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

Choose measurable representatives $u$ of $[f]$ and $v$ of $g$. Then $u\in {\mathcal{L}}^p(\mu )$; if $q<\mathrm{\infty }$ then $v\in {\mathcal{L}}^q(\mu )$, while if $q=\mathrm{\infty }$ then $v$ is an essentially bounded representative. Thus [L2] gives $uv\in L^1(\mu )$ and $\mid \int uvd\mu \mid \le \mathrm{\parallel }u{\mathrm{\parallel }}_p\mathrm{\parallel }v{\mathrm{\parallel }}_q=\mathrm{\parallel }[f]{\mathrm{\parallel }}_p\mathrm{\parallel }g{\mathrm{\parallel }}_q\mathrm{.}$

If $u^{\mathrm{\prime }}$ is another representative of $[f]$, then $u=u^{\mathrm{\prime }}$ almost everywhere. Hence $uv=u^{\mathrm{\prime }}v$ almost everywhere. Since both products are integrable by step 1.1, [L3] gives $\int uvd\mu =\int u^{\mathrm{\prime }}vd\mu ,$ so ${\mathrm{\Lambda }}_g([f])$ is well defined on the class $[f]$.

If $v^{\mathrm{\prime }}$ is another representative of $g$, then $v=v^{\mathrm{\prime }}$ almost everywhere. Hence $uv=u{v}^{\mathrm{\prime }}$ almost everywhere. Since both products are integrable by step 1.1, [L3] gives $\int uvd\mu =\int u{v}^{\mathrm{\prime }}d\mu ,$ so ${\mathrm{\Lambda }}_g([f])$ is independent of the chosen representative of $g$.

Let $[f],[h]\in L^p(\mu )$ and let $a,b\in \mathbb{R}$. Choose representatives $u,w$. By step 2.1 we may compute with representatives, and [L4] gives ${\mathrm{\Lambda }}_g(a[f]+b[h])=\int (au+bw)vd\mu =a\int uvd\mu +b\int wvd\mu =a{\mathrm{\Lambda }}_g([f])+b{\mathrm{\Lambda }}_g([h])\mathrm{.}$ The estimate from step 1.1 also shows $\mathrm{\mid }{\mathrm{\Lambda }}_g([f])\mathrm{\mid }\le \mathrm{\parallel }[f]{\mathrm{\parallel }}_p\mathrm{\parallel }g{\mathrm{\parallel }}_q,$ so ${\mathrm{\Lambda }}_g$ is linear and bounded.

The estimate in step 3.1 holds for every $[f]$ with $\mathrm{\parallel }[f]{\mathrm{\parallel }}_p\le 1$. Taking the supremum over the unit ball gives $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }\le \mathrm{\parallel }g{\mathrm{\parallel }}_q\mathrm{.}$ Thus ${\mathrm{\Lambda }}_g$ is a bounded linear functional on $L^p(\mu )$.

The upper bound $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }\le \mathrm{\parallel }g{\mathrm{\parallel }}_q$ is exactly [L1].

If $g=0$ in $L^q(\mu )$, then ${\mathrm{\Lambda }}_g=0$ by definition, so $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }=0=\mathrm{\parallel }g{\mathrm{\parallel }}_q$. Hence only the case $g\mathrm{\ne }0$ remains.

Assume $1<p<\mathrm{\infty }$. Choose a representative $u$ of $g$ and define $s(x):=\{\begin{array}{ll}u(x)\mathrm{/}\mathrm{\mid }u(x)\mathrm{\mid },& u(x)\mathrm{\ne }0,\\ 0,& u(x)=0,\end{array}f(x):=\frac{\mathrm{\mid }u(x){\mathrm{\mid }}^{q-1}s(x)}{\mathrm{\parallel }g{\mathrm{\parallel }}_q^{q-1}}\mathrm{.}$ Then $\mathrm{\mid }f{\mathrm{\mid }}^p=\mathrm{\mid }u{\mathrm{\mid }}^q\mathrm{/}\mathrm{\parallel }g{\mathrm{\parallel }}_q^q$ by [L2], so $\mathrm{\parallel }[f]{\mathrm{\parallel }}_p^p=\frac{1}{\mathrm{\parallel }g{\mathrm{\parallel }}_q^q}\int \mathrm{\mid }u{\mathrm{\mid }}^{q}d\mu =1.$ Also $fu=\mathrm{\mid }u{\mathrm{\mid }}^q\mathrm{/}\mathrm{\parallel }g{\mathrm{\parallel }}_q^{q-1}$, so ${\mathrm{\Lambda }}_g([f])=\frac{1}{\mathrm{\parallel }g{\mathrm{\parallel }}_q^{q-1}}\int \mathrm{\mid }u{\mathrm{\mid }}^{q}d\mu =\mathrm{\parallel }g{\mathrm{\parallel }}_q\mathrm{.}$ Therefore $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }\ge \mathrm{\parallel }g{\mathrm{\parallel }}_q$. Together with step 1.1, this gives $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }=\mathrm{\parallel }g{\mathrm{\parallel }}_q$.

Assume $p=1$, so $q=\mathrm{\infty }$, and assume $\mu$ is semifinite. Put $M:=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$. Step 2.1 leaves only $g\mathrm{\ne }0$, so $M>0$. Choose a representative $u$ of $g$, and define $s(x):=\{\begin{array}{ll}1,& u(x)>0,\\ -1,& u(x)<0,\\ 0,& u(x)=0.\end{array}$ For $0<\epsilon <M$, [L3] gives a measurable set $E_{\epsilon }:=\{\mathrm{\mid }u\mathrm{\mid }>M-\epsilon \}$ of positive measure. By [L4], choose $F_{\epsilon }\subseteq E_{\epsilon }$ with $0<\mu (F_{\epsilon })<\mathrm{\infty }$, and set $f_{\epsilon }:=\frac{s{\mathbf{1}}_{F_{\epsilon }}}{\mu (F_{\epsilon })}\mathrm{.}$ Then $\mathrm{\parallel }[f_{\epsilon }]{\mathrm{\parallel }}_1=1$ and ${\mathrm{\Lambda }}_g([f_{\epsilon }])=\frac{1}{\mu (F_{\epsilon })}{\int }_{F_{\epsilon }}\mathrm{\mid }u\mathrm{\mid }d\mu \ge M-\epsilon \mathrm{.}$ Hence $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }\ge M-\epsilon$ for every $0<\epsilon <M$, so $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }\ge M=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$. Step 1.1 gives the reverse inequality, and therefore $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$.

Step 2.2 proves the strict-exponent case, and step 3.1 proves the $q=\mathrm{\infty }$ endpoint under semifiniteness. Together with steps 1.1 and 2.1, this proves the proposition.

For every $[f]\in L^p(\mu )$, the assumption gives [given, algebra] ${\mathrm{\Lambda }}_{g-h}([f])=\int f(g-h)d\mu =0.$ So ${\mathrm{\Lambda }}_{g-h}$ is the zero functional on $L^p(\mu )$.

Applying [L1] to $u=g-h$ yields [L1, step 1.1] $\mathrm{\parallel }g-h{\mathrm{\parallel }}_q=\mathrm{\parallel }{\mathrm{\Lambda }}_{g-h}\mathrm{\parallel }=0.$

If $q<\mathrm{\infty }$, [L2] says that zero $L^q$ norm means the zero class, so [L2, L3, step 2.1] $g=h$ in $L^q(\mu )$. If $q=\mathrm{\infty }$, then step 2.1 and [L3] give $g=h$ almost everywhere. In either case the representing class is unique. ∎

Choose a measurable representative $u$ of $[f]$ and, for each $n\ge 1$, [L1, L2, given, choose, construct] set $E_n:=\{\mathrm{\mid }u\mathrm{\mid }\ge 1\mathrm{/}n\}=\{\mathrm{\mid }u{\mathrm{\mid }}^p\ge n^{-p}\}\mathrm{.}$ Since $u\in {\mathcal{L}}^p(\mu )$, [L2] gives $\mu (E_n)\le n^p\int \mathrm{\mid }u{\mathrm{\mid }}^{p}d\mu <\mathrm{\infty }\mathrm{.}$ So every $E_n$ has finite measure.

Put [L3, step 1.1] $S:={\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n\mathrm{.}$ By step 1.1 and [L3], the set $S$ is sigma-finite.

If $x\mathrm{\notin }S$, then $\mathrm{\mid }u(x)\mathrm{\mid }<1\mathrm{/}n$ for every $n\ge 1$, hence $u(x)=0$. [step 2.1, algebra] Therefore $u=0$ on $X\setminus S$, so the class $[f]$ vanishes almost everywhere outside the sigma-finite set $S$. ∎

For every measurable $E\subseteq X$, $\int \mathrm{\mid }{\mathbf{1}}_E{\mathrm{\mid }}^{p}d\mu =\mu (E)\le \mu (X)<\mathrm{\infty },$ so $[{\mathbf{1}}_E]\in L^p(\mu )$ and $\nu (E)$ is a finite scalar. Also $\nu (\mathrm{\varnothing })=\mathrm{\Lambda }(0)=0$.

Let $(E_n)$ be pairwise disjoint measurable sets, and put $E:={\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n,s_N:={\sum }_{n=1}^N{\mathbf{1}}_{E_n}={\mathbf{1}}_{{\bigcup }_{n=1}^N{E}_n}\mathrm{.}$ Then $s_N\to {\mathbf{1}}_E$ pointwise and $\mathrm{\mid }{s}_N-{\mathbf{1}}_E{\mathrm{\mid }}^p={\mathbf{1}}_{{\bigcup }_{n>N}{E}_n}\le {\mathbf{1}}_E\mathrm{.}$ Because $\mu (X)<\mathrm{\infty }$, the majorant ${\mathbf{1}}_E$ is integrable, so [L3] gives $\mathrm{\parallel }[s_N]-[{\mathbf{1}}_E]{\mathrm{\parallel }}_p^p=\int \mathrm{\mid }{s}_N-{\mathbf{1}}_E{\mathrm{\mid }}^{p}d\mu ⟶0.$

If $A,B\in \mathcal{A}$ are disjoint, then ${\mathbf{1}}_{A\cup B}={\mathbf{1}}_A+{\mathbf{1}}_B$. By linearity of $\mathrm{\Lambda }$, $\nu (A\cup B)=\mathrm{\Lambda }([{\mathbf{1}}_A+{\mathbf{1}}_B])=\mathrm{\Lambda }([{\mathbf{1}}_A])+\mathrm{\Lambda }([{\mathbf{1}}_B])=\nu (A)+\nu (B)\mathrm{.}$ So $\nu$ is finitely additive on disjoint measurable sets.

By continuity of $\mathrm{\Lambda }$ from [L1], finite additivity from step 2.1, and the convergence from step 1.2, $\nu (E)=\mathrm{\Lambda }([{\mathbf{1}}_E])={\lim }_{N\to \mathrm{\infty }}\mathrm{\Lambda }([s_N])={\lim }_{N\to \mathrm{\infty }}{\sum }_{n=1}^N\nu (E_n)\mathrm{.}$ So $\nu$ is countably additive. Together with step 1.1, [L2] shows that $\nu$ is a finite signed measure.

Let $E\in \mathcal{A}$ with $\mu (E)=0$. Then ${\mathbf{1}}_E=0$ almost [L2, given] everywhere, so [L2] gives $[{\mathbf{1}}_E]=[0]\text{in }{L}^p(\mu )\mathrm{.}$

Applying $\mathrm{\Lambda }$ and then [L3] yields [L3, step 1.1] $\nu (E)=\mathrm{\Lambda }([{\mathbf{1}}_E])=\mathrm{\Lambda }([0])=0.$ Therefore $\nu \ll \mu$.

By [L1] and [L2], choose $g\in L^1(\mu )$ such that $\nu (E)={\int }_{E}gd\mu (E\in \mathcal{A})\mathrm{.}$ Since $\nu (E)=\mathrm{\Lambda }([{\mathbf{1}}_E])$, this means $\mathrm{\Lambda }([{\mathbf{1}}_E])=\int {\mathbf{1}}_{E}gd\mu (E\in \mathcal{A})\mathrm{.}$

Let $u$ be bounded and measurable, with $\mathrm{\mid }u\mathrm{\mid }\le M$. By [L3], choose simple functions $(s_n)$ such that $\mathrm{\mid }{s}_n\mathrm{\mid }\le M$ for every $n$ and $s_n\to u$ pointwise. Since $\mu (X)<\mathrm{\infty }$, $\mathrm{\mid }{s}_n-u{\mathrm{\mid }}^p\le (2M{)}^p{\mathbf{1}}_X,$ and the majorant is integrable. Therefore [L4] gives $\mathrm{\parallel }[s_n]-[u]{\mathrm{\parallel }}_p^p=\int \mathrm{\mid }{s}_n-u{\mathrm{\mid }}^{p}d\mu ⟶0,$ so continuity of $\mathrm{\Lambda }$ yields $\mathrm{\Lambda }([s_n])⟶\mathrm{\Lambda }([u])\mathrm{.}$

For each simple approximant $s_n$, step 1.1 and linearity give $\mathrm{\Lambda }([s_n])=\int s_{n}gd\mu \mathrm{.}$ Thus the formula already holds for every simple function, and in particular for the chosen sequence $(s_n)$.

Because $g\in L^1(\mu )$ and $\mathrm{\mid }(s_n-u)g\mathrm{\mid }\le 2M\mathrm{\mid }g\mathrm{\mid }$ with $2M\mathrm{\mid }g\mathrm{\mid }\in L^1(\mu )$, [L4] gives $\int s_{n}gd\mu ⟶\int ugd\mu \mathrm{.}$ Combining this with step 1.2 and step 2.1 yields $\mathrm{\Lambda }([u])=\int ugd\mu \mathrm{.}$ So the Radon-Nikodym density represents $\mathrm{\Lambda }$ on every bounded measurable representative.

Assume first $1<p<\mathrm{\infty }$. Choose a representative $u$ of $g$ and, for $k\ge 1$, define $u_k:=u{\mathbf{1}}_{\{\mathrm{\mid }u\mathrm{\mid }\le k\}},f_k:=\mathrm{\mid }{u}_k{\mathrm{\mid }}^{q-1}\mathrm{sgn}u\mathrm{.}$ Each $f_k$ is bounded, so [L1] gives $\mathrm{\Lambda }([f_k])=\int f_{k}gd\mu =\int \mathrm{\mid }{u}_k{\mathrm{\mid }}^{q}d\mu \mathrm{.}$ Also $\mathrm{\mid }{f}_k{\mathrm{\mid }}^p=\mathrm{\mid }{u}_k{\mathrm{\mid }}^q$ by [L2], hence $\int \mathrm{\mid }{u}_k{\mathrm{\mid }}^{q}d\mu =\mathrm{\mid }\mathrm{\Lambda }([f_k])\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\mathrm{\parallel }[f_k]{\mathrm{\parallel }}_p=\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }{(\int \mathrm{\mid }{u}_k{\mathrm{\mid }}^{q}d\mu )}^{1\mathrm{/}p}\mathrm{.}$ Therefore $\int \mathrm{\mid }{u}_k{\mathrm{\mid }}^{q}d\mu \le \mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}^q(k\ge 1)\mathrm{.}$

Assume instead $p=1$, so $q=\mathrm{\infty }$. Put $M:=\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }$ and let $u$ be a representative of $g$. Fix $t>M$ and set $E_t:=\{\mathrm{\mid }u\mathrm{\mid }>t\}$. If $\mu (E_t)>0$, then $E_t$ has finite measure because $\mu (X)<\mathrm{\infty }$, and $f_t:=\mathrm{sgn}u{\mathbf{1}}_{E_t}$ is bounded with $\mathrm{\parallel }[f_t]{\mathrm{\parallel }}_1=\mu (E_t)$. By [L1], $\mathrm{\mid }\mathrm{\Lambda }([f_t])\mathrm{\mid }={\int }_{E_t}\mathrm{\mid }u\mathrm{\mid }d\mu >t\mu (E_t)=t\mathrm{\parallel }[f_t]{\mathrm{\parallel }}_1,$ contradicting the bound $\mathrm{\mid }\mathrm{\Lambda }([f_t])\mathrm{\mid }\le M\mathrm{\parallel }[f_t]{\mathrm{\parallel }}_1$. Hence $\mu (E_t)=0$ for every $t>M$, and [L4] gives $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le M$.

Step 1.1 and monotone convergence prove the strict-exponent case, while step 1.2 proves the endpoint case. Indeed, from step 1.1 the sequence $\mathrm{\mid }{u}_k{\mathrm{\mid }}^q$ increases pointwise to $\mathrm{\mid }u{\mathrm{\mid }}^q$, so [L3] gives $\int \mathrm{\mid }u{\mathrm{\mid }}^{q}d\mu ={\lim }_{k\to \mathrm{\infty }}\int \mathrm{\mid }{u}_k{\mathrm{\mid }}^{q}d\mu \le \mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}^q\mathrm{.}$ Thus in every case $g\in L^q(\mu )$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_q\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }$.

Let $[f]\in L^p(\mu )$ and choose a representative $w$. For each $n\ge 1$, set $w_n:=w{\mathbf{1}}_{\{\mathrm{\mid }w\mathrm{\mid }\le n\}}\mathrm{.}$ Then each $w_n$ is bounded and $\mathrm{\mid }{w}_n-w{\mathrm{\mid }}^p\le \mathrm{\mid }w{\mathrm{\mid }}^p$ with $\mathrm{\mid }{w}_n-w{\mathrm{\mid }}^p\to 0$ pointwise, so dominated convergence gives $\mathrm{\parallel }[w_n]-[f]{\mathrm{\parallel }}_p^p=\int \mathrm{\mid }{w}_n-w{\mathrm{\mid }}^{p}d\mu ⟶0.$ By step 2.1, $g\in L^q(\mu )$, so [L5] shows that $I_g([h]):=\int hgd\mu$ is a bounded linear functional on $L^p(\mu )$. Since [L1] gives $\mathrm{\Lambda }([w_n])=I_g([w_n])$ for every $n$, continuity of both functionals yields $\mathrm{\Lambda }([f])=I_g([f])=\int wgd\mu \mathrm{.}$ Thus the Radon-Nikodym density represents $\mathrm{\Lambda }$ on all of $L^p(\mu )$.

By [L3], choose an increasing sequence $(X_n)$ of measurable sets with [L1, L3, given, choose, construct] $X={\bigcup }_n{X}_n$ and $\mu (X_n)<\mathrm{\infty }$ for every $n$. For each $n$, let ${\mathrm{\Lambda }}_n$ be the bounded functional on $L^p(\mu {\mathrm{\mid }}_{X_n})$ defined by ${\mathrm{\Lambda }}_n([u]):=\mathrm{\Lambda }([\tilde{u}]),$ where $\tilde{u}$ is the extension of $u$ by $0$ outside $X_n$. Then $\mathrm{\parallel }{\mathrm{\Lambda }}_n\mathrm{\parallel }\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }$. Applying [L1] on $X_n$, choose $g_n\in L^q(\mu {\mathrm{\mid }}_{X_n})$ such that ${\mathrm{\Lambda }}_n([u])={\int }_{X_n}u{g}_{n}d\mu ([u]\in L^p(\mu {\mathrm{\mid }}_{X_n}))$ and $\mathrm{\parallel }{g}_n{\mathrm{\parallel }}_q\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }$.

If $m>n$, then both $g_n$ and $g_m{\mathrm{\mid }}_{X_n}$ represent the same functional [L2, L3, step 1.1] ${\mathrm{\Lambda }}_n$ on the finite measure space $X_n$. Since finite measure implies semifinite, [L2] gives $g_m=g_n\mu \text{-almost everywhere on }{X}_n\mathrm{.}$ Therefore the local densities are compatible on overlaps.

Define a measurable function $g$ by setting $g=g_n$ almost everywhere on [L6, step 1.1, step 2.1, algebra] each $X_n$. This is consistent by step 2.1. If $q<\mathrm{\infty }$, then ${\int }_{X_n}\mathrm{\mid }g{\mathrm{\mid }}^{q}d\mu ={\int }_{X_n}\mathrm{\mid }{g}_n{\mathrm{\mid }}^{q}d\mu \le \mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}^q,$ so monotone convergence gives ${\int }_X\mathrm{\mid }g{\mathrm{\mid }}^{q}d\mu ={\lim }_{n\to \mathrm{\infty }}{\int }_{X_n}\mathrm{\mid }g{\mathrm{\mid }}^{q}d\mu \le \mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}^q\mathrm{.}$ If $q=\mathrm{\infty }$, then for each $n$ there is a null set $N_n\subseteq X_n$ with $\mathrm{\mid }{g}_n\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }$ on $X_n\setminus N_n$; the union $N:={\bigcup }_n{N}_n$ is null, and $\mathrm{\mid }g\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }$ on $X\setminus N$. Hence in every case $g\in L^q(\mu )$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_q\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }$.

Let $[f]\in L^p(\mu )$ and choose a representative $w$. [L4, L5, step 1.1, step 3.1, given, construct] Set $w_n:=w{\mathbf{1}}_{X_n}$. Then $\mathrm{\mid }{w}_n-w{\mathrm{\mid }}^p=\mathrm{\mid }w{\mathrm{\mid }}^p{\mathbf{1}}_{X\setminus X_n}\le \mathrm{\mid }w{\mathrm{\mid }}^p,$ and the right-hand side is integrable. Since $w_n\to w$ pointwise, [L5] gives $\mathrm{\parallel }[w_n]-[f]{\mathrm{\parallel }}_p^p=\int \mathrm{\mid }{w}_n-w{\mathrm{\mid }}^{p}d\mu ⟶0.$ For each $n$, step 1.1 gives $\mathrm{\Lambda }([w_n])={\mathrm{\Lambda }}_n([w{\mathrm{\mid }}_{X_n}])={\int }_{X_n}w{g}_{n}d\mu =\int w_{n}gd\mu \mathrm{.}$ By step 3.1 and [L4], the map $I_g([h])=\int hgd\mu$ is a bounded linear functional on $L^p(\mu )$. Taking limits in the last display yields $\mathrm{\Lambda }([f])=I_g([f])=\int wgd\mu \mathrm{.}$ So $g$ represents $\mathrm{\Lambda }$ on all of $L^p(\mu )$.

Since $\mathrm{\Lambda }={\mathrm{\Lambda }}_g$, [L2, L3, L4, step 4.1] [L4] gives $\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }=\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }=\mathrm{\parallel }g{\mathrm{\parallel }}_q\mathrm{.}$ If $h\in L^q(\mu )$ also represents $\mathrm{\Lambda }$, then sigma-finite implies semifinite by [L3], so [L2] gives $h=g$ in $L^q(\mu )$. Thus the representing class is unique.

Let $\mathcal{S}$ be the family of sigma-finite measurable subsets of $X$. For each $E\in \mathcal{S}$, define the restricted functional ${\mathrm{\Lambda }}_E([u]):=\mathrm{\Lambda }([\tilde{u}])$ on $L^p(\mu {\mathrm{\mid }}_E)$, where $\tilde{u}$ is extended by $0$ outside $E$. By [L1], there is a unique $g_E\in L^q(\mu {\mathrm{\mid }}_E)$ such that ${\mathrm{\Lambda }}_E([u])={\int }_{E}u{g}_{E}d\mu ([u]\in L^p(\mu {\mathrm{\mid }}_E)),$ and $\mathrm{\parallel }{g}_E{\mathrm{\parallel }}_q=\mathrm{\parallel }{\mathrm{\Lambda }}_E\mathrm{\parallel }\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\mathrm{.}$