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Let ‎$‎X‎$ ‎be a‎ ‎Banach ‎space ‎and ‎‎$‎T: X‎\rightarrow ‎X‎$, ‎be ‎an ‎invertible ‎bounded ‎linear ‎operator ‎with ‎‎$‎\|T\|=\|T^{-1}\|=1‎$ ‎then can we conclude that ‎‎$‎T‎$ ‎is ‎an ‎isometry?

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  • $\begingroup$ Welcome to stackexchange. You're more likely to get answers if you edit the question to show what you tried and where you are stuck. $\endgroup$ – Ethan Bolker Feb 23 '18 at 22:00
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Hint: $\Vert x \Vert = \Vert T^{-1} T x \Vert \leq \Vert Tx \Vert \leq \Vert x \Vert$

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