Why Products Have a Different Integrability Exponent
Multiplying two measurable functions can make their singularities stronger, so membership of each factor in an \(L^p\) space does not by itself say that the product belongs to the same space. The right target exponent comes from Hölder’s inequality: the reciprocal exponents add. This gives both a precise membership statement and a quantitative bound for the product.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; no finiteness, sigma-finiteness, or completeness assumption is imposed. Functions may be real-valued or complex-valued and are measurable. For \(1\le s<\infty\), \(L^s(\mu)\) consists of almost-everywhere equivalence classes of functions with \(\int_X |f|^s\,d\mu<\infty\), with norm \(\|f\|_s=(\int_X|f|^s\,d\mu)^{1/s}\). The space \(L^\infty(\mu)\) consists of classes with finite essential supremum norm \(\|f\|_\infty\). All statements about representatives, bounds, and products below are understood up to \(\mu\)-a.e. equality.
The Hölder inequality used here says that if \(1\le a,b\le\infty\), \(1/a+1/b=1\), and \(u\in L^a(\mu)\), \(v\in L^b(\mu)\), then \(\int_X|uv|\,d\mu\le\|u\|_a\|v\|_b\). The endpoint convention is \(1/\infty=0\). The earlier material on Hölder’s inequality establishes this result.
The Product Theorem
Proof. Choose measurable representatives of \(f\) and \(g\). Their product is measurable. If either representative is changed on a null set, the product changes only on the union of the two null sets, so it defines the same almost-everywhere equivalence class. It remains to prove the norm bound and membership.
First suppose \(p,q<\infty\). The condition \(1/p+1/q\le1\) gives \(r\ge1\), and the reciprocal relation gives \(r\le p\) and \(r\le q\). If \(r<\infty\), set \(a=p/r\) and \(b=q/r\). These exponents are at least \(1\), and $$ \frac1a+\frac1b=\frac rp+\frac rq=r\left(\frac1p+\frac1q\right)=1. $$ Apply Hölder to \(u=|f|^r\in L^a(\mu)\) and \(v=|g|^r\in L^b(\mu)\). The required integrability follows from \(\int|u|^a\,d\mu=\int|f|^p\,d\mu<\infty\) and \(\int|v|^b\,d\mu=\int|g|^q\,d\mu<\infty\). Thus $$ \int_X|fg|^r\,d\mu \le \left(\int_X|f|^p\,d\mu\right)^{r/p} \left(\int_X|g|^q\,d\mu\right)^{r/q}. $$ Taking the \(r\)-th root proves the stated bound and shows that \(fg\in L^r(\mu)\).
If \(p=\infty\) and \(q<\infty\), the reciprocal relation gives \(r=q\). By the definition of essential supremum, \(|f|\le\|f\|_\infty\) a.e.; hence \(|fg|^q\le\|f\|_\infty^q|g|^q\) a.e. Integration and taking the \(q\)-th root give \(\|fg\|_q\le\|f\|_\infty\|g\|_q\). The case \(q=\infty\), \(p<\infty\) follows by the same pointwise estimate with the roles exchanged. If \(p=q=\infty\), then \(|fg|\le\|f\|_\infty\|g\|_\infty\) a.e., so \(fg\in L^\infty(\mu)\) and the same bound holds. These cases exhaust the possibilities and prove the theorem. \(\square\)
The finite-exponent proof explains the reciprocal rule: raising the product to the target power \(r\) produces \(|f|^r\) and \(|g|^r\), to which Hölder applies with conjugate exponents \(p/r\) and \(q/r\). At an infinite endpoint, the essentially bounded factor is controlled pointwise instead.
Worked Example: A Product of Two Power Singularities
On \((0,1)\) with Lebesgue measure \(m\), let \(f(x)=g(x)=x^{-1/8}\). Take \(p=q=4\). The reciprocal relation gives \(1/r=1/4+1/4=1/2\), so \(r=2\).
For each factor, evaluate the fourth-power integral: $$ \int_0^1|f(x)|^4\,dm(x)=\int_0^1x^{-1/2}\,dx =\left[2x^{1/2}\right]_0^1=2. $$ Therefore \(\|f\|_4=2^{1/4}\), and the same calculation gives \(\|g\|_4=2^{1/4}\). Their product is \(fg=x^{-1/4}\), so $$ \|fg\|_2=\left(\int_0^1x^{-1/2}\,dx\right)^{1/2} =2^{1/2}. $$ The theorem’s bound is \(\|f\|_4\|g\|_4=2^{1/4}2^{1/4}=2^{1/2}\), equal to the product norm. Check: squaring the directly computed norm gives \(\int_0^1|fg|^2\,dm=2\); squaring the bound also gives \((2^{1/2})^2=2\).
Worked Example: A Strict Product Norm Bound
On \([0,1]\) with Lebesgue measure, set \(f(x)=x\) and \(g(x)=1\). Choose \(p=2\) and \(q=4\). Then \(1/r=1/2+1/4=3/4\), so \(r=4/3\).
The factor norms are $$ \|f\|_2=\left(\int_0^1x^2\,dx\right)^{1/2} =\left(\frac13\right)^{1/2}, \qquad \|g\|_4=\left(\int_0^1 1\,dx\right)^{1/4}=1. $$ Since \(fg=x\), its product norm is $$ \|fg\|_{4/3} =\left(\int_0^1x^{4/3}\,dx\right)^{3/4} =\left(\frac37\right)^{3/4}. $$ This is strictly below the Hölder bound \(1/\sqrt{3}\). To check the comparison exactly, both sides are positive, so raising them to the fourth power preserves their order: $$ \left(\frac37\right)^3=\frac{27}{343} \quad\hbox{and}\quad \left(\frac1{\sqrt3}\right)^4=\frac19. $$ The first is smaller because \(27\cdot9=243<343\). Thus \((3/7)^{3/4}<1/\sqrt3\), as claimed. The direct integral calculation and the theorem’s bound agree with the required inequality.
Worked Example: Products on a Counting Measure Space
Let \(X=\mathbb{N}\) with the power-set sigma-algebra and counting measure \(\mu\). Define \(f(1)=3\), \(f(2)=4\), and \(f(n)=0\) for \(n\ge3\); define \(g(1)=4\), \(g(2)=3\), and \(g(n)=0\) for \(n\ge3\). With \(p=q=2\), the relation gives \(r=1\).
Counting measure turns each integral into a sum. Thus $$ \|f\|_2=\left(3^2+4^2\right)^{1/2}=\sqrt{9+16}=5, \qquad \|g\|_2=\left(4^2+3^2\right)^{1/2}=\sqrt{16+9}=5. $$ The product has values \((fg)(1)=3\cdot4=12\), \((fg)(2)=4\cdot3=12\), and \((fg)(n)=0\) for \(n\ge3\). Therefore $$ \|fg\|_1=\sum_{n=1}^\infty|f(n)g(n)|=12+12=24. $$ The bound is \(\|f\|_2\|g\|_2=5\cdot5=25\), so \(24\le25\). Check: summing the two nonzero product values gives \(24\) directly, while the squared factor norms are each \(25\), giving the bound \(25\).
Worked Example: The Infinite-Exponent Endpoint
On \([0,1]\) with Lebesgue measure, set \(f(x)=2\) and \(g(x)=x\), and take \(p=\infty\), \(q=1\). The reciprocal relation gives \(r=1\). Direct integration yields $$ \|f\|_\infty=2,\qquad \|g\|_1=\int_0^1x\,dx=\frac12,\qquad \|fg\|_1=\int_0^1 2x\,dx=1. $$ The endpoint estimate gives \(\|fg\|_1\le2\cdot\frac12=1\), with equality. Check: the integrand \(fg=2x\) integrates to \([x^2]_0^1=1\), exactly the product of the two factor norms.
When the Exponent Relation Is Misused
The reciprocal relation is part of the theorem, not an optional way to guess a convenient target space. For example, with \(p=q=2\), the theorem gives \(r=1\), not \(r=2\). Here is a concrete reason that the distinction matters. On \((0,1)\) with Lebesgue measure, take \(f(x)=g(x)=x^{-1/3}\). Each factor belongs to \(L^2(m)\), because $$ \int_0^1|f(x)|^2\,dm(x)=\int_0^1x^{-2/3}\,dx =\left[3x^{1/3}\right]_0^1=3. $$ The same calculation applies to \(g\). But the product is \(fg=x^{-2/3}\), and its squared integral diverges: $$ \int_\varepsilon^1|f(x)g(x)|^2\,dm(x) =\int_\varepsilon^1x^{-4/3}\,dx =3\bigl(\varepsilon^{-1/3}-1\bigr)\longrightarrow\infty \quad\text{as }\varepsilon\downarrow0. $$ Thus \(fg\notin L^2(m)\). In contrast, \(fg\in L^1(m)\), since \(\int_0^1x^{-2/3}\,dm=3\), which matches \(1/r=1/2+1/2\), or \(r=1\).
More generally, if \(1/p+1/q>1\), the reciprocal equation would require \(r<1\). That is outside the range \(r\ge1\) of the \(L^r\) spaces and norm estimate considered here. It does not imply that a particular product fails to be integrable; it means this theorem supplies no general \(L^r\) conclusion in that range. For instance, on a finite measure space some products may still be integrable, while the singular example above shows that \(L^2\) membership of both factors does not guarantee \(L^2\) membership of their product.
The theorem also gives a well-defined multiplication operation on equivalence classes: changing either input on a null set changes the product only on a null set. Its norm estimate controls the output without asserting equality in general. The examples show both possibilities: the singular power example and the endpoint example attain the bound, whereas the \(x\) and \(1\) example and the counting-measure example give strict bounds.
Check Your Understanding
Use the reciprocal relation first, and check that every proposed exponent lies in the theorem’s range.
- If \(f\in L^3(\mu)\) and \(g\in L^6(\mu)\), what is the target exponent \(r\), and what norm bound follows?
- On \([0,1]\) with Lebesgue measure, let \(f(x)=x\) and \(g(x)=x\). For \(p=q=4\), calculate the target exponent and \(\|fg\|_r\).
- Explain why \(f\in L^\infty(\mu)\) and \(g\in L^2(\mu)\) imply \(fg\in L^2(\mu)\), including the norm estimate.
- For \(p=q=2\), identify the exponent guaranteed for \(fg\). Why does this not guarantee \(fg\in L^2(\mu)\)?
- Give the reciprocal condition on \(p,q\) that ensures the theorem’s target exponent satisfies \(r\ge1\), using \(1/\infty=0\).