Lecture 17 - Lp Spaces

Let $(X,\mu )$ be a measure space. For $p>0$ and a measurable function $f:X\to \mathbb{C}$, then we define $\Vert f{\Vert }_p=(\int \mid f{\mid }^{p}d\mu {)}^{\frac{1}{p}}$ .

Note

$\mid f+g\mid \le \mid f\mid +\mid g\mid$
$\Vert f+g{\Vert }_p\le \Vert f{\Vert }_p+\Vert g{\Vert }_p$

A pair of conjugate exponents is $(1,\infty )$, $(\infty ,1)$, or $(p,q)$ with $p,q>0$ and $\frac{1}{p}+\frac{1}{q}=1$. Note that $(2,2)$ is okay (it also gives you a Hilbert space).

Proposition

Let $f,g:X\to [0,\infty ]$ be measurable on $(X,\mu )$. Then $\Vert f\times g{\Vert }_1\le \Vert f{\Vert }_p\times \Vert g{\Vert }_q$ (Hölder’s Inequality) and $\Vert f+g{\Vert }_p\le \Vert f{\Vert }_p+\Vert g{\Vert }_p$ (Minkowski’s Inequality) for any pair of Conjugate Exponents $(p,q)$ with $p,q>0$.

Proof

Hölder’s Inequality

To show Hölder’s Inequality, we may assume that $\Vert f{\Vert }_p$, $\Vert g{\Vert }_q$ are positive real numbers

Need:

${\Pi }_{i=1}^n{y}_i^{a_i}{\le }^{(\ast )}{\sum }_{i=1}^n{a}_i\times y_i$ when $a_i,y_i>0$ and ${\sum }_{i=1}^n{a}_i=1$.

Use (* from the note above) with $a_1=\frac{1}{p}$, $a_2=\frac{1}{q}$, $y_1={\left(\frac{\mid f(x)\mid }{\Vert f{\Vert }_p}\right)}^p$, $y_2={\left(\frac{\mid g(x)\mid }{\Vert g{\Vert }_q}\right)}^q$ for $x$ such that $y_i>0$.

Then
$\frac{\mid f(x)\times g(x)\mid }{\Vert f{\Vert }_p\times \Vert g{\Vert }_q}\le \frac{\mid f(x){\mid }^p}{p\times \Vert f{\Vert }_p^p}+\frac{\mid g(x){\mid }^q}{q\times \Vert g{\Vert }_q^q}$
Integration gives
$\frac{\Vert f\times g{\Vert }_1}{\Vert f{\Vert }_p\times \Vert g{\Vert }_q}\le \frac{1}{p}+\frac{1}{q}=1⟹\text{Ho¨lder’s Inequality}$

Minkowski’s Inequality

To get Minkowski’s Inequality, note that $\Vert f\times (f+g{)}^{p-1}{\Vert }_1\le \Vert f{\Vert }_p\times \Vert (f+g{)}^{p-1}{\Vert }_q$ by Hölder’s Inequality.

And similarly
$\Vert g\times (f+g{)}^{p-1}{\Vert }_1\le \Vert g{\Vert }_p\times \Vert (f+g{)}^{p-1}{\Vert }_q$
Then combine the two
$\Vert f+g{\Vert }_p^p=\Vert (f+g{)}^p{\Vert }_1=\Vert (f+g)\times (f+g{)}^{p-1}{\Vert }_1\le \Vert f\times (f+g{)}^{p-1}{\Vert }_1+\Vert g\times (f+g{)}^{p-1}{\Vert }_1$
$\le \Vert f{\Vert }_p\times \Vert (f+g{)}^{p-1}{\Vert }_q+\Vert g{\Vert }_p\times \Vert (f+g{)}^{p-1}{\Vert }_q$
$=\Vert (f+g{)}^{p-1}{\Vert }_q\times \left(\Vert f{\Vert }_p+\Vert g{\Vert }_p\right)$
Then we have
$\frac{\Vert f+g{\Vert }_p^p}{\Vert (f+g{)}^{p-1}{\Vert }_q}=\Vert f+g{\Vert }_p$
(This is something that you can check)

And then we get the Minkowski’s Inequality.
Provided the denominator is $\ne 0,\infty$, this can be assumed: Minkowski’s Inequality holds if $\text{LHS}=0$. If $\text{RHS}<\infty$, then $\Vert f+g{\Vert }_p<\infty$ since ${\left|\frac{(f+g)}{2}\right|}^p\le \frac{1}{2}\mid f{\mid }^p+\frac{1}{2}\mid g{\mid }^p$.

QED.

$L^p(\mu )$-spaces

Let $p\in [1,\infty ⟩$. By Minkowski’s Inequality the set of measure $f:X\to \mathbb{C}$ with $\Vert f{\Vert }_p<\infty$ on $(X,\mu )$ is a complex vector space $V$. Note $\Vert a\times f{\Vert }_p=(\int \underset{\mid a{\mid }^p\times \mid f{\mid }^p}{\underbrace{\mid af{\mid }^p}}d\mu {)}^{\frac{1}{p}}=\mid a\mid \times {\left(\int \mid f{\mid }^{p}d\mu \right)}^{\frac{1}{p}}=\mid a\mid \times \Vert f{\Vert }_p$.

Note

$\Vert f{\Vert }_p={\left(\int \mid f{\mid }^{p}d\mu \right)}^{\frac{1}{p}}$
$\Vert f+g{\Vert }_p\le \Vert f{\Vert }_p+\Vert g{\Vert }_p$

We have almost a norm $\Vert \cdot {\Vert }_p$ on $V$, except $\Vert f{\Vert }_p=0/⟹f=0$ almost everywhere.

In fact $f=0$ almost everywhere, in that $\exists$measure $A\subset X$ such that $f{\upharpoonright }_A=0$ and $\mu (A^{\complement })=0$.

Example

$\int fd\mu ={\int }_{A\cup A^{\complement }}fd\mu$
$={\int }_{A}fd\mu +{\int }_{A^{\complement }}fd\mu$

Example

$f{\upharpoonright }_A=f$ on $A$.
$g(x)=\{\begin{array}{ll}f(x),& \forall x\in A\\ 0,& \forall x\in A^{\complement },A=f^{-1}(\mathbb{C}\setminus \{0\})\end{array}$

Example

$\int gd\mu ={\int }_{A}gd\mu +{\int }_{A^{\complement }}gd\mu ={\int }_{A}gd\mu =\int fd\mu$

Define equivalence relation on $V$ by $f\sim g$ if $f-g=0$ almost everywhere.
Then $V\setminus \sim$ will be a vector space and with norm $\Vert [f]{\Vert }_p\equiv \Vert f{\Vert }_p$. Then $\Vert [f]{\Vert }_p=0⟹f=0$ almost everywhere, so $[f]=[0]=0$.

Theorem

$L^p(\mu )=V\setminus \sim$ is a Banach Space.
$L^2(\mu )$ is a Hilbert space with $(f\mid g)=\int f\times \bar{g}d\mu$.

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Exercises

This is the exercise part of the lecture.

Exercise 3.3.6 Question 2

$(X,\mu )$, $f:X\to [0,\infty ⟩$ measurable such that $\int fd\mu =0$.
Show that $f=0$ almost everywhere.
Hint: $A\equiv \{x\mid f(x)>0\}={\cup }_{n=1}^{\infty }{A}_n$, where $A_n=\underset{f^{-1}\left(⟨\frac{1}{n},\infty ⟩\right)}{\underbrace{\left\{x\mid f(x)>\frac{1}{n}\right\}}}\in \sigma$.

Proof

If $\mu (A)>0$, then $0<\mu (A)<{\sum }_{n=1}^{\infty }\mu (A_n)$, so $\mu (A_m)>0$ for some $m$.

Then $\int fd\mu \ge {\int }_{A}fd\mu \ge {\int }_{A_m}fd\mu \ge {\int }_{A_m}\frac{1}{m}d\mu =\frac{1}{m}\times \mu (A_m)>0$, which is a contradiction.

So $\mu (A)=0$ and $f=0$ almost everywhere.

QED.

Exercise 3.4.5 Question 3

What is $L^p(\mu )$ for $p\in [1,\infty ⟩$ when $\mu$ is the counting measure?

On "counting measure"

I thing “counting measure” is the same way that has been said in short hand with “measure” for all the previous lectures.

Proof

$f:X\to \mathbb{C}$ all functions. $\sigma =\wp (X)$.

What is $L^p(\mu )?$

$\Vert f{\Vert }_p=(\int \mid f{\mid }^{p}d\mu {)}^{\frac{1}{p}}$
$V=\left\{f:X\to \mathbb{C}\mid \int \mid f{\mid }^{p}d\mu <0\right\}$
$L^p(\mu )=V\setminus \sim$

Say $g:X\to [0,\infty ⟩$.
Then (define: $s={\sum }_{i=1}^n{a}_i{X}_{A_i}$) $\int gd\mu ={\sup }_{0\le s\le g}\int sd\mu ={\sup }_{0\le s\le g}{\sum }_{i=1}^n\stackrel{\ne 0}{\overbrace{a_i}}\times \mu (A_i)=\infty$ if any $A_i$ is infinite.

For any finite subset $F$ of $X$, define
$S_F(x)=\{\begin{array}{ll}g(x),& x\in F\\ 0,& x\notin F\end{array}$
Then $S_F$ is a simple function, and $0\le S_F\le g$. Have $\int S_{F}d\mu ={\int }_{F}gd\mu ={\sum }_{x\in F}g(x)$

$S_F={\sum }_{x\in F}g(x)X_{\{x\}}$ since $\int S_{F}d\mu ={\sum }_{x\in F}g(x)\mu \underset{1}{\underbrace{(\{x\})}}$
${\sum }_{x\in F}g(x)\times \underset{\delta x,y}{\underbrace{X_{\{x\}}(y)}}=g(y)=S_F(y)$

$X=\mathbb{N}$
$F=\{1,\dots ,n\}$

Then $\int gd\mu ={\sup }_{0\le s\le g}\int sd\mu \ge {\sup }_{\text{F finite}}{\sum }_{x\in F}g(x)\stackrel{x=\mathbb{N}}{\overbrace{=}}{\sum }_{n=1}^{\infty }g(n)$
($\mathbb{N}$, $\mu$ continuous measure) $⟹l^p(\mathbb{N})=L^p(\mu )=\left\{\{x_n\}\mid {\sum }_{n=1}^{\infty }|x_n{|}^p<\infty \right\}$

$\Vert f{\Vert }_p=0⟹f=0$ almost everywhere $⟺f=0$ since $\mu (A)>0\forall A\ne \varnothing$
$V\setminus \sim =V$