The Duality of $L^p$ and $L^q$ - Examples

Since $aq<1$, one has $1-aq>0$ and therefore ${\int }_0^1{x}^{-aq}dx={\left[\frac{x^{1-aq}}{1-aq}\right]}_0^1=(1-aq{)}^{-1}\mathrm{.}$ Hence $g\in L^q(0,1)$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_q^q={\int }_0^1{x}^{-aq}dx=(1-aq{)}^{-1},\mathrm{\parallel }g{\mathrm{\parallel }}_q=(1-aq{)}^{-1\mathrm{/}q}\mathrm{.}$

The chosen function $f$ satisfies $\mathrm{\mid }f(x){\mathrm{\mid }}^p=(1-aq)x^{-aq}\mathrm{.}$ Step 1.1 therefore gives $\mathrm{\parallel }[f]{\mathrm{\parallel }}_p^p=(1-aq){\int }_0^1{x}^{-aq}dx=1,$ so $\mathrm{\parallel }[f]{\mathrm{\parallel }}_p=1$.

Also $f(x)g(x)=(1-aq{)}^{1\mathrm{/}p}{x}^{-aq},$ so another use of step 1.1 gives ${\int }_0^{1}f(x)g(x)dx=(1-aq{)}^{1\mathrm{/}p}{\int }_0^1{x}^{-aq}dx=(1-aq{)}^{1\mathrm{/}p-1}=(1-aq{)}^{-1\mathrm{/}q}\mathrm{.}$ By step 1.1 and [L1], this equals $\mathrm{\parallel }g{\mathrm{\parallel }}_q=\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }$.

Let $g:={\mathbf{1}}_{[0,1\mathrm{/}2]}$. [given, construct] Then $\mathrm{\Lambda }([f])={\int }_0^{1}f(x)g(x)dx={\mathrm{\Lambda }}_g([f]),$ so $\mathrm{\Lambda }$ is exactly the pairing functional associated to $g$.

By [L1], if $q<\mathrm{\infty }$ then $\mathrm{\parallel }g{\mathrm{\parallel }}_q^q={\int }_0^1\mathrm{\mid }g{\mathrm{\mid }}^{q}dx={\int }_0^{1\mathrm{/}2}1dx=\frac{1}{2},$ so $\mathrm{\parallel }g{\mathrm{\parallel }}_q=2^{-1\mathrm{/}q}\mathrm{.}$

If $q=\mathrm{\infty }$, then $\mathrm{\mid }g\mathrm{\mid }\le 1$ everywhere and $g=1$ on a set of positive measure, so $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}=1=2^{-1\mathrm{/}\mathrm{\infty }}$.

Applying [L2] to step 1.1 and steps 1.2-1.3 gives $\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }=\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }=\mathrm{\parallel }g{\mathrm{\parallel }}_q=2^{-1\mathrm{/}q}\mathrm{.}$

Apply [L1] with $p=q=2$. Then there is a unique sequence $b\in {\mathrm{\ell }}^2$ such that $\mathrm{\Lambda }(a)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{b}_n$ for every $a\in {\mathrm{\ell }}^2$, and $\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }=\mathrm{\parallel }b{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\mathrm{.}$

The summation formula is exactly the counting-measure integral pairing from [L2], so this is the concrete ${\mathrm{\ell }}^2$ instance of the A-page theorem.

Let $u:=0$ and $v:={\mathbf{1}}_{\{0\}}$ on $[0,1]$. [given, construct] Then $u(0)=0,v(0)=1,$ so point evaluation at $0$ distinguishes these two representatives.

By [L2], the set $\{0\}$ is null, so $u=v$ almost everywhere. [L1, L2, step 1.1] Therefore [L1] says that $u$ and $v$ define the same element of $L^p([0,1])$.

A well-defined map on $L^p([0,1])$ cannot assign two different values to [L1, step 1.1, step 2.1] the same class. Steps 1.1 and 2.1 show that evaluation at $0$ would have to do exactly that, so it does not descend to $L^p([0,1])$.

The family $\mathcal{A}$ contains $\mathrm{\varnothing }$ and is closed under complements by definition. Let $(E_n)$ lie in $\mathcal{A}$. If every $E_n$ is at most countable, then ${\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n$ is at most countable by [L1]. If some $E_j$ is cocountable, then $X\setminus {\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n\subseteq X\setminus E_j,$ so [L1] makes the complement at most countable. Thus $\mathcal{A}$ is a sigma-algebra on $X$.

The function $\mu$ satisfies $\mu (\mathrm{\varnothing })=0$. Let $(E_n)$ be pairwise disjoint in $\mathcal{A}$. If two members $E_i,E_j$ were cocountable, then $X=(X\setminus E_i)\cup (X\setminus E_j)$ would be at most countable by [L1], contradicting that $X$ is uncountable. So at most one $E_n$ is cocountable. If every $E_n$ is at most countable, then ${\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n$ is at most countable by [L1], and both sides of countable additivity are $0$. If some $E_j$ is cocountable, then $X\setminus {\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n\subseteq X\setminus E_j$ is at most countable by [L1], so the union is cocountable and both sides of countable additivity are $+\mathrm{\infty }$. Hence $\mu$ is a measure on $(X,\mathcal{A})$.

By definition of $\mu$, a measurable set has finite $\mu$-measure exactly when it is at most countable, and then its $\mu$-measure is $0$. In particular, $X$ is measurable and $\mu (X)=+\mathrm{\infty }$ because $X\setminus X=\mathrm{\varnothing }$ is at most countable.

Let $[f]\in L^1(\mu )$ and choose an integrable representative $u$. For each $n\ge 1$, set $E_n:=\{\mathrm{\mid }u\mathrm{\mid }\ge 1\mathrm{/}n\}$. By [L4], $\mu (E_n)\le n\int \mathrm{\mid }u\mathrm{\mid }d\mu <\mathrm{\infty }\mathrm{.}$ Step 3.1 therefore gives $\mu (E_n)=0$, so each $E_n$ is at most countable. By [L1], $\{u\mathrm{\ne }0\}={\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n$ is at most countable and hence null. Thus every element of $L^1(\mu )$ is the zero class.

The constant functions $0$ and $1$ are both in $L^{\mathrm{\infty }}(\mu )$, but they do not define the same $L^{\mathrm{\infty }}$ class because $\{1\mathrm{\ne }0\}=X,$ and step 3.1 gives that $X$ is not null. Thus $L^{\mathrm{\infty }}(\mu )$ contains at least two distinct classes.

Since $L^1(\mu )$ has only the zero class by step 4.1, its only bounded linear functional is the zero functional. Both $0$ and $1$ represent that zero functional, because $\int f\cdot 0d\mu =0=\int f\cdot 1d\mu$ for the unique class $f=0$ in $L^1(\mu )$. By step 4.2, those two $L^{\mathrm{\infty }}$ classes are distinct, so uniqueness fails. Therefore the $p=1$ representation theorem is false on this measure space.