$L^p$-norm is dual to $L^q$

If $(X,𝔐,\mu )$ is any measure space and $1\le p,q\le \mathrm{\infty }$ are Hölder conjugates (http://planetmath.org/ConjugateIndex) then, for $f\in L^p$, the following linear function can be defined

${\mathrm{\Phi }}_f:L^q\to ℂ,$

$g\mapsto {\mathrm{\Phi }}_f(g)\equiv {\displaystyle \int fg𝑑\mu }.$

The Hölder inequality (http://planetmath.org/HolderInequality) shows that this gives a well defined and bounded linear map. Its operator norm is given by

$$\parallel {\mathrm{\Phi }}_f\parallel =\{{\parallel fg\parallel }_1:g\in L^q,{\parallel g\parallel }_q=1\}.$$

The following theorem shows that the operator norm of ${\mathrm{\Phi }}_f$ is equal to the $L^p$-norm of $f$.

Theorem.

Let $\mathrm{(}X\mathrm{,}\mathrm{M}\mathrm{,}\mu \mathrm{)}$ be a $\sigma$-finite measure space and $p\mathrm{,}q$ be Hölder conjugates. Then, any measurable function $f\mathrm{:}X\mathrm{\to }\mathrm{C}$ has $L^p$-norm

$${\parallel f\parallel }_p=sup\{{\parallel fg\parallel }_1:g\in L^q,{\parallel g\parallel }_q=1\}.$$ (1)

Furthermore, if either and or $p\mathrm{=}\mathrm{1}$ then $\mu$ is not required to be $\sigma$-finite.

Note that the $\sigma$-finite condition is required, except in the cases mentioned. For example, if $\mu$ is the measure satisfying $\mu (A)=\mathrm{\infty }$ for every nonempty set $A$, then $L^p(\mu )=\{0\}$ for and it is easily checked that equality (1) fails whenever $f=1$ and $p>1$.

Title	$L^p$-norm is dual to $L^q$
Canonical name	LpnormIsDualToLq
Date of creation	2013-03-22 18:38:13
Last modified on	2013-03-22 18:38:13
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	5
Author	gel (22282)
Entry type	Theorem
Classification	msc 28A25
Classification	msc 46E30
Related topic	LpSpace
Related topic	HolderInequality
Related topic	BoundedLinearFunctionalsOnLinftymu
Related topic	BoundedLinearFunctionalsOnLpmu