# bounded linear functionals on ${L}^{p}(\mu )$

If $\mu $ is a positive measure^{} on a set $X$, $1\le p\le \mathrm{\infty}$, and $g\in {L}^{q}(\mu )$, where $q$ is the Hölder conjugate of $p$, then Hölder’s inequality^{} implies that the map $f\mapsto {\int}_{X}fg\mathit{d}\mu $ is a bounded linear functional^{} on ${L}^{p}(\mu )$. It is therefore natural to ask whether or not all such functionals^{} on ${L}^{p}(\mu )$ are of this form for some $g\in {L}^{q}(\mu )$. Under fairly mild hypotheses, and excepting the case $p=\mathrm{\infty}$, the Radon-Nikodym Theorem^{} answers this question affirmatively.

###### Theorem.

Let $\mathrm{(}X\mathrm{,}\mathrm{M}\mathrm{,}\mu \mathrm{)}$ be a $\sigma $-finite measure space, $$, and $q$ the Hölder conjugate of $p$. If $\mathrm{\Phi}$ is a bounded linear functional on ${L}^{p}\mathit{}\mathrm{(}\mu \mathrm{)}$, then there exists a unique $g\mathrm{\in}{L}^{q}\mathit{}\mathrm{(}\mu \mathrm{)}$ such that

$$\mathrm{\Phi}(f)={\int}_{X}fg\mathit{d}\mu $$ | (1) |

for all $f\mathrm{\in}{L}^{p}\mathit{}\mathrm{(}\mu \mathrm{)}$. Furthermore, $\mathrm{\parallel}\mathrm{\Phi}\mathrm{\parallel}\mathrm{=}{\mathrm{\parallel}g\mathrm{\parallel}}_{q}$. Thus, under the stated hypotheses, ${L}^{q}\mathit{}\mathrm{(}\mu \mathrm{)}$ is isometrically isomorphic to the dual space^{} of ${L}^{p}\mathit{}\mathrm{(}\mu \mathrm{)}$.

If $$, then the assertion of the theorem remains valid without the assumption^{} that $\mu $ is $\sigma $-finite; however, even with this hypothesis, the result can fail in the case that $p=\mathrm{\infty}$. In particular, the bounded linear functionals on ${L}^{\mathrm{\infty}}(m)$, where $m$ is Lebesgue measure^{} on $[0,1]$, are not all obtained in the above manner via members of ${L}^{1}(m)$. An explicit example illustrating this is constructed as follows: the assignment $f\mapsto f(0)$ defines a bounded linear functional on $C([0,1])$, which, by the Hahn-Banach Theorem, may be extended to a bounded linear functional $\mathrm{\Phi}$ on ${L}^{\mathrm{\infty}}(m)$. Assume for the sake of contradiction^{} that there exists $g\in {L}^{1}(m)$ such that $\mathrm{\Phi}(f)={\int}_{[0,1]}fg\mathit{d}m$ for every $f\in {L}^{\mathrm{\infty}}(m)$, and for $n\in {\mathbb{Z}}^{+}$, define ${f}_{n}:[0,1]\to \u2102$ by ${f}_{n}(x)=\mathrm{max}\{1-nx,0\}$. As each ${f}_{n}$ is continuous^{}, we have $\mathrm{\Phi}({f}_{n})=\phi ({f}_{n})=1$ for all $n$; however, because ${f}_{n}\to 0$ almost everywhere and $|{f}_{n}|\le 1$, the Dominated Convergence Theorem, together with our hypothesis on $g$, gives

$$1=\underset{n\to \mathrm{\infty}}{lim}\mathrm{\Phi}({f}_{n})=\underset{n\to \mathrm{\infty}}{lim}{\int}_{[0,1]}{f}_{n}g\mathit{d}m=0\text{,}$$ |

a contradiction. It follows that no such $g$ can exist.

Title | bounded linear functionals on ${L}^{p}(\mu )$ |

Canonical name | BoundedLinearFunctionalsOnLpmu |

Date of creation | 2013-03-22 18:32:57 |

Last modified on | 2013-03-22 18:32:57 |

Owner | azdbacks4234 (14155) |

Last modified by | azdbacks4234 (14155) |

Numerical id | 15 |

Author | azdbacks4234 (14155) |

Entry type | Theorem |

Classification | msc 28B15 |

Related topic | LpSpace |

Related topic | HolderInequality |

Related topic | ContinuousLinearMapping |

Related topic | BanachSpace |

Related topic | DualSpace |

Related topic | ConjugateIndex |

Related topic | RadonNikodymTheorem |

Related topic | BoundedLinearFunctionalsOnLinftymu |

Related topic | LpNormIsDualToLq |