# Regarding norm attaining functions

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?

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.