Let $X$ and $Y$ be Banach spaces.Let $L(X,Y)$ denote the space of all bounded linear map from $X$ to $Y$. $T:X\longrightarrow Y$ is said to be norm attaining if there exists a $x\in S_X$(the closed unit circle in X) such that $$\|T(x)\|=\|T\|.$$Let $NA(X,Y)$ denote the set of all norm attaining maps in $L(X,Y)$. Let ${NA}_1(X,Y)=\{T\in L(X,Y): T^*\in NA(Y^*,X^*)\}$ and ${NA}_2(X,Y)=\{T\in L(X,Y): T^{**}\in NA(X^{**}, Y^{**})\}$. There is a result by Zizler that states that ${NA}_1(X,Y)$ is dense in $L(X,Y)$. A corrolary to the result is that if $X$ is reflexive, then $$NA(X,Y)=NA_1(X,Y)=NA_2(X,Y).$$ Can anyone tell hoe the corollary comes easily?

## 1 Answer

Let $T\in NA(X,Y)$ so there is $x\in X, \|x\|=1, \|T(x)\| = \|T\|$. By Hahn-Banach there is $f\in Y^*, \|f\|=1, f(T(x)) = \|T(x)\| = \|T\|$. But $f(T(x)) = T^*(f)(x)$ so $$ \|T^*\| = \|T\| = |T^*(f)(x)| \leq \|T^*(f)\| \|x\| \leq \|T^*\| \|f\| \|x\| =\|T^*\|. $$ Hence we have equality throughout, in particular, $\|T^*(f)\| = \|T^*\|$. So $T^* \in NA(Y^*,X^*)$. Similarly, $T^* \in NA(Y^*,X^*) \implies T^{**} \in NA(X^{**}, Y^{**})$. We conclude that $$ NA(X,Y) \subseteq NA_1(X,Y) \subseteq NA_2(X,Y). $$

For the canonical map $\kappa_X:X\rightarrow X^{**}$, for $T\in L(X,Y)$, we have that $T^{**}\kappa_X = \kappa_Y T$. If now $X$ is reflexive, so that $\kappa_X$ is onto, if $T\in NA_2(X,Y)$ then there is $x\in X, \|x\|=1$ with $\|\kappa_YT(x)\| = \|T^{**}\kappa_X(x)\| = \|T^{**}\| = \|T\|$ so $\|T(x)\| = \|T\|$ so $T\in NA(X,Y)$. So in this case $$ NA_2(X,Y) \subseteq NA(X,Y), $$ and all the spaces are hence equal.