# If $T$ is a bounded linear operator between Hilbert Spaces and $\lVert{T}\rVert = \lVert{T^{-1}}\rVert =1$, is $T$ unitary?

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!).

• Can you deduce that $T$ is isometric? – Daniel Fischer Feb 7 '17 at 16:53
• you can first prove that T is isometric (which emanates directly from assumptions ). then make use of the polarization formula to get further. – Guy Fsone Feb 7 '17 at 16:58
• Yes, @user160738 by unitary operator I mean a bijective, linear isometry. Thanks for the help! – uk62116 Feb 7 '17 at 17:00

$$\|x\|=\|TT^{-1}x\|\le\|Tx\|\le\|x\|.$$ 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}