Hölder's Inequality

Hölder’s Inequality as a Product Estimate

The \(L^p\) norm measures the size of a function by integrating a power of its absolute value. Hölder’s inequality explains how two different kinds of size control combine: if one function has finite \(p\)th-power integral and another has finite \(q\)th-power integral, their product is integrable when the exponents are conjugate. The theorem was stated and proved among the Properties of the \(L^p\) Norm earlier in the course. Here we use it as a tool, paying particular attention to choosing exponents to fit the quantity we want to estimate.

For \(1<p<\infty\) and its conjugate \(q\), Hölder’s inequality says that if \(f\in L^p(X,\mu)\) and \(g\in L^q(X,\mu)\), then \(fg\in L^1(X,\mu)\) and $$ \int_X |fg|\,d\mu\leq \|f\|_p\|g\|_q. $$ The absolute value is important: the estimate controls the integral of the product’s magnitude, not just the possibly smaller magnitude of its signed integral. At the endpoint, if \(f\in L^1\) and \(g\in L^\infty\), the corresponding estimate is \(\int_X|fg|\,d\mu\leq\|f\|_1\|g\|_\infty\), using an almost-everywhere bound for \(g\).

A practical way to use the inequality is to identify the expression to be integrated, then choose exponents that match the powers of its factors. Normalization can also make the estimate easier to interpret. If neither norm is zero, define $$ F=\frac{|f|}{\|f\|_p}, \qquad G=\frac{|g|}{\|g\|_q}. $$ Then \(\int_X F^p\,d\mu=1\) and \(\int_X G^q\,d\mu=1\), while Hölder gives \(\int_X FG\,d\mu\leq1\). If either norm is zero, the Zero Integral Criterion implies that function is zero almost everywhere, so the product integral is zero. This normalization highlights that the product bound compares the functions’ sizes relative to their respective norms.

Products in a Further \(L^r\) Space

Hölder’s inequality does more than show that a product is integrable. By applying it to powers of the factors, we can find an \(L^r\) space containing the product itself. The exponents must satisfy a balance condition: the reciprocal exponent for the product is the sum of the reciprocal exponents for its factors.

Proof. First take \(p,q\) finite. The exponent relation gives $$ \frac{r}{p}+\frac{r}{q}=1. $$ Since \(r<p\) and \(r<q\), the numbers \(p/r\) and \(q/r\) are both greater than \(1\); the displayed relation says that they are conjugate. Apply Hölder’s inequality to \(|f|^r\) and \(|g|^r\) with these exponents. It gives $$ \int_X |fg|^r\,d\mu \leq \left(\int_X |f|^p\,d\mu\right)^{r/p} \left(\int_X |g|^q\,d\mu\right)^{r/q} =\|f\|_p^r\|g\|_q^r. $$ The right-hand side is finite, so \(fg\in L^r\). Taking the \(r\)th root proves the claimed norm estimate. For the endpoint \(p=\infty,\ q=r\), we have \(|f|\leq\|f\|_\infty\) almost everywhere by the Least Almost-Everywhere Bound theorem. Therefore $$ \int_X|fg|^r\,d\mu \leq \|f\|_\infty^r\int_X|g|^r\,d\mu, $$ which gives the result after taking the \(r\)th root. Interchanging \(f\) and \(g\) proves the other endpoint case.

Interpolation Between \(L^p\) Norms

Another useful consequence of Hölder’s inequality is an estimate for an intermediate \(L^p\) norm. If a function belongs to both \(L^{p_0}\) and \(L^{p_1}\), with \(p_0<p_1\), Hölder can bound its norm at an exponent strictly between them. The relationship among the three exponents determines the powers on the two endpoint norms.

Proof. First suppose \(p_1<\infty\). Since \(p_0<p<p_1\), the exponent relation implies $$ 0<\frac{\theta p}{p_0}<1, \qquad 0<\frac{(1-\theta)p}{p_1}<1, \qquad \frac{\theta p}{p_0}+\frac{(1-\theta)p}{p_1}=1. $$ Thus \(a=p_0/(\theta p)\) and \(b=p_1/((1-\theta)p)\) are conjugate exponents greater than \(1\). Factor the integrand as $$ |f|^p=|f|^{\theta p}|f|^{(1-\theta)p}. $$ Hölder’s inequality applied with exponents \(a,b\) gives $$ \int_X|f|^p\,d\mu \leq \left(\int_X|f|^{p_0}\,d\mu\right)^{\theta p/p_0} \left(\int_X|f|^{p_1}\,d\mu\right)^{(1-\theta)p/p_1} =\|f\|_{p_0}^{\theta p}\|f\|_{p_1}^{(1-\theta)p}. $$ This is finite, so \(f\in L^p\); taking the \(p\)th root proves the estimate.

If \(p_1=\infty\), the relation \(1/p=\theta/p_0\) gives \(\theta p=p_0\). The Least Almost-Everywhere Bound theorem yields \(|f|\leq\|f\|_\infty\) almost everywhere, and hence $$ \int_X|f|^p\,d\mu \leq \|f\|_\infty^{(1-\theta)p}\int_X|f|^{\theta p}\,d\mu =\|f\|_\infty^{(1-\theta)p}\|f\|_{p_0}^{p_0}. $$ Taking the \(p\)th root and using \(p_0=\theta p\) proves the stated inequality. If one of the norms on the right-hand side is zero, \(f=0\) almost everywhere, and the same conclusion holds directly.

Choosing Exponents and Reading the Bound

In applications, exponent selection is part of the argument. For a product estimate, start with the desired target exponent \(r\), then check whether \(1/r=1/p+1/q\). For an integral estimate, the target is \(r=1\), so the condition becomes the conjugacy relation. For interpolation, solve the reciprocal-exponent equation for \(\theta\) before applying the theorem. In each case, the relation is a necessary part of the estimate, not an optional convenience.

A common mistake is to infer that \(f\in L^p\) and \(g\in L^q\) automatically imply \(fg\in L^1\), without checking the exponents. Hölder gives that conclusion when \(p\) and \(q\) are conjugate; the Product Estimate gives a different conclusion when their reciprocal exponents sum to \(1/r\). The measure space need not have finite measure for either estimate, but the assumed norms must be finite. Also, an \(L^\infty\) bound is an almost-everywhere bound, which is sufficient for integral estimates; values on a null set do not change the \(L^p\) element.