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If $T:K \rightarrow L$ is a bounded linear operator between two Hilbert Spaces $K$ and $L$, then we have automatically that if $T$ is unitary, then $\lVert{T}\rVert = \lVert{T^{-1}}\rVert = 1$ by the following:

$\lVert{Tx}\rVert^{2}_{L} = \langle Tx, Tx\rangle_{L} = \langle x, x\rangle_{K} = \lVert{x}\rVert^{2}_{K} \Rightarrow \lVert{Tx}\rVert_{L} = \lVert{x}\rVert_{K}$

Then immediately from the definition of the operator norm we get $\lVert{T}\rVert = 1$, similarly we can obtain $\lVert{T^{-1}}\rVert = 1$.

However, I get a little confused when going the other way, proving or disproving the converse... (Any insight or hints are much appreciated!).

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    $\begingroup$ Can you deduce that $T$ is isometric? $\endgroup$ – Daniel Fischer Feb 7 '17 at 16:53
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    $\begingroup$ you can first prove that T is isometric (which emanates directly from assumptions ). then make use of the polarization formula to get further. $\endgroup$ – Guy Fsone Feb 7 '17 at 16:58
  • $\begingroup$ Yes, @user160738 by unitary operator I mean a bijective, linear isometry. Thanks for the help! $\endgroup$ – uk62116 Feb 7 '17 at 17:00
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Let suppose we work in the Real vector space and complex case will be simsilar.

$$ \|x\|=\|TT^{-1}x\|\le\|Tx\|\le\|x\|. $$ ie $\|x\|=\|Tx\|$. Therefore, this with the polarization formula imply \begin{split} \langle Tx,y \rangle &=&\frac{1}{4}(\|Tx+y\|^2-\|Tx-y\|^2) \\ &=&\frac{1}{4}(\|Tx+TT^{-1}y\|^2-\|Tx-TT^{-1}y\|^2)\\ &=&\frac{1}{4}(\|x+T^{-1}y\|^2-\|x-T^{-1}y\|^2)\\ &=&\langle x,T^{-1}y \rangle \end{split}

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    $\begingroup$ A note for anyone who asks, the proof of the case of complex scalars is the same (just two more terms to consider in the last equation). Nice work. $\endgroup$ – Aweygan Feb 7 '17 at 17:13
  • $\begingroup$ Yes this holds true only for the real case. the complex case is similar. thank for emphasizing. I fixed it $\endgroup$ – Guy Fsone Feb 7 '17 at 17:17
  • $\begingroup$ Thanks again! Although, just to clarify, what is the polarization formula actually telling us? $\endgroup$ – uk62116 Feb 7 '17 at 17:18
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    $\begingroup$ see the first line formula in blcok of equations in the answer above $\endgroup$ – Guy Fsone Feb 7 '17 at 17:20
  • $\begingroup$ sorry the complex case is totally different. you have check beforehand. small exercise for you. $\endgroup$ – Guy Fsone Feb 7 '17 at 17:25

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