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Math Help - Eigenspace of \lambda = 1 for a bounded linear operator and its adjoint

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    Eigenspace of \lambda = 1 for a bounded linear operator and its adjoint

    H is a Hilbert space and T \in B(H) is such that ||T||_{H} \leq 1. Show that Tx = x \iff T^{*}x = x .

    I have tried looking at this problem in several ways and I just cannot seem to see the significance of ||A|| \leq 1 nor how to go about proving it. I understand how the adjoint is defined for a Hilbert space as a consequence of Reisz representation. I have tried to attack it using fixed point theorems. Any nudge in the right direction would be appreciated.
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    If \Vert T \Vert <1 then I-T is invertible and so the only possible eigenvector of T would be 0. For the case \Vert T\Vert =1 I don't know how to prove this yet.
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    Quote Originally Posted by jprobst View Post
    H is a Hilbert space and T \in B(H) is such that ||T||_{H} \leq 1. Show that Tx = x \iff T^{*}x = x .
    The important thing to know is that \|T^*\|=\|T\|. Therefore \|T^*x\|\leqslant\|x\|.

    If Tx=x then \|T^*x-x\|^2 = \langle T^*x-x,T^*x-x\rangle = ... (multiply out the inner product and use the fact that \langle T^*x,x\rangle = \langle x,Tx\rangle) ... = \|T^*x\|^2 - \|x\|^2\leqslant0. Hence T^*x-x=0.
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