slot makes them complex conjugates. – Zach L. Feb 8 '13 at 1:27
Pull the $i$'s out of the inner products, but pay attention to the comment from @ZachL. – Dominique Feb 8 '13 at 1:30
$\langle Tx,Tx\rangle\pm i \langle x,Tx\rangle\mp i \langle Tx,x\rangle\pm \langle x,x\rangle$ – Berci Feb 8 '13 at 1:31
Please continue yourself. Is it clear, that $\bar i=-i$? And, the other one is $\bar i i=?$ instead of $i^2$. You will also need that $T$ is selfadjoint, $T^*=T$, or more usefully now, $\langle Tu,v\rangle=\langle u,Tv\rangle$ for any $u,v$. – Berci Feb 8 '13 at 1:37
up vote 3 down vote accepted

You have $$\| T(x) + ix\|^2 = \|T(x)\|^2 + \|x\|^2 + \langle T(x), ix\rangle + \langle ix, T(x)\rangle = \|T(x)\|^2 + \|x\|^2 - i\langle T(x), x\rangle + i \langle x, T(x)\rangle$$

Now we know that $\langle x, T(x)\rangle = \langle T(x), x\rangle$, so our conclusion follows forthwith.

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