# $L^{p}$-norm is dual to $L^{q}$

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

 $\displaystyle\Phi_{f}\colon L^{q}\rightarrow\mathbb{C},$ $\displaystyle g\mapsto\Phi_{f}(g)\equiv\int fg\,d\mu.$

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

 $\|\Phi_{f}\|=\left\{\|fg\|_{1}:g\in L^{q},\|g\|_{q}=1\right\}.$

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

###### Theorem.

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

 $\|f\|_{p}=\sup\left\{\|fg\|_{1}:g\in L^{q},\|g\|_{q}=1\right\}.$ (1)

Furthermore, if either $p<\infty$ and $\|f\|_{p}<\infty$ or $p=1$ 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 $\mu(A)=\infty$ for every nonempty set $A$, then $L^{p}(\mu)=\{0\}$ for $p<\infty$ and it is easily checked that equality (1) fails whenever $f=1$ and $p>1$.

Title $L^{p}$-norm is dual to $L^{q}$ LpnormIsDualToLq 2013-03-22 18:38:13 2013-03-22 18:38:13 gel (22282) gel (22282) 5 gel (22282) Theorem msc 28A25 msc 46E30 LpSpace HolderInequality BoundedLinearFunctionalsOnLinftymu BoundedLinearFunctionalsOnLpmu