The transpose is bounded with the same norm

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 the norming-functional supplier [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 Every nonzero vector has a norming functional, with its stated hypotheses: Let $X$ be a normed space over $\mathbb{R}$ or $\mathbb{C}$, and let $x\in X$ be nonzero. Then there exists $f\in X^{\ast }$ such that $\mathrm{\parallel }f\mathrm{\parallel }=1\text{and}f(x)=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{.}$