Dual spaces record all continuous scalar observations of a vector. Annihilators and quotient dualities describe precisely which observations survive when one passes to subspaces or quotients.
Continuous duals
The dual space collects all continuous scalar measurements of vectors. Continuity is essential; algebraic linear functionals alone are usually too wild.
Definition. The continuous dual \(X^*\) of a normed space \(X\) is \(\mathcal L(X,\mathbb F)\), with norm \(\|f\|=\sup_{\|x\|\le1}|f(x)|\). The canonical embedding is \(J_X:X\to X^{**}\), \(J_Xx(f)=f(x)\).
Proposition. The map \(J_X\) is a linear isometry.
Proof. For \(\|f\|\le1\), \(|J_Xx(f)|=|f(x)|\le\|x\|\), so \(\|J_Xx\|\le\|x\|\). Hahn-Banach gives, for \(x\ne0\), a functional \(f\in X^*\) with \(\|f\|=1\) and \(f(x)=\|x\|\), hence \(\|J_Xx\|\ge\|x\|\). Linearity is immediate.
Annihilators
Annihilators describe which measurements vanish on a subspace. They are the basic language for translating subspace geometry into dual geometry.
Definition. For \(M\subset X\) and \(N\subset X^*\), define \(M^\perp=\{f\in X^*:f(m)=0\ \forall m\in M\}\) and \({}^\perp N=\{x\in X:f(x)=0\ \forall f\in N\}\).
Proposition. \({}^\perp(M^\perp)=\overline{\operatorname{span}M}\). In particular, if \(M\) is a closed subspace, then \({}^\perp(M^\perp)=M\).
Proof. The inclusion \(\overline{\operatorname{span}M}\subset{}^\perp(M^\perp)\) is clear by continuity. If \(x_0\notin\overline{\operatorname{span}M}\), apply Hahn-Banach separation to the closed subspace \(\overline{\operatorname{span}M}\): there is \(f\in X^*\) vanishing on it but with \(f(x_0)\ne0\). Thus \(f\in M^\perp\) and \(x_0\notin{}^\perp(M^\perp)\).
Dual of a quotient
A functional on a quotient is exactly a functional on the original space that ignores the quotient subspace. This is the cleanest way to remember the quotient-dual formula.
Theorem. If \(M\) is a closed subspace of \(X\), then \((X/M)^*\) is isometrically isomorphic to \(M^\perp\) by \(g\mapsto g\circ q\), where \(q:X\to X/M\) is the quotient map.
Proof. If \(g\in(X/M)^*\), then \(g\circ q\in X^*\) and vanishes on \(M\). Also \(\|g\circ q\|\le\|g\|\). Conversely, if \(f\in M^\perp\), define \(g(x+M)=f(x)\); this is well-defined because \(f\) vanishes on \(M\). For any representative, \(|g(x+M)|=|f(x)|\le\|f\|\|x\|\), and taking the infimum over representatives gives \(|g(x+M)|\le\|f\|\|x+M\|\). Hence \(\|g\|\le\|f\|\). Since \(q\) maps the unit ball of \(X\) densely enough onto the quotient unit ball by definition of the quotient norm, the reverse norm inequality follows, so the map is isometric and onto.
Dual of a subspace
Subspace duality goes in the opposite direction: functionals on the subspace extend to the ambient space by Hahn-Banach, but not uniquely.
Theorem. If \(M\subset X\) is a subspace, then restriction \(R:X^*\to M^*\), \(R(f)=f|_M\), is a quotient map and \(M^*\cong X^*/M^\perp\) isometrically.
Proof. Hahn-Banach makes \(R\) onto and norm preserving in the quotient sense: every \(g\in M^*\) has an extension \(f\in X^*\) with \(\|f\|=\|g\|\). The kernel of \(R\) is \(M^\perp\). Therefore the induced map \(X^*/M^\perp\to M^*\) is bijective. Its norm is at most one by definition, and Hahn-Banach gives the reverse inequality by choosing norm-preserving representatives.
Adjoints of bounded operators
The Banach-space adjoint is defined by precomposition. It moves information backward from observations on the range to observations on the domain.
Definition. If \(T\in\mathcal L(X,Y)\), its Banach-space adjoint is \(T^*:Y^*\to X^*\), \(T^*g=g\circ T\).
Proposition. \(T^*\in\mathcal L(Y^*,X^*)\) and \(\|T^*\|=\|T\|\).
Proof. The inequality \(\|T^*g\|\le\|g\|\|T\|\) gives \(\|T^*\|\le\|T\|\). For the reverse inequality, choose \(x\) with \(\|x\|\le1\) and \(\|Tx\|>\|T\|-\varepsilon\). Hahn-Banach gives \(g\in Y^*\), \(\|g\|=1\), with \(g(Tx)=\|Tx\|\). Then \(\|T^*g\|\ge |T^*g(x)|>\|T\|-\varepsilon\). Let \(\varepsilon\downarrow0\).