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?



Regarding norm attaining functions
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.


