Bounded below is equivalent to surjectivity of the transpose

Given: The Axiom of Choice and 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.

From Banach closed-range theorem, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Assume AC and let $T:X\to Y$ be bounded linear between Banach spaces. The following are equivalent: $\mathrm{ran}T$ is norm closed; $\mathrm{ran}{T}^{\ast }$ is norm closed; and there is $C>0$ such that $\mathrm{dist}(x,\ker T)\le C\mathrm{\parallel }Tx\mathrm{\parallel }$ for all $x\in X$. In that case $\mathrm{ran}T={}^{\perp }(\ker T^{\ast }),\mathrm{ran}{T}^{\ast }=(\ker T{)}^{\perp }\mathrm{.}$

The Axiom of Choice is used exactly for the closed-range implication [F1] in step 1.2; it also implies the DC premise of [F4].

From Elementary kernel and range annihilator identities, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. For a bounded linear $T:X\to Y$ between normed spaces, $(\mathrm{ran}T{)}^{\perp }=\ker T^{\ast },{}^{\perp }(\mathrm{ran}{T}^{\ast })=\ker T,{\overline{\mathrm{ran}T}}^{\mathrm{\parallel }\cdot \mathrm{\parallel }}={}^{\perp }(\ker T^{\ast })\mathrm{.}$ The closure in the last identity is in $Y$.

From Membership in the transpose range by an operator estimate, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $T:X\to Y$ be bounded linear between normed spaces and $f\in X^{\ast }$. Then $f\in \mathrm{ran}{T}^{\ast }⟺\mathrm{\exists }C\ge 0\mathrm{\forall }x\in X:\mathrm{\mid }f(x)\mathrm{\mid }\le C\mathrm{\parallel }Tx\mathrm{\parallel }\mathrm{.}$ For any such $C$, a representing $g\in Y^{\ast }$ can be chosen with $\mathrm{\parallel }g\mathrm{\parallel }\le C$.

From Under Dependent Choice, a bounded operator between Banach spaces is bounded below exactly when it is injective with closed range, with its stated hypotheses: Assume the Axiom of Dependent Choice (def-dependent-choice). Let $X$ and $Y$ be Banach spaces over the same scalar field, and let $T:X\to Y$ be a bounded linear operator. Then $T$ is bounded below if and only if it is injective and has closed range.