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Suppose $X$ and $Y$ are Banach spaces and $T:X\to Y$ is a bounded linear operator. Show that $T$ is an isometric isomorphism if and only if its adjoint $T^*$ is also an isometric isomorphism. Given an example where $T$ is isometric while $T^*$ is not.

I manage to prove that if $T$ is an isometric isomorphism, then $T^*$ is an isomorphism. However, I don't know how to show $T^*$ is isometric. Can someone please help me?

For the other direction, can I just show $T^{**}=T$ (reflexive) and conclude; or do I have to use something else to prove?

Also, what would be the example? Can I use the shift in $l^2$?

Thank you.

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  • $\begingroup$ In general it is not true that $T^{\ast \ast} = T$. Are $X,Y$ reflexive spaces? $\endgroup$ – Christopher A. Wong Apr 23 '15 at 1:05
  • $\begingroup$ No, they are only Banach spaces. $\endgroup$ – Ovl0422 Apr 23 '15 at 1:08
  • $\begingroup$ The most trivial counterexample would be the zero operator $T = 0$ from $X = \{0\}$ into an arbitrary Banach space $Y$. $\endgroup$ – gerw Apr 23 '15 at 6:27
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The defining property of the adjoint is that for all $x\in X$ and $y^*\in Y^*$ then

$$(Tx,y^*)=(x,T^*y^*)$$

We have that $$\|T^*y^*\|=\sup_{x\neq0}\frac{(x,T^*y^*)}{\|x\|}=\sup_{x\neq0}\frac{(Tx,y^*)}{\|x\|}=\sup_{x\neq0}\left(\frac{Tx}{\|x\|},y*\right)=\sup_{\|y\|=1}(y,y^*)=\|y^*\|$$

The second-to-last equality is because $\frac{Tx}{\|x\|}$ returns all norm-one vector of $Y$ as $x$ moves along the non-zero elements of $X$. This is because $T$ is an isometric isomorphism.

The other direction is the same story, just begin with $\|Tx\|=\sup_{y^*\neq0}\frac{(Tx,y^*)}{\|y^*\|}$.


Ah, for the example, yes use the unilateral shift operator. The adjoint is the backwards shift, which kills the first component and therefore a little bit of the norm in some cases.

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  • $\begingroup$ How are you using scalar products in a Banach space $\endgroup$ – badatmath Jul 10 at 18:19

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