Invertibility and the inverse of the transpose

For every bounded $A:E\to F$, composition gives $\mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }\le \mathrm{\parallel }A\mathrm{\parallel }$. Conversely, for $Ax\mathrm{\ne }0$, [F2] supplies a unit functional $g$ with $g(Ax)=\mathrm{\parallel }Ax\mathrm{\parallel }$, so $\mathrm{\parallel }Ax\mathrm{\parallel }=\mathrm{\mid }(A^{\ast }g)(x)\mathrm{\mid }\le \mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$. The same inequality is immediate if $Ax=0$. Taking suprema gives $\mathrm{\parallel }A\mathrm{\parallel }=\mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }$, including zero spaces. This is the first use of HB.

If $T$ is bijective, [F1] gives its bounded inverse $U$. Transposing $TU=I_Y$ and $UT=I_X$ gives $U^{\ast }{T}^{\ast }=I_{Y^{\ast }}$ and $T^{\ast }{U}^{\ast }=I_{X^{\ast }}$. Thus $T^{\ast }$ is bijective and $(T^{\ast }{)}^{-1}=U^{\ast }$.

Suppose $T^{\ast }$ is bijective. Its inverse $S:X^{\ast }\to Y^{\ast }$ is bounded by [F1] and [F4]. Set $C=\max (1,\mathrm{\parallel }S\mathrm{\parallel })$. For every unit-ball $f\in X^{\ast }$, $g=Sf$ satisfies $\mathrm{\parallel }g\mathrm{\parallel }\le C$ and $f(x)=g(Tx)$; hence [F2] gives $\mathrm{\parallel }x\mathrm{\parallel }\le C\mathrm{\parallel }Tx\mathrm{\parallel }$. This proves injectivity. It also proves closedness of $M=T(X)$: if $T{x}_j$ converges, $(x_j)$ is Cauchy by this estimate, so completeness and continuity give a preimage of its limit. Sequential closedness implies closedness in this metric setting, using DC to select approximants at distances less than $1\mathrm{/}j$ when needed.

If some $y\mathrm{\notin }M$ existed, closedness would give $\delta =\mathrm{dist}(y,M)>0$. On the scalar-linear subspace $M+\mathbb{K}y$, the formula $\mathrm{\ell }(m+ay)=a$ is well-defined and linear. For $a\mathrm{\ne }0$, $\mathrm{\parallel }m+ay\mathrm{\parallel }\ge \mathrm{\mid }a\mathrm{\mid }\delta$, so $\mathrm{\mid }\mathrm{\ell }(z)\mathrm{\mid }\le {\delta }^{-1}\mathrm{\parallel }z\mathrm{\parallel }$, also when $a=0$. By the second use of HB, [F2] extends $\mathrm{\ell }$ to $g\in Y^{\ast }$ with $g{\mathrm{\mid }}_M=0$ and $g(y)=1$. Then $T^{\ast }g=0$ contradicts injectivity of $T^{\ast }$. Thus $T$ is onto and bijective.

For any bounded bijection $A$, being an isometry is equivalent to $\mathrm{\parallel }A\mathrm{\parallel }\le 1$ and $\mathrm{\parallel }{A}^{-1}\mathrm{\parallel }\le 1$: the two inequalities give $\mathrm{\parallel }Ax\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }Ax\mathrm{\parallel }$. Steps 1.1 and 1.2 transfer these two bounds between $T$ and $T^{\ast }$. Together with the equivalence of bijectivity this proves the surjective-isometry equivalence, including the unique bijection of zero spaces.