---
tag: Q1V7ZNS332
kind: statement
role: theorem
name: Dual of Lp
msc: ["46E30", 46B10]
uses: [SASP14MA1Z, 07JBSZE0V4, 0ZXSF2YQV2, WEBBT18D4X]
about: [SASP14MA1Z, 07JBSZE0V4]
sources:
  - url: https://en.wikipedia.org/wiki/Lp_space
---

## Statement

Let $(X, \Sigma, \mu)$ be a measure space, let $1 \le p < \infty$, and let $q \in (1, \infty]$ be the conjugate exponent, with $1/p + 1/q = 1$.
If $p = 1$, assume moreover that $\mu$ is [[σ-finite|WEBBT18D4X]].
For $g \in L^q(\mu)$, let $\Lambda_g$ be the functional on [[$L^p(\mu)$|SASP14MA1Z]] given by
$$\Lambda_g(f) = \int_X f g \, d\mu.$$
Then $g \mapsto \Lambda_g$ is an [[isometric isomorphism|0ZXSF2YQV2]] of $L^q(\mu)$ onto the [[dual space|07JBSZE0V4]] $L^p(\mu)^*$.

## 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|1ZSZS5G1Q4]] 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 $\ell^1$.

## Explanations

### L1 human

Every continuous linear way of turning a function in $L^p$ into a number is integration against a fixed function $g$.
The function $g$ lives in the partner space $L^q$, where $1/p + 1/q = 1$.
So $L^2$ is its own partner, $L^1$ is paired with $L^\infty$, and $L^{3/2}$ is paired with $L^3$.

### L2 human

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.