Continuous dual space

The continuous dual $X^{\ast }$ of a topological vector space is the vector space of its continuous scalar-valued linear functionals. For a normed space these are exactly the bounded linear functionals.

A vector subspace $Z\subseteq X^{\ast }$ is norming if some $c>0$ satisfies the displayed inequality. It retains enough continuous linear functionals to control the norm of every vector. Every such vector subspace is dense for the weak-star topology: otherwise a nonzero evaluation linear functional would annihilate its weak-star closure, contradicting the inequality. In the real case its symmetric unit ball makes the absolute-value and signed supremum versions equivalent.

Finite truncations of the sign or phase sequence of $x\in {\ell }^1$ belong to $c_0$ and attain increasing partial sums of ${\sum }_n|x_n|$. Thus the space of sequences converging to zero is 1-norming for the absolutely summable sequence space. Its codimension in ${\ell }^{\infty }$ is infinite: indicators of pairwise disjoint infinite subsets have linearly independent classes modulo $c_0$. Norming does not require finite codimension.

A norm-closed finite-codimensional vector subspace $Y\subseteq X^{\ast }$ that is weak-star dense is a norming subspace. Its annihilator $F=Y^{\perp }\subseteq X^{\ast \ast }$ is finite-dimensional and disjoint from $JX$. The positive distance $d$ of the unit sphere from $F$ gives a triple-dual linear functional vanishing on $F$ and taking a value at least $d$ at a unit vector. Apply the finite-dimensional interpolation form of Goldstine's theorem to the Banach space $X^{\ast }$, then normalize and take a supremum to obtain norming constant $d$.

In a Hausdorff locally convex space, a separating family of continuous seminorms has $p(x)>0$ for any chosen nonzero $x$ and some $p$. Define a linear functional on its one-dimensional span attaining $p(x)$ and extend it under that seminorm by the real Hahn-Banach theorem. The extension is continuous because it is seminorm-bounded, and distinguishes $x$ from zero. The Hausdorff condition is essential.

For a locally convex space $E$, the strong dual topology on its continuous dual space $E^{\prime }$ is generated by seminorms $p_B(T)={\sup }_{\phi \in B}|⟨T,\phi ⟩|$, where $B$ runs through the bounded sets in a topological vector space $E$. Convergence therefore means uniform convergence of pairings on every such set. For the Schwartz space, the defining boundedness is boundedness of every Schwartz seminorm, rather than one norm or one common compact support.

The dual pairing evaluates a linear functional on a vector: $⟨f,x⟩=f(x)$. It is the basic pairing between a topological vector space and its continuous dual.

For a bounded linear operator $T:X\to Y$, its transpose or dual map is $T^{\ast }:Y^{\ast }\to X^{\ast }$ defined byIt satisfies $\Vert T^{\ast }\Vert =\Vert T\Vert$.

A linear functional $\Lambda$ on an ordered vector space of functions is positive when $f\ge 0$ implies $\Lambda (f)\ge 0$. Positivity implies monotonicity: $f\le g$ gives $\Lambda (f)\le \Lambda (g)$.

A real positive linear functional on a vector subspace of $C(K)$ containing $1$ extends positively to all of $C(K)$. Its norm is $\varphi (1)$ by order bounds. If this is positive, apply the Hahn-Banach theorem to obtain an extension of the same norm, normalize it to value one at $1$, and use the unital contraction positivity criterion. If $\varphi (1)=0$, the functional is zero and its zero extension suffices.

For a real positive linear functional on the space of continuous functions on a compact space, $-\Vert f{\Vert }_{\infty }1\le f\le \Vert f{\Vert }_{\infty }1$ gives $|\varphi (f)|\le \varphi (1)\Vert f{\Vert }_{\infty }$. Evaluation at $1$ gives equality of the operator norm and $\varphi (1)$. Thus positivity alone already implies continuity in the supremum norm.

A real-linear contraction on $C(X)$ that preserves $1$ is positive: for $0\le f\le 1$, $\Vert 1-Tf{\Vert }_{\infty }\le 1$ forces $Tf\ge 0$, then scale. In the complex case, a unital norm-one functional is real on real functions, since $|1+it\varphi (g)|\le \Vert 1+itg{\Vert }_{\infty }=1+O(t^2)$ for both signs of $t$. The same positivity argument applies pointwise to $\varphi (f)=Tf(x)$. This turns a unital contraction semigroup into a Feller semigroup.

The operator norm is $\Vert T\Vert ={\sup }_{\Vert x\Vert \le 1}\Vert Tx\Vert$.

For bounded operators between Hilbert spaces, write the norm as the supremum of $|⟨Ax,y⟩|$ over unit vectors $x,y$. Moving the operator to the other slot proves equality of its norm with that of its adjoint operator. Applied to the exponential matrix, this makes the primal and dual analytic large sieve inequalities equivalent.

The Euclidean logarithmic norm is the largest eigenvalue of a matrix's Hermitian part. Differentiating $\Vert y{\Vert }_2^2$ along $y^{\prime }=Ay$ gives $\Vert e^{tA}{\Vert }_2\le e^{t{\mu }_2(A)}$. It is the least exponent for such a bound with prefactor one, as a first-order expansion at zero shows. It equals the spectral abscissa for a normal matrix, but need not do so otherwise. A scalar rational stability function does not generally inherit a bound by its value at this real number.

For composable bounded linear operators, the operator norm satisfies $\Vert AB\Vert \le \Vert A\Vert \Vert B\Vert$. Indeed, $\Vert ABx\Vert \le \Vert A\Vert \Vert Bx\Vert \le \Vert A\Vert \Vert B\Vert \Vert x\Vert$, and taking the supremum over unit vectors proves the bound. It implies $\Vert [A,B]\Vert \le 2\Vert A\Vert \Vert B\Vert$ by the triangle inequality.

If $A_j$ and $B_j$ have operator norm at most one, then inserting and subtracting one factor at a time gives

Insert one factor difference at a time: the $j$th term is $U_m\cdots U_{j+1}(U_j-V_j)V_{j-1}\cdots V_1$. Multiplication by the surrounding unitary operators preserves the spectral norm, so the triangle inequality gives the displayed dimension-independent bound. No commutation assumption is required.

The matrix 2-norm is the operator norm induced by the Euclidean norm:It equals the largest singular value of $A$ and is invariant under multiplication by orthogonal or unitary matrices.

The continuous dual of every normed space is Banach because an operator-norm Cauchy sequence converges pointwise to a bounded linear functional and then uniformly on the unit ball.

Coordinate pairing gives $({\ell }^p{)}^{\ast }={\ell }^q$ for $1<p<\infty$, $({\ell }^1{)}^{\ast }={\ell }^{\infty }$, and $(c_0{)}^{\ast }={\ell }^1$.

For $y\in {\ell }^{\infty }$, the formula ${\varphi }_y(x)={\sum }_n{x}_n{y}_n$ defines a bounded functional on ${\ell }^1$ with $\Vert {\varphi }_y\Vert =\Vert y{\Vert }_{\infty }$. Conversely, every $\varphi \in ({\ell }^1{)}^{\ast }$ is represented this way by the bounded sequence $y_n=\varphi (e_n)$. This is an isometric isomorphism of normed spaces.

A Banach space has the Schur property when every weakly convergent sequence converges with respect to the norm topology.

A gliding hump argument selects a subsequence and successive finite coordinate blocks so that each selected vector has little mass before and after its assigned block. A bounded dual vector can then align its signs independently on those disjoint blocks.

The l-p sequence space ${\ell }^1$ has the Schur property. If a weakly null sequence stayed bounded below in norm, coordinatewise convergence and summability would select disjoint blocks containing almost all of successive terms. A sequence in ${\ell }^{\infty }$ matching their signs on those blocks would pair uniformly positively with a subsequence, contradicting weak convergence through the duality of l1 and l infinity.

For $1\le p<\infty$ and conjugate exponent $q$, every bounded linear functional on $L^p$ over a sigma-finite measure space has the formfor a unique $\omega \in L^q$, and its operator norm is $\Vert \omega {\Vert }_q$. At the endpoint $p=1$, this says $(L^1{)}^{\ast }=L^{\infty }$.

For conjugate exponents $p,q\in (1,\infty )$, every bounded linear functional on an Lp space is uniquely $h\mapsto \int hgd\mu$ for $g\in L^q$, with functional norm $\Vert g{\Vert }_q$. This holds on arbitrary measure spaces. Apply the Radon-Nikodym theorem on finite-measure pieces, then use support localization of an Lp functional to obtain one global density without assuming sigma-finiteness of the entire measure.

For $1<p<\infty$, local Radon-Nikodym derivatives $g_E$ representing an $L^p$ functional on finite-measure sets have uniformly bounded $L^q$ mass. Choose a sequence of finite-measure sets approaching the supremum of that mass, and glue their compatible densities on their union $S$. Any finite-measure set outside $S$ must have zero local density, since otherwise it would increase the supremum. Every Lp space function is approximable by simple functions of finite-measure support, so the glued density represents the functional on the whole space.

On ${\mathbb{R}}^n$, integration embeds $L^1$ isometrically into $(L^{\infty }{)}^{\ast }$, but this is not surjective. For example, the functional taking the limit at infinity on $\mathbb{C}1\oplus C_0({\mathbb{R}}^n)$ extends to $L^{\infty }$ by the Hahn-Banach theorem; it vanishes on compactly supported functions and is one on the constant function, so no $L^1$ density represents it.

For $1<p<\infty$, the identifications $(L^p{)}^{\ast }=L^q$ and $(L^q{)}^{\ast }=L^p$ make $L^p$ reflexive. On a non-atomic space such as ${\mathbb{R}}^n$, neither $L^1$ nor $L^{\infty }$ is reflexive.

On a finite measure space, a positive functional $\Lambda \in (L^p{)}^{\ast }$ defines the finite measure $\nu (E)=\Lambda (1_E)$. The Radon-Nikodym theorem gives $d\nu =\omega d\mu$ with $\omega \ge 0$. The boundand $L^p$-$L^q$ norm duality imply $\omega \in L^q$, after which density extends the integral representation to every $f\in L^p$.