Prove Hölder's Inequality

Why Products Need Their Own Bound

The Dominated Convergence Theorem controls integrals when a sequence is bounded by one integrable function. A different question arises when the function to be integrated is a product: how can the sizes of two factors control the integral of that product? Hölder’s inequality provides the answer. Its proof has a useful structure: first prove a numerical inequality for two nonnegative numbers, then normalize functions so that their powers have integral one, and finally integrate the numerical inequality.

Throughout, \((X,\mathcal A,\mu)\) is a measure space, and all functions are measurable and real-valued. For \(1<p<\infty\), the exponent conjugate to \(p\) is the number \(q\) satisfying \(1/p+1/q=1\). Equivalently, \(q=p/(p-1)\). The restriction \(1<p<\infty\) ensures that both exponents are finite and greater than one.

The essential supremum ignores values on sets of measure zero, just as integrals do. The theorem below concerns \(f\in L^p(X)\) and \(g\in L^q(X)\), so the two norms on its right-hand side are finite. It asserts, in particular, that the product \(fg\) is integrable.

The Numerical Inequality Behind the Proof

The normalization step will produce numbers whose \(p\)-th and \(q\)-th powers each have integral one. The numerical estimate needed at each point is Young’s inequality.

Proof. If \(b=0\), the inequality reads \(0\leq a^p/p\), and equality holds exactly when \(a=0\), which is exactly the condition \(a^p=b^q\). Now suppose \(b>0\), and define \( \displaystyle \phi(a)=\frac{a^p}{p}-ab+\frac{b^q}{q}\qquad(a\geq0). \) For \(a>0\), its derivative is \(\phi'(a)=a^{p-1}-b\). This derivative is negative when \(0<a<b^{1/(p-1)}\), zero when \(a=b^{1/(p-1)}\), and positive when \(a>b^{1/(p-1)}\). Thus \(\phi\) has its minimum at \(a=b^{1/(p-1)}=b^{q-1}\). At that value \(a^p=b^q\) and \(ab=b^q\), so \( \displaystyle \phi(a)=\frac{b^q}{p}-b^q+\frac{b^q}{q}=0, \) because \(1/p+1/q=1\). Therefore \(\phi(a)\geq0\) for every \(a\geq0\), proving the inequality. The minimum is attained only at \(a=b^{q-1}\), which is equivalent to \(a^p=b^q\). This also covers \(a=0\), since for \(b>0\) the minimum value zero occurs at a positive \(a\). \(\square\)

The equality condition is important: the two terms being compared are balanced precisely when their powers agree. In the functional inequality, this condition will apply to the normalized absolute values of the functions.

Hölder’s Inequality

Proof. Write \(A=\|f\|_p\) and \(B=\|g\|_q\). Both are finite and nonnegative. First consider the case \(A=0\) or \(B=0\). If \(A=0\), then \(\int_X|f|^p\,d\mu=0\), so \(f=0\) almost everywhere. To see the last implication, for each positive integer \(n\), the set \(E_n=\{x:|f(x)|\geq1/n\}\) satisfies \( \displaystyle 0=\int_X|f|^p\,d\mu\geq\frac{\mu(E_n)}{n^p}, \) so each \(E_n\) is null. Since \(\{x:|f(x)|>0\}=\bigcup_{n=1}^{\infty}E_n\), \(f=0\) almost everywhere. The same argument applies if \(B=0\). In either case, \(|fg|=0\) almost everywhere, and the asserted inequality follows.

It remains to consider \(A>0\) and \(B>0\). At every point where the functions are finite, apply Young’s inequality with \(a=|f(x)|/A\) and \(b=|g(x)|/B\). These functions are finite almost everywhere because their finite power integrals rule out an infinite value on a set of positive measure. We obtain, almost everywhere, \( \displaystyle \frac{|f(x)g(x)|}{AB}\leq\frac{1}{p}\frac{|f(x)|^p}{A^p}+\frac{1}{q}\frac{|g(x)|^q}{B^q}. \) The right-hand side is integrable. Integrating this nonnegative inequality gives \( \displaystyle \frac{1}{AB}\int_X|fg|\,d\mu\leq\frac{1}{pA^p}\int_X|f|^p\,d\mu+\frac{1}{qB^q}\int_X|g|^q\,d\mu=\frac1p+\frac1q=1. \) Thus \(fg\in L^1(X)\) and \(\int_X|fg|\,d\mu\leq AB\). Finally, the absolute-value bound for an integral gives \( \displaystyle \left|\int_Xfg\,d\mu\right|\leq\int_X|fg|\,d\mu\leq AB. \) This proves both claims. \(\square\)

The proof does not require the functions to be bounded or the measure space to have finite measure. The essential ingredients are the two finite power integrals and the relation between the exponents. The final absolute-value bound for an integral is the same general estimate used earlier in this course.

Worked Applications

The Endpoint Exponents

The conjugate-exponent relation also has endpoint cases: \(p=1\), \(q=\infty\), and \(p=\infty\), \(q=1\). These do not follow by substituting an infinite exponent into the normalization proof, but their bounds have a direct proof from the definition of essential supremum.

Proof. Let \(C=\|g\|_\infty\). By the definition of essential supremum, \(|g|\leq C\) almost everywhere. Therefore \(|fg|\leq C|f|\) almost everywhere, and monotonicity of the integral gives \( \displaystyle \int_X|fg|\,d\mu\leq C\int_X|f|\,d\mu=\|g\|_\infty\|f\|_1. \) The product is measurable and this bound is finite, so it is integrable. Interchanging the roles of \(f\) and \(g\) proves the other endpoint case. \(\square\)

Equality and a Common Proof Pitfall

The normalization also explains the equality condition in the standard case. When both norms are positive and finite, equality in Hölder can occur only if Young’s inequality is an equality almost everywhere; that is, \( \displaystyle \left(\frac{|f|}{\|f\|_p}\right)^p=\left(\frac{|g|}{\|g\|_q}\right)^q \quad\text{almost everywhere}. \) Conversely, if this relation holds almost everywhere, equality holds in the integrated Young inequality and hence in the bound for \(\int_X|fg|\,d\mu\). This criterion concerns the integral of the absolute product. Equality in the signed bound \(|\int_Xfg\,d\mu|\leq\|f\|_p\|g\|_q\) also requires that cancellation of signs not reduce the absolute value of the integral.

A common mistake is to apply Young’s inequality directly to \(|f|\) and \(|g|\), then integrate, without first normalizing. That yields a bound involving \(\int|f|^p\) and \(\int|g|^q\), but it does not give the product of the desired norms. Dividing by the norms first is what makes the two power integrals equal to one. The zero-norm cases must be handled separately because division by a zero norm is not defined.