From Young's Inequality to an Integral Bound
Young's inequality controls a product of two numbers by a weighted sum of their powers. Hölder's inequality turns that pointwise estimate into a bound on the integral of a product. Its key step is normalization: divide each function by its own norm so that the integrals of the relevant powers both equal one. The resulting estimate applies on arbitrary measure spaces; the measure need not be finite or sigma-finite.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space, with no assumption of completeness, finiteness, or sigma-finiteness. Functions are measurable, finite-valued real- or complex-valued representatives. For \(1\le r<\infty\), \(L^r(\mu)\) consists of the almost-everywhere equivalence classes for which \(\int_X |h|^r\,d\mu<\infty\), and \(\|h\|_r=(\int_X|h|^r\,d\mu)^{1/r}\). The integral of a complex-valued function is defined when its absolute value has finite integral.
Hölder's Inequality
Proof. Write \(A=\|f\|_p\) and \(B=\|g\|_q\), both finite by hypothesis. First handle the cases where either norm is zero. If \(A=0\), then \(\int_X|f|^p\,d\mu=0\). For each positive integer \(n\), the measurable set \(E_n=\{|f|>1/n\}\) satisfies $$ 0\le n^{-p}\mu(E_n)\le\int_X|f|^p\,d\mu=0. $$ Thus every \(E_n\) is null, and \(\{|f|>0\}=\bigcup_{n=1}^\infty E_n\) is null. Hence \(f=0\) almost everywhere and \(|fg|=0\) almost everywhere, so the claimed integral bound holds with both sides zero. If \(B=0\), the same argument applied to \(g\) gives the same conclusion. This covers all cases in which at least one norm vanishes.
Now suppose \(A>0\) and \(B>0\), and define the nonnegative measurable functions \(u=|f|/A\) and \(v=|g|/B\). By the definitions of \(A\) and \(B\), with the constants pulled out of the nonnegative integrals, $$ \int_X u^p\,d\mu=A^{-p}\int_X|f|^p\,d\mu=1,\qquad \int_X v^q\,d\mu=B^{-q}\int_X|g|^q\,d\mu=1. $$ Young's inequality applies at each point of \(X\), since \(u,v\ge0\). It gives \(uv\le u^p/p+v^q/q\). Integrating this pointwise inequality and using additivity and homogeneity of the nonnegative Lebesgue integral gives $$ \int_X uv\,d\mu\le\frac1p\int_Xu^p\,d\mu+\frac1q\int_Xv^q\,d\mu =\frac1p+\frac1q=1. $$ In particular, \(uv\) has finite integral. Since \(|fg|=ABuv\), homogeneity now yields $$ \int_X|fg|\,d\mu=AB\int_Xuv\,d\mu\le AB. $$ Thus \(fg\) is integrable, including when \(f,g\) are complex-valued.
For completeness, if \(h\) is an integrable complex-valued function, then \(\left|\int_Xh\,d\mu\right|\le\int_X|h|\,d\mu\). If \(z=\int_Xh\,d\mu\ne0\), set \(c=\overline z/|z|\). Then \(|c|=1\) and $$ \left|\int_Xh\,d\mu\right|=|z|=\operatorname{Re}(cz) =\int_X\operatorname{Re}(ch)\,d\mu \le\int_X|h|\,d\mu. $$ The last inequality holds pointwise because \(\operatorname{Re}(w)\le|w|\) for every complex number \(w\). If \(z=0\), the inequality holds because its left side is zero and its right side is nonnegative. Apply this fact to \(h=fg\) and use the bound just proved. \(\square\)
The proof uses no interchange of a limit and an integral. It uses Young's inequality pointwise, then integrates a nonnegative inequality whose right-hand side has finite integral. In particular, no assumption that the whole space has finite measure is needed.
Worked Applications
Worked Example: A Power Function on the Unit Interval
Take \([0,1]\) with its Borel sigma-algebra and Lebesgue measure \(m\), let \(p=3\), \(q=3/2\), \(f(x)=1\), and \(g(x)=x\). These are measurable real-valued functions, and \(1/3+2/3=1\), so the exponents are conjugate.
First compute the norms: $$ \|f\|_3=\left(\int_0^1 1\,dx\right)^{1/3}=1,\qquad \|g\|_{3/2}=\left(\int_0^1x^{3/2}\,dx\right)^{2/3} =\left(\frac{2}{5}\right)^{2/3}. $$ The product integral is \(\int_0^1|fg|\,dx=\int_0^1x\,dx=1/2\). Hölder's bound is therefore \(1/2\le(2/5)^{2/3}\). To verify this comparison exactly, both sides are nonnegative, so cube them: \(1/8\le4/25\) is equivalent to \(25\le32\), which is true. Thus the computed integral satisfies the bound.
Check: The direct integral is \(1/2\), while the cube of the right-hand side is \(((2/5)^{2/3})^3=4/25\); the comparison \(1/8\le4/25\) confirms the inequality without a decimal approximation.
Worked Example: Finite Counting Measure
Let \(X=\{1,2,3\}\), let \(\mathcal{M}\) be its power set, and let \(\mu\) be counting measure, so \(\int_X h\,d\mu=h(1)+h(2)+h(3)\). Choose \(p=q=2\), \(f=(1,2,2)\), and \(g=(2,1,0)\). These are real-valued measurable functions and the exponents are conjugate.
The product integral is computed entry by entry: $$ \int_X|fg|\,d\mu=|1\cdot2|+|2\cdot1|+|2\cdot0|=2+2+0=4. $$ The two norms are $$ \|f\|_2=(1^2+2^2+2^2)^{1/2}=\sqrt9=3,\qquad \|g\|_2=(2^2+1^2+0^2)^{1/2}=\sqrt5. $$ The claimed bound is \(4\le3\sqrt5\). Both sides are nonnegative; squaring reduces it to \(16\le45\), which is true.
Check: The integral uses all three points, including the zero contribution at \(3\), and the squared comparison \(16\le45\) verifies the exact norm bound.
Worked Example: Overlapping Indicator Functions
On \([0,1]\) with Lebesgue measure, take \(p=4\), \(q=4/3\), \(f=2\chi_{[0,1/4]}\), and \(g=3\chi_{[0,1/2]}\). These are real-valued measurable functions. The exponents are conjugate because \(1/4+3/4=1\), and both functions are supported on sets of finite measure.
The intersection of their supports is \([0,1/4]\), of measure \(1/4\). Thus $$ \int_0^1|fg|\,dx=2\cdot3\cdot\frac14=\frac32. $$ For the first norm, \(\int_0^1|f|^4\,dx=2^4(1/4)=4\), so \(\|f\|_4=4^{1/4}=\sqrt2\). For the second, \(\int_0^1|g|^{4/3}\,dx=3^{4/3}(1/2)\), so $$ \|g\|_{4/3}=\left(3^{4/3}\frac12\right)^{3/4} =3\left(\frac12\right)^{3/4}. $$ The right-hand side of Hölder's inequality is therefore \(3\sqrt2(1/2)^{3/4}=3\cdot2^{-1/4}\). To check \(3/2\le3\cdot2^{-1/4}\), divide by \(3\) and raise both nonnegative sides to the fourth power: \(1/16\le1/2\). The bound holds.
Check: Only the overlap of the two supports contributes to the product integral; using its length \(1/4\) gives \(3/2\), the value compared above with the product of the separately computed norms.
A Common Exponent Mistake
The exponents must be conjugate; choosing two exponents greater than one is not enough. For a concrete failure, use Lebesgue measure on \([0,2]\) and let \(f=g=1\). If one incorrectly tries to use \(p=q=3\), then $$ \int_0^2|fg|\,dx=2,\qquad \|f\|_3\|g\|_3 =\left(\int_0^2 1\,dx\right)^{1/3} \left(\int_0^2 1\,dx\right)^{1/3} =2^{2/3}. $$ Since \(2^{2/3}<2\) (cubing gives \(4<8\)), this would assert a false inequality. The error is that \(1/3+1/3=2/3\ne1\). With the correct conjugate pair \(p=3,q=3/2\), the same functions have norms \(2^{1/3}\) and \(2^{2/3}\); their product is \(2\), matching the integral.
Hölder's inequality is useful because it converts separate integrability information into absolute integrability of a product and a quantitative bound for its integral. The theorem's hypotheses are specific: the first function must have finite \(p\)-norm, the second finite \(q\)-norm, and the exponents must be conjugate. It does not assert that arbitrary measurable functions have integrable products. The zero-norm case is also part of the statement: a function with zero \(L^p\) norm vanishes almost everywhere, so its product with any measurable function vanishes almost everywhere.
Check Your Understanding
For each question, identify the measure space and verify the exponent condition before applying the theorem.
- On \([0,1]\) with Lebesgue measure, take \(p=2\), \(q=2\), \(f(x)=x\), and \(g(x)=1\). Compute both norms and the product integral, then check Hölder's bound exactly.
- Explain why the proof treats \(\|f\|_p=0\) separately before dividing by \(\|f\|_p\).
- For counting measure on \(\{1,2\}\), compute the Hölder bound for \(f=(1,3)\), \(g=(2,2)\), with \(p=3\) and \(q=3/2\).
- In the proof, where is Young's inequality applied, and which normalized integrals equal one?
- Give a pair of exponents greater than one that is not conjugate, and explain why the theorem cannot be applied to that pair.