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

Let $(X,\U0001d510,\mu )$ be a $\sigma $-finite measure space and $p,q$ be Hölder conjugates. Then, we show that a measurable function^{} $f:X\to \mathbb{R}$ 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 $p=1$ then $\mu $ is not required to be $\sigma $-finite.

If ${\parallel f\parallel}_{p}=0$ then $f$ is zero almost everywhere, and both sides of equality (1) are zero. So, we only need to consider the case where ${\parallel f\parallel}_{p}>0$.

Let $K$ be the right hand side of equality (1).
For any $g\in {L}^{q}$ with ${\parallel g\parallel}_{q}=1$, the Hölder inequality^{} gives ${\parallel fg\parallel}_{1}\le {\parallel f\parallel}_{p}$, so $K\le {\parallel f\parallel}_{p}$. Only the reverse inequality remains to be shown.

If $$ and $$ then, setting $g={|f|}^{p-1}$ gives

$$ |

Therefore, $g\in {L}^{q}$ and,

$$K\ge {\parallel f(g/{\parallel g\parallel}_{q})\parallel}_{1}={\parallel {|f|}^{p}\parallel}_{1}/{\parallel g\parallel}_{q}={\parallel f\parallel}_{p}^{p}/{\parallel f\parallel}_{p}^{p-1}={\parallel f\parallel}_{p}.$$ |

On the other hand, if $p=1$ so that $q=\mathrm{\infty}$, then setting $g=1$ gives ${\parallel g\parallel}_{q}=1$ and

$$K\ge {\parallel fg\parallel}_{1}={\parallel f\parallel}_{1}.$$ |

So, we have shown that $K={\parallel f\parallel}_{p}$ when $$ and $$, and when $p=1$. From now on, it is assumed that the measure^{} is $\sigma $-finite. Then there is a sequence ${A}_{n}\in \U0001d510$ increasing to the whole of $X$ and such that $$.

Now consider the case where $$ and ${\parallel f\parallel}_{p}=\mathrm{\infty}$. Let ${f}_{n}$ be the sequence of functions

$${f}_{n}={1}_{{A}_{n}}{1}_{|f|\le n}f$$ |

then, $|{f}_{n}|\le |f|$ and monotone convergence gives ${\parallel {f}_{n}\parallel}_{p}\to {\parallel f\parallel}_{p}=\mathrm{\infty}$. Therefore,

$$K\ge sup\{{\parallel {f}_{n}g\parallel}_{1}:g\in {L}^{q},{\parallel g\parallel}_{q}=1\}={\parallel {f}_{n}\parallel}_{p}.$$ |

and letting $n$ go to infinity gives $K=\mathrm{\infty}$.

We finally consider $p=\mathrm{\infty}$. Then, for any $$ there exists a set $A\in \U0001d510$ with $\mu (A)>0$ such that $|f|\ge L$ on $A$. Also, monotone convergence gives $\mu (A\cap {A}_{n})\to \mu (A)$ and, therefore, $\mu (A\cap {A}_{n})>0$ eventually. Replacing $A$ by $A\cap {A}_{n}$ if necessary, we may suppose that $$. So, setting $g={1}_{A}/\mu (A)$ gives ${\parallel g\parallel}_{1}=1$ and,

$$K\ge {\parallel fg\parallel}_{1}={\int}_{A}|f|\mathit{d}\mu /\mu (A)\ge L.$$ |

Letting $L$ increase to ${\parallel f\parallel}_{p}$ gives $K\ge {\parallel f\parallel}_{p}$ as required.

Title | proof of ${L}^{p}$-norm is dual to ${L}^{q}$ |
---|---|

Canonical name | ProofOfLpnormIsDualToLq |

Date of creation | 2013-03-22 18:38:16 |

Last modified on | 2013-03-22 18:38:16 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 4 |

Author | gel (22282) |

Entry type | Proof |

Classification | msc 46E30 |

Classification | msc 28A25 |