9 LP Spaces

Young

If $a,b>0$, $p,q>1$ with $\frac{1}{p}+\frac{1}{q}=1$, then $ab⩽\frac{1}{p}{a}^p+\frac{1}{q}{b}^q$.

\begin{proof} Let $f=x^p/p-xb$ and consider its critical points. \end{proof}

Holder and Minkowski

Let $f,g:X\to [0,\infty ]$ be measurable functions. Then $f$ and $g$ satisfy the Hölder inequality

$${\int }_{X}fgd\mu ⩽{\left({\int }_X{f}^{p}d\mu \right)}^{1/p}{\left({\int }_X{g}^{q}d\mu \right)}^{1/q}$$

and the Minkowski inequality

\left(\int_X(f+g)^p d \mu\right)^{1 / p} \leqslant\left(\int_X f^p d \mu\right)^{1 / p}+\left(\int_X g^p d \mu\right)^{1 / p}. $$ ^cef2ff

\begin{proof} For Holder inequality, assume that $\Vert f{\Vert }_p=\Vert g{\Vert }_q=1$ and use ^bbb4ba.

For Minkowski inequality, we can suppose that $\Vert f{\Vert }_p<\infty$ and $\Vert g{\Vert }_q<\infty$. It deduces that $\Vert f+g{\Vert }_p<\infty$ by $f,g⩽(f^p+g^p{)}^{1/p}$ and $(f+g{)}^p⩽2^p(f^p+g^p)$. Then by Holder inequality, there is

$$\begin{array}{rl}{C}^p& ={\int }_{X}f(f+g{)}^{p-1}d\mu +{\int }_{X}g(f+g{)}^{p-1}d\mu \\ & \le {\left({\int }_X{f}^{p}d\mu \right)}^{1/p}{\left({\int }_X(f+g{)}^{(p-1)q}d\mu \right)}^{1/q}+{\left({\int }_X{g}^{p}d\mu \right)}^{1/p}{\left({\int }_X(f+g{)}^{(p-1)q}d\mu \right)}^{1/q}\\ & =(A+B){\left({\int }_X(f+g{)}^{p}d\mu \right)}^{1/q}=(A+B)C^{p-1},\end{array}$$

where $C={\left({\int }_X(f+g{)}^{p}d\mu \right)}^{1/p}$, $A={\left({\int }_X{f}^{p}d\mu \right)}^{1/p}$ and $B={\left({\int }_X{g}^{p}d\mu \right)}^{1/p}$. Now we finish the proof.

Also see here. \end{proof}

Definition of $L^p$

Definition

Let $(X,\mathcal{A},\mu )$ be a measure space and let $1⩽p<\infty$. Let $f$ be a measurable function on $X$. The $L^p$-norm of $f$ is the number

$$\Vert f{\Vert }_p={\left({\int }_X|f{|}^{p}d\mu \right)}^{1/p}.$$

A function $f:X\to \mathbb{R}$ is called $p$-integrable or an $L^p$-function if it is measurable and $\Vert f{\Vert }_p<\infty$. The space of $L^p$-functions is denoted by

$${\mathfrak{L}}^p(\mu ):=\left\{f:X\to \mathbb{R}\mid f\text{ is }\mathcal{A}\text{-measurable and }\Vert f{\Vert }_p<\infty \right\}.$$

We say $f\sim g$ if $f=g$ a.e. respect to $\mu$.

Remark. By Minkowski inequality, there is $\Vert f+g{\Vert }_p⩽\Vert f{\Vert }_p+\Vert g{\Vert }_p$. Hence $\Vert \cdot {\Vert }_p$ is indeed a norm.

Definition

Let $(X,\mathcal{A},\mu )$ be a measure space and let $f:X\to [0,\infty ]$ be a measurable function and set

$$G(f)=\{c:\mu (\{x:|f(x)|>c\})=0\},$$

and define the essential supremum of $f$, denoted by $\Vert f{\Vert }_{\infty }$, as

$$\Vert f{\Vert }_{\infty }=\mathrm{ess}\sup f=\{\begin{array}{cc}\inf G(f),& \text{ if }G(f)\ne \varnothing \\ \infty ,& \text{ if }G(f)=\varnothing \end{array}$$

The set of $L^{\infty }$-functions on $X$ will be denoted by

$${\mathfrak{L}}^{\infty }(\mu ):=\left\{f:X\to \mathbb{R}\mid f\text{ is }\mathcal{A}\text{ - measurable and }\Vert f{\Vert }_{\infty }<\infty \right\}.$$

Lemma

Let $||f|{|}_{\infty }<\infty$. Then:

• if $\alpha \in G(f)$, then $[\alpha ,\infty )\subseteq G(f)$;
• $||f|{|}_{\infty }\in G(f)$;
• $\mu (\{x:|f(x)|>c\})=0$ iff that $c⩾||f|{|}_{\infty }$ and $\mu (\{x:f(x)|>c\})>0$ iff that $c<||f|{|}_{\infty }$.

\begin{proof} i) is easy.

ii) Let $\{t_n\}\subseteq G(f)$ with $t_n\downarrow ||f|{|}_{\infty }$. Let $E_n=\{x:|f(x)|>t_n\}$, then $\mu (E_n)=0$. Since $\{x:|f(x)|>||f|{|}_{\infty }\}={\cup }_{n=1}^{\infty }{E}_n$, we have that $\mu (\{x:|f(x)|>||f|{|}_{\infty }\})=0$ and so $||f|{|}_{\infty }\in G(f)$.

iii) is also easy. \end{proof}

Lemma

For a measurable function $f$,

• $||f|{|}_{\infty }⩽M$ iff $|f|⩽M$ a.e.;
• $f\in {\mathfrak{L}}^{\infty }$ iff there exists a bounded function $g$ such that $f=g$ almost everywhere.

\begin{proof} Easy. \end{proof}

Denote the equivalence class of a function $f\in {\mathfrak{L}}^{\infty }(\mu )$ under the equivalence relation $f\sim g⟺f=g$ almost everywhere. Then define $L^{\infty }(\mu )={\mathfrak{L}}^{\infty }(\mu )/\sim$ and it is a norm space where $||\cdot |{|}_{\infty }$ is the norm.

Completeness

Recall that a metric space $M$ is called complete (or a Cauchy space) if every Cauchy sequence of points in $M$ has a limit that is also in $M$.

Lemma

A normed vector space $V$ is complete iff every absolutely convergent series in $V$ converges.

\begin{proof} If $V$ is complete and ${\sum }_{n=1}^{\infty }\Vert x_n\Vert <\infty$, let $S_N={\sum }_{n=1}^N{x}_n$. Then for $N>M$, we have

$$\Vert S_N-S_M\Vert \le \sum _{n=M+1}^N\Vert x_n\Vert \to 0\text{, as }M,N\to \infty$$

so the sequence $\left\{S_N\right\}$ is Cauchy and hence convergent.

Conversely, suppose that every absolutely convergent series converges, and let $\left\{x_n\right\}$ be a Cauchy sequence. We can choose $n_1<n_2<\cdots$ such that $\Vert x_n-x_m\Vert <2^{-j}$ for $m,n\ge n_j$. Let $y_1=x_{n_1},y_j=x_{n_j}-x_{n_{j-1}}$ for $j>1$. Then ${\sum }_{j=1}^k{y}_j=x_{n_k}$, and

$$\sum _{j=1}^{\infty }\Vert y_j\Vert \le \Vert y_1\Vert +\sum _{j=1}^{\infty }\frac{1}{2^j}<\infty$$

so ${\lim }_{k\to \infty }{x}_{n_k}={\sum }_{j=1}^{\infty }{y}_j$ exists. But since $\left\{x_n\right\}$ is Cauchy, we know that $\left\{x_n\right\}$ converges to the same limit as $\left\{x_{n_k}\right\}$. \end{proof}

Theorem

For $1⩽p⩽\infty$, $L^p(\mu )$ is complete. Hence, $L^p(\mu )$ is a Banach space.

\begin{proof} i) When $p<\infty$, define $\{f_n\}\subseteq L^p$ and ${\sum }_{n=1}^{\infty }||f|{|}_p<\infty$. We aim to show there exists $f\in L^p$ such that $||{\sum }_{n=1}^{\infty }{f}_n-f||\to 0$.

Consider $F_n={\sum }_{k=1}^n|f_k|$, then $F_n\in L^p$ and $F_n\uparrow F$. By MCT, ${\int }_X|F{|}^p={\lim }_{n\to \infty }{\int }_X|F_n{|}^p⩽M^p<\infty$ and so $F$ is a.e. finite. Therefore, $\sum |f_k|$ converges almost everywhere and so $\sum f_k$ converges almost everywhere. Let $f=\sum f_k$. Then ${\int }_X|f{|}^p⩽{\int }_X{F}^{p}d\mu <\infty$ and so $f\in L^p$. Therefore, $||F-{\sum }_{k=1}^n{f}_i|{|}_p^p=\int |f-{\sum }_{i=1}^n{f}_i{|}^p\to 0$ and so $\sum f_n$ converges to $f$ in the $L^p$ norm.

ii) Suppose now that $p=\infty$ and that $\left\{f_k\right\}$ is a sequence of functions that belong to $L^{\infty }(\mu )$ and satisfy ${\sum }_{k=1}^{\infty }{\Vert f_k\Vert }_{\infty }<\infty$. For each positive integer $k$, let $E_k=\left\{x\in X:\left|f_k(x)\right|>{\Vert f_k\Vert }_{\infty }\right\}$, and $E={\cup }_{k=1}^{\infty }{E}_k$.

We have $\mu \left(E_k\right)=0$ and so $\mu (E)=0$. Hence ${\sum }_{k=1}^{\infty }\left|f_k(x)\right|\le {\sum }_{k=1}^{\infty }{\Vert f_k\Vert }_{\infty }<\infty$ for all $x\in E^c$, which implies that ${\sum }_{k=1}^{\infty }{f}_k(x)$ converges for all $x\in E^c$. Define

$$f(x)=\{\begin{array}{ll}{\sum }_{k=1}^{\infty }{f}_k(x),& \text{ if }x\notin E,\\ 0,& \text{ if }x\in E.\end{array}$$

Then

$$\left|f(x)-\sum _{k=1}^n{f}_k(x)\right|=\left|\sum _{k=n+1}^{\infty }{f}_k(x)\right|⩽\sum _{k=n+1}^{\infty }\left|f_k(x)\right|⩽\sum _{k=n+1}^{\infty }{\Vert f_k\Vert }_{\infty }\text{ a.e. }$$

Thus ${\Vert f-{\sum }_{k=1}^n{f}_k\Vert }_{\infty }\le {\sum }_{k=n+1}^{\infty }{\Vert f_k\Vert }_{\infty }$. Letting $n\to \infty$, we have ${\sum }_{k=n+1}^{\infty }{\Vert f_k\Vert }_{\infty }\to 0$, so ${\Vert f-{\sum }_{k=1}^n{f}_k\Vert }_{\infty }\to 0$, i.e., ${\sum }_{l^n}^n{f}_k\to f$ in $L^{\infty }$. Therefore, $L^{\infty }$ is complete. \end{proof}

Remark. 虽然在 $L^p$ 意义下 Cauchy 不一定收敛，但在一般的求和下 Cauchy 一定收敛。所以用 $\sum |f_k|<\infty$ almost everywhere 就能说明 $\sum f_k$ 存在了。

Dense Subspace of $L^p$

Proposition

The follows hold.

• $f_n\to f$ in $L^{\infty }\iff f_n\to f$ uniformly a.e..
• For any $1⩽p⩽\infty$, simple measurable functions determine a dense subspace of $L^p$.

\begin{proof} i) Assume that $f_n\to f$ in $L^{\infty }$. Let $E_n=\{x:|f_n(x)-f(x)|>||f_n-f|{|}_{\infty }\}$ and let $E={\cup }_{n=1}^{\infty }{E}_n$. Then $\mu (E)=0$ and $f_n\to f$ uniformly on $X\setminus E$. Conversely, if $f_n\to f$ uniformly almost everywhere, then for any $ϵ>0$ there exists $N$ such that $|f_n(x)-f(x)|<ϵ$ a.e. and so $f_n\to f$ in $L^{\infty }$.

ii) Assume that $p<\infty$. There exists simple functions $f_n$ such that $f_n\to f$ pointwise satisfying $|f_n|\uparrow |f|$. Note that $|f_n-f|⩽2|f|$, and it follows that $f_n\in L^p$ and $\lim \Vert f_n-f{\Vert }_p\to 0$ as $n\to \infty$ by DCT.

Now assume that $p=\infty$. Define $E=\{x:|f(x)|>||f|{|}_{\infty }\}$, then $\mu (E)=0$. On $E^c$, $|f(x)|⩽||f|{|}_{\infty }<\infty$ and so $f$ is bounded on $E^c$. Therefore, a series of simple functions $\{f_n\}$ converges to $f$ uniformly on $E^c$ and so $f_n\to f$ in $L^{\infty }$. \end{proof}

Remark. We will prove that the set of continuous functions with compact support is also a dense subspace of $L^p$ for $1⩽p<\infty$. See ^fe0685.

Dual Space & Riesz Representation Theorem

dual

Let $(X,\mathcal{A},\mu )$ be a measure space and let $1⩽p⩽\infty$. Let $q$ be the conjugate exponent of $p$, i.e., $1/p+1/q=1$. Let $g\in L^q(\mu )$ and define $T_g:L^p(\mu )\to \mathbb{R}$ by

$$T_g(f)={\int }_{X}fgd\mu ,f\in L^p(\mu ).$$

Since $|T_g(f)|⩽||f|{|}_p||g|{|}_q$ by ^cef2ff, we have $||T_g||⩽||g|{|}_q$ and so $T_g\in L^p(\mu {)}^{\ast }$.

Lemma

Let $(X,\mathcal{A},\mu )$ be a $\sigma$-finite measure space. Let $1⩽p<\infty$. If $g_1,g_2$ are in $L^q(\mu )$, where $q$ is the conjugate exponent of $p$, such that $T_{g_1}=T_{g_2}$, then $g_1=g_2$ a.e.

\begin{proof} It is easy. See here. \end{proof}

Theorem

Let $(X,\mathcal{A},\mu )$ be a finite measure space. Let $1⩽p<\infty$. Let $T$ be a continuous linear functional on $L^p(\mu )$. Then, there exists a unique $g\in L^q(\mu )$ such that $T=T_g$ and $\Vert T\Vert =\Vert g{\Vert }_q$.

\begin{proof} Let $\lambda :\mathcal{A}\to \mathbb{R},E\mapsto T({\chi }_E)$ for all ${\chi }_E\in L^p$.

Claim that it $\{E_i\}\subseteq \mathcal{A}$ and $E_i\cap E_j=\varnothing$, then $\lambda ({\cup }_{i=1}^{\infty }{E}_i)={\sum }_{i=1}^{\infty }\lambda (E_i)$. In fact, let $F_n={\cup }_{i=1}^n{E}_i$, then $||{\chi }_E-{\chi }_{F_n}|{|}_p^p={\int }_X|{\chi }_{E-F_n}{|}^p=\mu (E-F_n)\to 0$. Then ${\chi }_{F_n}\to {\chi }_E$ in $L^p$. It deduces that $T({\chi }_{F_n})\to T({\chi }_E)$. Therefore, $\lambda (E)={\lim }_{n\to \infty }{\sum }_{i=1}^{n}T({\chi }_E)={\sum }_{i=1}^{\infty }\chi (E_i)$ and so $\lambda$ is a signed measure on $\mathcal{A}$.

If $\mu (E)=0$, then $\lambda (E)=T({\chi }_E)=0$ and so $\lambda \ll \mu$. By ^bdb1bb there exists $g\in L^1(\mu )$ such that $\lambda (E)={\int }_{E}gd\mu ={\int }_{X}g{\chi }_{E}d\mu$ for all $E\in \mathcal{A}$. Now we proved that for all simple measurable $s$, $T(s)={\int }_{X}gsd\mu$.

If $f\in L^{\infty }(\mu )$, $||f|{|}_{\infty }⩽M<\infty$ yields that $f\in L^p$. Let $s_n$ be simple measurable functions with $|s_n|⩽|f|$ and $s_n\to f$ as $n\to \infty$. Since $|s_n-f{|}^p⩽(2|f|{)}^p\in L^1$ and $|s_n-f{|}^p\to \infty$, we have that $||s_n-f|{|}_p\to 0$. As $T(s_n)={\int }_X{s}_{n}gd\mu$, we have that $T(f)={\int }_{X}fgd\mu$.

When $p>1$, claim that $g\in L^q(\mu )$. Let $E_n=\{x:|g(x)|⩽n\}$ and let $h=\pm 1$ so that $hg=|g|$. Let $f_n=|g{|}^{q-1}h{\chi }_{E_n}$. Then we have $|f_n{|}^p=|g{|}^{(q-1)p}{\chi }_{E_n}=|g{|}^q{\chi }_{E_n}$. Since $f_n\in L^{\infty }$, there is $|T(f_n)|=|{\int }_X{f}_{n}g|={\int }_{E_n}|g{|}^q$. It follows that

$${\int }_{E_n}|g{|}^q⩽||T||||f_n|{|}_p=|T||({\int }_{E_n}|g{|}^{q}d\mu {)}^{1/p}$$

and so $({\int }_{E_n}|g{|}^{q}d\mu {)}^{1/q}⩽||T||$.

When $p=1$, $\lambda (E)={\int }_{E}gd\mu ={\int }_{X}g{\chi }_{E}d\mu =T({\chi }_E)$. Then $|{\int }_{E}gd\mu |=|T({\chi }_E)|⩽||T||||{\chi }_E|{|}_q=||T||\mu (E)$ and so $\frac{1}{\mu (E)}|{\int }_{E}gd\mu |⩽||T||$. It yields that ${\int }_E(g-||T||)⩽0$ and ${\int }_E(g+||T||)⩾0$. Hence, $|g|⩽||T||$ a.e. and so $||g|{|}_{\infty }⩽||T||$. Now we proved that $g\in L^q(\mu )$.

Since $T_g,T\in L^p(\mu {)}^{\ast }$ and $T(s)=T_g(s)$ for any simple measurable function $s$, we have $T=T_g$ on a dense subspace of $L^p$ and $T=T_g$. Note that $||T||=||T_g||⩽||g|{|}_q⩽||T||$ yields $||g|{|}_q=||T||$, then we have done. \end{proof}

Riesz representation theorem

Let $(X,\mathcal{A},\mu )$ be a $\sigma$-finite measure space. Let $1⩽p<\infty$. Let $T$ be a continuous linear functional on $L^p(\mu )$. Then, there exists a unique $g\in L^q(\mu )$ such that $T(f)={\int }_{X}fgd\mu$ for all $f\in L^p$, and $\Vert T\Vert =\Vert g{\Vert }_q$.

\begin{proof} Let $\{E_i\}\subseteq \mathcal{A}$ with $\mu (E_i)<\infty$ and $E_i\uparrow X$. We identity $L^p(E_n)$ as a subspace of $L^p(X)$. By ^051f47, there exists $g_n\in L^q(E_n)\subseteq L^q(X)$ such that $T(f)={\int }_{E_n}f{g}_n$ for all $f\in L^p(E_n)$. Note that $g_{n+1}=g_n$ almost everywhere on $E_n$. Define $g:X\to R$ such that $g(x)=g_n(x)$ for $x\in E_n$. Claim that $T(f)={\int }_{X}fgd\mu$ for all $f\in L^p(X)$.

Note that $f{\chi }_{E_n}\to f$ in $L^p$, $T(f{\chi }_{E_n})\to T(f)$ and so $T(f)={\int }_{X}fgd\mu$. \end{proof}

Remark. 复习烦的要死，忽然发现这个证明不考。

Reflexivity & Weakly Convergence

Definition

A sequence $\left\{f_n\right\}$ in a Banach space $X$ weakly converges to an element $f$, if $\ell \left(f_n\right)\to \ell (f)$ for all $\ell \in X^{\ast }$, where $X^{\ast }$ is the dual space of $X$.

Corollary

Let $(X,\mathcal{A},\mu )$ be a $\sigma$-finite measure space with $1<p<\infty$; then $L^p(\mu )$ is reflexive. That is, $L^p(\mu {)}^{\ast \ast }\simeq L^p(\mu )$.

Recall that a Banach space is reflexive if and only if every bounded sequence in the space has a weakly convergent subsequence. It deduces the following theorem. Furthermore, 如果一个函数列 $\{f_n\}$ 在 $L^p(\mu )$ 空间中有界（也就是说，这些函数的 p-范数都小于某个固定的常数），那么我们总能找到一个子列 $\{f_{n_k}\}$，使得它在几乎处处的意义下收敛（也就是除了在一个零测集上之外，$\{f_{n_k}(x)\}$ 会对每个 x 收敛）.

The Riesz Weak Compactness Theorem

Let $(X,\mathcal{A},\mu )$ be a $\sigma$-finite measure space and $1<p<\infty$. Then every bounded sequence in $L^p(\mu )$ has a weakly convergent subsequence; that is, if $\left\{f_n\right\}$ is a bounded sequence $L^p(\mu )$, then there is a subsequence $\left\{f_{n_k}\right\}$ of $\left\{f_n\right\}$ and a function $f$ in $L^p(\mu )$ for which

$$\underset{k\to \infty }{\lim }{\int }_X{f}_{n_k}gd\mu ={\int }_{X}fgd\mu ,\forall g\in L^q(\mu )\text{, where }\frac{1}{p}+\frac{1}{q}=1.$$

Interpolation

Lemma

If $1⩽p<q<r⩽\infty$, then $L^q\subset L^p+L^r$; that is, each $f\in L^q$ is the sum of a function in $L^p$ and a function in $L^r$.

\begin{proof} If $f\in L^q$, let $E=\{x:|f(x)|>1\}$ and set $g=f{\chi }_E$ and $h=f{\chi }_{E^c}$. Then $|g{|}^p=|f{|}^p{\chi }_E⩽|f{|}^q{\chi }_E$, so $g\in L^p$, and $|h{|}^r=|f{|}^r{\chi }_{E^c}⩽|f{|}^q{\chi }_{E^c}$, so $h\in L^r$. (For $r=\infty$, obviously $\Vert h{\Vert }_{\infty }⩽1$.) \end{proof}

Theorem

If $1⩽p<q<r⩽\infty$, then $L^p\cap L^r\subset L^q$. For $f\in L^p\cap L^r$, we have $\Vert f{\Vert }_q⩽\Vert f{\Vert }_p^{\lambda }\Vert f{\Vert }_r^{1-\lambda }$, where $\lambda \in (0,1)$ is given by

$$\frac{1}{q}=\frac{\lambda }{p}+\frac{1-\lambda }{r}$$

\begin{proof} If $r=\infty$, then $|f{|}^q⩽\Vert f{\Vert }_{\infty }^{q-p}|f{|}^p$ where $\lambda =p/q$ and so $\Vert f{\Vert }_q⩽\Vert f{\Vert }_p^{\lambda }\Vert f{\Vert }_r^{1-\lambda }$.

If $r<\infty$, we use Holder’s inequality, taking the pair of conjugate exponents to be $\frac{p}{\lambda q}$ and $\frac{r}{(1-\lambda )q}$

$$\int |f{|}^q=\int |f{|}^{\lambda q}\cdot |f{|}^{(1-\lambda )q}⩽{\Vert {\left|f\right|}^{\lambda q}\Vert }_{\frac{p}{\lambda q}}\cdot {\Vert |f{|}^{(1-\lambda )q}\Vert }_{\frac{r}{(1-\lambda )q}}=\Vert f{\Vert }_p^{\lambda q}\cdot \Vert f{\Vert }_r^{(1-\lambda )q}.$$

Also see here. \end{proof}