The Representation Problem
Let \(1<p<\infty\), and write \(q=p/(p-1)\), so \(1/p+1/q=1\). Hölder's inequality shows that every real-valued \(g\in L^q(\mu)\) defines a bounded linear functional on \(L^p(\mu)\) by integration. The converse is the substantive direction: given a bounded linear functional \(T\), we must construct a measurable function \(g\), prove that \(g\in L^q(\mu)\), and show that \(T(f)=\int fg\,d\mu\) for every \(f\in L^p(\mu)\).
Here \(L^p(\mu)\) consists of equivalence classes, modulo equality \(\mu\)-a.e., of measurable real-valued functions with \(\int |f|^p\,d\mu<\infty\), and \(\|f\|_p=(\int |f|^p\,d\mu)^{1/p}\). We use Hölder's inequality and the Radon–Nikodym theorem for finite signed measures, both standard results. Sigma-finiteness supplies increasing measurable sets \(X_n\) with \(X=\bigcup_{n\ge1}X_n\) and \(\mu(X_n)<\infty\).
From the Functional to Local Measures
Let \(C=\|T\|\), so \(|T(f)|\le C\|f\|_p\) for every \(f\in L^p(\mu)\). On each finite-measure set \(X_n\), the indicator \(\chi_E\) belongs to \(L^p(\mu)\) whenever \(E\in\mathcal{M}\) and \(E\subseteq X_n\). Define \(\nu_n(E)=T(\chi_E)\) for such \(E\). This is a finite signed measure on the measurable subsets of \(X_n\), as the following argument verifies.
First, \(\nu_n(\varnothing)=0\). If \(E_j\subseteq X_n\) are pairwise disjoint and \(E=\bigcup_{j\ge1}E_j\), then the indicators of the partial unions converge to \(\chi_E\) in \(L^p\): indeed, $$ \left\|\chi_E-\sum_{j=1}^N\chi_{E_j}\right\|_p^p =\mu\!\left(E\setminus\bigcup_{j=1}^N E_j\right)\longrightarrow 0, $$ by continuity from below of measure and \(\mu(E)\le\mu(X_n)<\infty\). Linearity and continuity of \(T\) therefore give \(\nu_n(E)=\sum_{j\ge1}\nu_n(E_j)\).
The variation of \(\nu_n\) is finite. For any finite measurable partition \(E_1,\ldots,E_k\) of a measurable subset of \(X_n\), choose \(\epsilon_j\in\{-1,1\}\) so that \(\epsilon_j\nu_n(E_j)=|\nu_n(E_j)|\). Then $$ \sum_{j=1}^k|\nu_n(E_j)| =T\!\left(\sum_{j=1}^k\epsilon_j\chi_{E_j}\right) \le C\left(\sum_{j=1}^k\mu(E_j)\right)^{1/p} \le C\mu(X_n)^{1/p}. $$ Taking the supremum over finite partitions bounds the total variation by \(C\mu(X_n)^{1/p}\). Also, \(\nu_n\ll\mu\): if \(\mu(E)=0\), then \(\|\chi_E\|_p=0\), hence \(\nu_n(E)=0\).
The Radon–Nikodym theorem for finite signed measures now gives a real-valued \(g_n\in L^1(X_n,\mu)\) such that \(\nu_n(E)=\int_Eg_n\,d\mu\) for every measurable \(E\subseteq X_n\). In particular, linearity gives \(T(s)=\int_{X_n}sg_n\,d\mu\) for simple functions \(s\) supported in \(X_n\).
These local densities agree where their domains overlap. If \(m\ge n\), then for every measurable \(E\subseteq X_n\), \(\int_Eg_m\,d\mu=T(\chi_E)=\int_Eg_n\,d\mu\). Uniqueness in the Radon–Nikodym theorem implies \(g_m=g_n\) \(\mu\)-a.e. on \(X_n\). Define \(D_1=X_1\) and \(D_n=X_n\setminus X_{n-1}\) for \(n\ge2\), and set \(g=g_n\) on \(D_n\). This defines a measurable function on \(X\). The countably many local agreement statements imply that \(g=g_n\) a.e. on each \(X_n\); changing values on the resulting null set if needed, take \(g\) finite everywhere.
Truncated Power Tests Establish \(g\in L^q\)
To use the functional bound to control \(g\), test it with bounded functions supported on finite-measure sets. The identity \(T(h)=\int hg\,d\mu\) holds for every bounded measurable real-valued \(h\) supported in \(X_n\): approximate \(h\) uniformly by simple functions supported in \(X_n\). The approximants converge in \(L^p\) because \(\mu(X_n)<\infty\), and their integrals against \(g\) converge because \(g\in L^1(X_n)\).
For \(n,N\ge1\), put \(A_{n,N}=X_n\cap\{|g|\le N\}\), and define \(h_{n,N}=\operatorname{sgn}(g)|g|^{q-1}\chi_{A_{n,N}}\), where \(\operatorname{sgn}(0)=0\). This function is bounded and supported in \(X_n\), so the identity just established applies. Let \(I_{n,N}=\int_{A_{n,N}}|g|^q\,d\mu\), which is finite since \(|g|\le N\) there and \(\mu(X_n)<\infty\). Because \((q-1)p=q\), the functional bound gives $$ I_{n,N}=T(h_{n,N}) \le C\|h_{n,N}\|_p =C I_{n,N}^{1/p}. $$ If \(I_{n,N}=0\), then \(I_{n,N}^{1/q}\le C\) holds. Otherwise divide by \(I_{n,N}^{1/p}>0\) to obtain the same inequality. Thus \(I_{n,N}\le C^q\) for every \(n,N\). The sets \(A_{n,N}\) increase to \(X\), and the Monotone Convergence Theorem, applied to the nonnegative functions \(|g|^q\chi_{A_{n,N}}\), yields \(\int_X|g|^q\,d\mu\le C^q\). Consequently \(g\in L^q(\mu)\) and \(\|g\|_q\le\|T\|\).
Extending the Identity to Every \(L^p\) Function
First consider a simple function \(s\) supported on an arbitrary measurable set \(A\) with \(\mu(A)<\infty\). The set \(A\) need not be contained in any one \(X_n\), so the local identity cannot be applied directly. Instead, \(s\chi_{A\cap X_n}\) is supported in \(X_n\), and $$ \|s-s\chi_{A\cap X_n}\|_p^p =\int_{A\setminus X_n}|s|^p\,d\mu\longrightarrow0. $$ The convergence follows from continuity from below (or continuity from above on the finite-measure set \(A\)) and the fact that \(\bigcup_nX_n=X\). For each \(n\), the local identity gives \(T(s\chi_{A\cap X_n})=\int s\chi_{A\cap X_n}g\,d\mu\). Continuity of \(T\) sends the left side to \(T(s)\); Hölder's inequality sends the right side to \(\int sg\,d\mu\), since \(\left|\int (s-s\chi_{A\cap X_n})g\,d\mu\right|\le\|s-s\chi_{A\cap X_n}\|_p\|g\|_q\). Therefore \(T(s)=\int sg\,d\mu\).
Such simple functions are dense in \(L^p(\mu)\). To see the role of sigma-finiteness, for \(f\in L^p(\mu)\) truncate to \(f\chi_{X_n\cap\{|f|\le n\}}\). These truncations converge to \(f\) in \(L^p\), since their \(p\)-th power errors are bounded by \(|f|^p\) and tend to zero a.e.; the Dominated Convergence Theorem applies with the integrable dominating function \(|f|^p\). Each truncation is bounded and supported on a finite-measure set, and can be approximated uniformly by simple functions there. Finally, if simple \(s_j\to f\) in \(L^p\), then \(T(s_j)\to T(f)\) by boundedness, while \(\int s_jg\,d\mu\to\int fg\,d\mu\) by Hölder's inequality. This proves the representation for every \(f\in L^p(\mu)\).
Proof. Hölder's inequality and the representation give \(|T(f)|\le\|f\|_p\|g\|_q\), so \(\|T\|\le\|g\|_q\). The truncated power tests above, followed by monotone convergence, gave \(\|g\|_q\le\|T\|\). The two inequalities prove equality. \(\square\)
Proof. Let \(d=g-h\). For each \(n,k\ge1\), the set \(E_{n,k}=X_n\cap\{|d|>1/k\}\) has finite measure. The function \(f=\operatorname{sgn}(d)\chi_{E_{n,k}}\) belongs to \(L^p(\mu)\). The two representations imply \(0=\int fd\,d\mu=\int_{E_{n,k}}|d|\,d\mu\). Since \(|d|>1/k\) on \(E_{n,k}\), this forces \(\mu(E_{n,k})=0\). The set where \(d\ne0\) is \(\bigcup_{n,k\ge1}E_{n,k}\), so it is null. \(\square\)
Worked Examples and a Common Pitfall
Worked Example: A Functional on Three Points
Take \(X=\{1,2,3\}\) with counting measure, \(p=3/2\), and \(q=3\). Define \(T(f)=2f(1)-f(2)\). Its representing function is \(g=(2,-1,0)\), since \(\int fg\,d\mu=2f(1)-f(2)=T(f)\). The proposed dual norm is $$ \|g\|_3=(|2|^3+|-1|^3+|0|^3)^{1/3}=(8+1+0)^{1/3}=9^{1/3}. $$ To verify the functional norm, use \(u=(4,-1,0)\), which is \(\operatorname{sgn}(g)|g|^{q-1}\). Its \(L^{3/2}\) norm is \(\|u\|_{3/2}=(4^{3/2}+1^{3/2})^{2/3}=(8+1)^{2/3}=9^{2/3}\). Also \(T(u)=2(4)-(-1)=9\), so \(T(u)/\|u\|_{3/2}=9/9^{2/3}=9^{1/3}\). Hölder's inequality gives the matching upper bound \(9^{1/3}\), hence \(\|T\|=9^{1/3}=\|g\|_3\). Substitution into the representation recovers \(T(f)\) for every \(f\).
Worked Example: A Lebesgue Integral Functional
On \(([0,1],\mathcal{L},m)\), let \(p=q=2\) and \(g(x)=2x-1\). Define \(T(f)=\int_0^1 f(x)(2x-1)\,dx\). Direct calculation gives $$ \|g\|_2^2=\int_0^1(2x-1)^2\,dx =\left[\frac{4x^3}{3}-2x^2+x\right]_0^1 =\frac43-2+1=\frac13, $$ so \(\|g\|_2=1/\sqrt3\). Cauchy–Schwarz gives \(|T(f)|\le\|f\|_2/\sqrt3\). For the reverse bound, take \(f(x)=\sqrt3(2x-1)\). Then \(\|f\|_2^2=3(1/3)=1\), and \(T(f)=\sqrt3\int_0^1(2x-1)^2\,dx=\sqrt3/3=1/\sqrt3\). Thus \(\|T\|=\|g\|_2=1/\sqrt3\). Substituting this \(g\) into the pairing gives exactly the defining formula for \(T\).
Worked Example: A Functional on a Counting-Measure Space
On \(\mathbb{N}=\{1,2,\ldots\}\) with counting measure, take \(p=q=2\) and \(g(n)=2^{-n}\). The geometric series gives $$ \|g\|_2^2=\sum_{n=1}^{\infty}4^{-n} =\frac{1/4}{1-1/4}=\frac13. $$ For \(f\in\ell^2\), the series \(\sum_{n\ge1}f(n)2^{-n}\) converges absolutely by Cauchy–Schwarz, and defines \(T(f)\) with \(|T(f)|\le\|f\|_2/\sqrt3\). Set \(f(n)=\sqrt3\,2^{-n}\). Then \(\|f\|_2^2=3(1/3)=1\), while \(T(f)=\sqrt3\sum_{n\ge1}4^{-n}=\sqrt3/3=1/\sqrt3\). Therefore \(\|T\|=\|g\|_2=1/\sqrt3\). This also checks the representation by direct substitution into the defining series.
Common pitfall: assuming every finite-measure set lies inside one exhaustion set. In \(\mathbb{R}\), the set \(A=\bigcup_{k\ge1}[k,k+2^{-k}]\) has finite measure but is unbounded, so it is not contained in any \(X_n=[-n,n]\). For \(s=\chi_A\), use \(s_n=\chi_{A\cap X_n}\), not the local identity on \(A\) itself. Since \(m(A\setminus X_n)\to0\), \(\|s-s_n\|_p^p=m(A\setminus X_n)\to0\). Continuity gives \(T(s_n)\to T(s)\), and Hölder gives \(\int s_ng\,dm\to\int sg\,dm\). Thus the identity on \(A\) follows by truncation; it does not follow merely by declaring \(A\) to be contained in some \(X_n\).
Check Your Understanding
Use the hypotheses and steps of the representation proof to organize each answer.
- Why does sigma-finiteness allow the construction to begin on finite-measure sets \(X_n\), even when \(\mu(X)=\infty\)?
- For disjoint \(E_j\subseteq X_n\), explain why the partial-union indicators converge in \(L^p\) to the indicator of their union.
- In the truncated power test, verify the exponent identity \((q-1)p=q\) from \(q=p/(p-1)\).
- Why is the representation on simple functions supported in a finite-measure set insufficient by itself, and how is it extended to all of \(L^p(\mu)\)?
- State the two inequalities that together give \(\|T\|=\|g\|_q\).