Dual of Lp

Remarks

Hölder's inequality gives $\abs{\Lambda_g(f)} \le \norm{f}_p \norm{g}_q$, so $\norm{\Lambda_g} \le \norm{g}_q$. For $1 < p < \infty$, the choice $f = \abs{g}^{q-1} \overline{\sgn g}$ gives equality, so the map is an isometry. Surjectivity is the main point. Let $\mu$ be $\sigma$-finite and let $\Lambda$ be a bounded functional. Then $E \mapsto \Lambda(\mathbf{1}_E)$, on sets of finite measure, is a set function absolutely continuous with respect to $\mu$. The Radon–Nikodym theorem gives $g$. For $1 < p < \infty$ and a general measure space, one reduces to the $\sigma$-finite case, or uses the uniform convexity of $L^p$.

Example: for counting measure on $\N$, the theorem says that $(\ell^p)^* = \ell^q$ for $1 \le p < \infty$.

The $\sigma$-finiteness for $p = 1$ cannot be dropped. Let $X = \set{x}$ with $\mu(\set{x}) = \infty$. Then $L^1(\mu) = \set{0}$, whose dual is $\set{0}$, but $L^\infty(\mu)$ is the one-dimensional space of constants.

The case $p = \infty$ is excluded. The map $g \mapsto \Lambda_g$ from $L^1$ into $(L^\infty)^*$ is an isometry, but in general it is not onto. For example, take the functional on convergent sequences that gives the limit. By the Hahn–Banach theorem it extends to a bounded functional on $\ell^\infty$, and this extension is not given by any element of $\ell^1$.

Going deeper

For a student

The theorem identifies $(L^p)^* = L^q$ for $1 \le p < \infty$ under the stated hypotheses. For $1 < p < \infty$ it follows that $L^p$ is reflexive, since $(L^p)^{**} = (L^q)^* = L^p$. Weak convergence in $L^p$ then means convergence of $\int f_n g \, d\mu$ for each $g \in L^q$, and bounded sequences have weakly convergent subsequences. Neither $L^1$ nor $L^\infty$ is reflexive in general, which is the source of many technical difficulties.