Membership in the transpose range by an operator estimate

Given: The Axiom of Choice, The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

AC is used through norm-preserving extension [F2] in step 2.1.

From The transpose of a bounded operator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $T:X\to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{\ast }:Y^{\ast }⟶X^{\ast },(T^{\ast }g)(x)=g(Tx)\mathrm{.}$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $\mathbb{K}$. No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

From A bounded linear functional on an arbitrary subspace extends with the same norm, without assuming the subspace is closed, with its stated hypotheses: Let $X$ be a normed space over $\mathbb{R}$ or $\mathbb{C}$, let $M\subseteq X$ be a linear subspace, and let $f_0:M\to \mathbb{R}$ or $f_0:M\to \mathbb{C}$ be a bounded linear functional over the ambient scalar field. Then there exists a bounded linear extension $F$ of $f_0$ to all of $X$ such that $\mathrm{\parallel }F\mathrm{\parallel }=\mathrm{\parallel }{f}_0\mathrm{\parallel }$. No closedness hypothesis on $M$ is needed.