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Thread: 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

    $\displaystyle H$ is a Hilbert space and $\displaystyle T \in B(H)$ is such that $\displaystyle ||T||_{H} \leq 1$. Show that $\displaystyle 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 $\displaystyle ||A|| \leq 1$ 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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  2. #2
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    If $\displaystyle \Vert T \Vert <1$ then $\displaystyle I-T$ is invertible and so the only possible eigenvector of $\displaystyle T$ would be $\displaystyle 0$. For the case $\displaystyle \Vert T\Vert =1$ I don't know how to prove this yet.
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  3. #3
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    Quote Originally Posted by jprobst View Post
    $\displaystyle H$ is a Hilbert space and $\displaystyle T \in B(H)$ is such that $\displaystyle ||T||_{H} \leq 1$. Show that $\displaystyle Tx = x \iff T^{*}x = x $.
    The important thing to know is that $\displaystyle \|T^*\|=\|T\|$. Therefore $\displaystyle \|T^*x\|\leqslant\|x\|$.

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