Prove that the double dual of isometry between normed spaces is also isometry by Hahn-Banach Theorem

Question

let $T^{**}: X^{**} \rightarrow Y^{**}$ be the double dual of isometry T between normed spaces $X$ and $Y$; $T: X \rightarrow Y$. let $x^{**}$ be any element in $X^{**}$. I could show $\|T^{**}(x^{**})\| \le \|x^{**}\|$ no problem. What I would like to ask is how could I show the other direction: $\|x^{**}\| \le \|T^{**}(x^{**})\|$. I have no idea of how to apply the Hahn-Banach Theorem.

Answer 1

The conclusion holds if $T$ is an isomorphism. Indeed, the converse operator $S:Y\to X$ is also an isometry. Then $T^*$ is an isomorphism from $Y^*$ to $X^*$ and $$\|T^*y^*\|=\sup_{\|x\|=1}|(T^*y^*)(x)|=\sup_{\|x\|=1}|y^*(Tx)|=\sup_{\|y\|=1}|y^*(y)|=\|y^*\|$$ Hence $T^*$ is an isometry. Thus $T^{**}$ is an isometry as well, basing on the first step of the reasoning.

Assume now that $Y_0:=T(X)\subsetneq Y.$ Then $Y_0$ is closed in $Y.$ Let $$M^*=\{y^*\in Y^*\,:\, y^*\mid_{Y_0}=0\}$$ The space $Y^*/M^*$ is isometrically isomorphic to $Y_0^*$ through the correspondence $$[y^*](y)=y^*(y),\qquad y\in Y_0 $$ Indeed let $[y_1^*]=[y_2^*].$ Then $y_1^*-y_2^*\in M^*.$ Hence $y_1^*(y)-y_2^*(y)=0$ for $y\in Y_0.$ Moreover $$\|y^*\|_{Y_0^*}=\sup_{y\in Y_0,\ \|y\|=1}|(y^*+m^*)(y)|\le \|y^*+m^*\| $$ Hence $$\|y^*\|_{Y_0^*}\le \|[y^*]\|$$ On the other hand let $y_0^*$ be a bounded linear functional on $Y_0.$ There exists a linear functional $y^*$ on $Y$ such that $\|y^*\|_{Y^*}=\|y_0^*\|_{Y_0^*}$ and $y^*\mid_{Y_0}=y_0^*.$ Then $$\|[y^*]\|=\|y_0^*\|=\|y^*\|_{Y_0^*}$$ Hence $$\|[y^*]\|=\|y^*\|_{Y_0^*}\qquad (*)$$ Denote $T_0:X\to Y_0.$ As $T_0$ is an isometry from $X$ onto $Y_0,$ the dual map $T_0^*$ is an isometry from $Y_0^*=Y^*/M^*$ onto $X^*,$ given by $$T_0^*([y^*])=T^*y^*$$ Hence $T^{**}$ is an isometry from $X^{**}$ onto $(Y^*/M^*)^*.$ The space $(Y^*/M^*)^*$ is isometrically isomorphic with the subspace $M^{**}\subset X^{**}$ $$M^{**}=\{y^{**}\in Y^{**}\,:\, y^{**}\mid_{M^*}=0\}$$