Elementary kernel and range annihilator identities

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 norming separation [F4] in step 1.2 and annihilator closure [F3] 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 Annihilator notation and the preannihilator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. For a normed $X$ and arbitrary subsets $M\subseteq X$, $N\subseteq X^{\ast }$, define $M^{\perp }=\{f\in X^{\ast }:f(m)=0\text{ for all }m\in M\},{}^{\perp }N=\{x\in X:f(x)=0\text{ for all }f\in N\}\mathrm{.}$ Here $X^{\ast }$ is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on $\mathrm{span}M$, since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$, not in $X^{\ast \ast }$. Empty sets impose no conditions: ${\mathrm{\varnothing }}^{\perp }=X^{\ast }$ and ${}^{\perp }\mathrm{\varnothing }=X$.

From The annihilator detects the closure of a subspace, with its stated hypotheses: For every linear subspace $M\subseteq X$, $\bar{M}={\bigcap }_{f\in M^{\perp }}\ker f\mathrm{.}$

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 $\mathrm{\parallel }f\mathrm{\parallel }=1\text{and}f(x)=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{.}$