Injective transpose does not imply surjectivity

Given: 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 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 Counting measure specializes the representation theorem to ${\mathrm{\ell }}^p$ and ${\mathrm{\ell }}^q$, with its stated hypotheses: Let $1\le p<\mathrm{\infty }$ and let $q$ be conjugate to $p$. Every bounded linear functional $\mathrm{\Lambda }:{\mathrm{\ell }}^p\to \mathbb{R}$ is of the form $\mathrm{\Lambda }(a)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{b}_n$ for a unique sequence $b\in {\mathrm{\ell }}^q$. Moreover, $\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }=\mathrm{\parallel }b{\mathrm{\parallel }}_{{\mathrm{\ell }}^q}\mathrm{.}$

From ${\mathrm{\ell }}^p$ is the $L^p$ space of counting measure, with its stated hypotheses: On $(\mathbb{N},\mathcal{P}(\mathbb{N}),\mathrm{\#})$ with counting measure, every function $f:\mathbb{N}\to \mathbb{R}$ is measurable. Writing $a_k:=f(k)$, one has $\int \mathrm{\mid }f{\mathrm{\mid }}^{p}d\mathrm{\#}={\sum }_{k=0}^{\mathrm{\infty }}\mathrm{\mid }{a}_k{\mathrm{\mid }}^p(0<p<\mathrm{\infty }),$ by the counting-measure integral dictionary, so ${\mathcal{L}}^p(\mathrm{\#})$ is exactly the usual sequence class ${\mathrm{\ell }}^p$. Also $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{k\in \mathbb{N}}\mathrm{\mid }{a}_k\mathrm{\mid },$ because a subset of $\mathbb{N}$ has counting measure zero only when it is empty. Hence the quotient by almost-everywhere equality does nothing: for counting measure on $\mathbb{N}$, equality almost everywhere means equality everywhere.