Duality of Lp spaces ($(L^p{)}^{\ast }=L^q$)

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 $(L^1{)}^{\ast }=L^{\infty }$.

This identifies the continuous dual space of an Lp space; it is not an alternative definition of the entire Lp-space topic.

Let $1<p<\infty$ and $q=p/(p-1)$. For a nonzero bounded linear functional $\ell$ on an Lp space, choose $f_0$ with $\ell (f_0)=1$ and a closest point $h$ in its closed kernel. Put $z=f_0-h$. The variational condition is $\int |z{|}^{p-2}\overline{z}k=0$ for every kernel element $k$. Since $f-\ell (f)z$ lies in the kernel, $\ell (f)=\int fg$ with $g=|z{|}^{p-2}\overline{z}/\Vert z{\Vert }_p^p\in L^q$. Hölder's inequality gives the upper norm bound; testing against $|g{|}^{q-2}\overline{g}$ gives equality. Thus the representation is an isometric linear isomorphism when the pairing is bilinear.

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 $f\in L^p$.