$T\in B(H)$. Why is $\forall x\in H$ $Tx=x \iff T^*x=x$?

Question

Let $H$ be a Hilbert space, $T:H\to H$ linear and bounded, $\|T\|\le 1$. Prove that: $\forall x\in H$ $Tx=x \iff T^*x=x$.

I tried it but there must be a mistake, because I haven't used $\|T\|\le 1$. For all $\forall x\in H$: $\langle Tx-x,x\rangle =\langle x,T^*x\rangle- \langle x,x\rangle =\langle x, T^*x-x\rangle$. I want to conclude from here that $Tx=x \iff T^*x=x$, but this seems to be wrong because I don't use $\|T\|\le 1$, right?

Alternatively, my guess is to do this with Cauchy-Schwarz somehow and then I have to use $\|T\|\le 1$, but how exactly?

Answer 1

It holds $$ \begin{split} \|T^*x-x\|^2 &= \|T^*x\|^2 - \langle x, T^*x\rangle - \langle T^*x, x\rangle + \|x\|^2\\ & = \|T^*x\|^2 - \langle T x, x\rangle - \langle x, T x\rangle + \|x\|^2\\ & = \|T^*x\|^2-\|x\|^2 \le 0 \end{split}$$ because of $\|T^*\|=\|T\|\le 1$. This implies $T^*x=x$.

Answer 2

Use $||z||^2=<z,z>$ for $z \in H$.

Then compute $||T^*x-x||^2=.....= $

$||T^*x||^2-||x||^2 \le ||T^*||||x||^2-||x||^2=||T||||x||^2-||x||^2 \le||x||^2-||x||^2 =0$.

For $.....$ use $Tx=x$ and observe that $||T^*||=||T||$.

Answer 3

I am unable to follow Fred's argument so I will give another (very simple) proof: suppose $Tx=x$. Then $||T^{*}x-Tx||^{2}=||T^{*}x||^{2}+||Tx||^{2}-2\Re \langle T^{*}x,Tx \rangle$ Now observe that $\langle T^{*}x,Tx \rangle=\langle x,T(T(x)) \rangle =||x||^{2}$ because $Tx=x$. Also $||T^{*}||=||T|| \leq 1$ so $||T^{*}x-Tx||^{2} \leq ||x||^{2}+||x||^{2}-2||x||^{2}=0$. Hence $T^{*}x=Tx=x$. The reverse implication follows from the fact that $T^{**}=T$.