Computing Hölder Bounds

Turning Hölder’s Inequality into a Calculation

Hölder’s inequality is useful not only as a general estimate but also as a way to produce a specific numerical bound for an integral. The main calculation is to identify the two factors, choose conjugate exponents, and compute the corresponding norms accurately. The choice of exponents can determine whether the estimate is sharp or substantially larger than the integral.

For the general statement below, let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite, complete, or sigma-finite. Functions are measurable and real-valued. Write \(L^r(\mu)\) for the functions with finite \(r\)-norm, where \(1\le r<\infty\), and use the essential-supremum norm for \(L^\infty(\mu)\). Norms and equalities between functions are understood up to \(\mu\)-almost-everywhere equality.

A practical version appears when the factors are powers. For positive real numbers \(\alpha,\beta\), apply Hölder to \(|f|^\alpha\) and \(|g|^\beta\). If \(|f|^\alpha\in L^p(\mu)\) and \(|g|^\beta\in L^q(\mu)\), then \(\int_X |f|^\alpha|g|^\beta\,d\mu\le \||f|^\alpha\|_p\||g|^\beta\|_q\). The norms on the right are \((\int_X|f|^{\alpha p}\,d\mu)^{1/p}\) and \((\int_X|g|^{\beta q}\,d\mu)^{1/q}\). Thus, choosing \(p,q\) is also choosing which powers of the individual functions must have finite integrals.

Choosing Exponents for Powers

On \(([0,1],\mathcal{B}_{[0,1]},m)\), the monomials \(x^a\) and \(x^b\), for \(a,b>0\), have an especially transparent calculation. For \(1<p<\infty\) and \(q=p/(p-1)\), direct integration gives \(\int_0^1x^{a+b}\,dm=1/(a+b+1)\), \(\|x^a\|_p=(ap+1)^{-1/p}\), and \(\|x^b\|_q=(bq+1)^{-1/q}\). The Hölder bound is the product of the last two quantities. The next proposition identifies exactly when that bound is sharp.

Proof. Both functions have positive finite norms. By the equality criterion in Hölder’s inequality, equality holds precisely when the normalized powers \(|x^a|^p/\|x^a\|_p^p\) and \(|x^b|^q/\|x^b\|_q^q\) agree almost everywhere. Their denominators are positive constants, so this is equivalent to \(x^{ap}=C x^{bq}\) almost everywhere on \((0,1)\), for some constant \(C>0\). If \(ap=bq\), the powers are the same and the condition holds. Conversely, if \(ap\ne bq\), their ratio is \(x^{ap-bq}\), which is not constant on any interval in \((0,1)\); it cannot equal the fixed constant \(C\) almost everywhere. This proves both directions. \(\square\)

The example also illustrates why a convenient-looking choice such as \(p=q=2\) need not be sharp. It still gives a valid bound, but the powers of the two factors then differ. When the goal is a sharp estimate, compare the powers \(ap\) and \(bq\), rather than selecting exponents before examining the integrand.

Indicators: The Overlap Controls the Integral

For a measurable set \(A\in\mathcal{M}\), its indicator \(\chi_A\) equals \(1\) on \(A\) and \(0\) on \(A^c\). If \(A,B\in\mathcal{M}\) have finite measure in a measure space \((X,\mathcal{M},\mu)\), then \(\chi_A\chi_B=\chi_{A\cap B}\) everywhere. Consequently, for \(1<p<\infty\) and conjugate \(q\), the exact integral and Hölder bound are \(\int_X\chi_A\chi_B\,d\mu=\mu(A\cap B)\) and \(\|\chi_A\|_p\|\chi_B\|_q=\mu(A)^{1/p}\mu(B)^{1/q}\). When both sets have positive measure, equality holds precisely when they agree up to a null set: the normalized indicators in the equality criterion must agree almost everywhere. This makes the amount of overlap, not just the separate sizes of the sets, central to the calculation.

Simple Functions: Compute Each Coordinate

On a finite set with counting measure, integration is a finite sum. For simple functions specified by their values, it is safest to compute the product at each point before applying Hölder. This avoids overlooking a zero or confusing a norm’s exponent with the exponent inside the integral.

A Singular Power and a Common Norm Error

A common error is to raise a norm to the wrong power. For example, with the first example’s \(f(x)=x^2\) and \(p=3/2\), one might incorrectly write \(\|x^2\|_{3/2}=(\int_0^1x^3\,dm)^{1/2}=1/2\). The definition instead raises the integral to \(1/p=2/3\), so the correct value is \((1/4)^{2/3}\). With \(g(x)=x\) and \(q=3\), the other norm is \((1/4)^{1/3}\). Multiplying the correct values gives \(1/4\), the exact integral; multiplying the incorrectly computed first norm by the second would give \(1/2\), not a sharp bound.

A Reliable Checklist for Hölder Bounds

For a particular integral, first express the integrand as a product of two factors. Choose \(p,q\) with \(1/p+1/q=1\), then check that each factor has the required finite norm. Compute the integral of each factor’s absolute value to the appropriate power, and only then take the \(p\)-th or \(q\)-th root. Finally, compare the resulting bound with the exact integral when it is available. For nonzero factors with \(1<p,q<\infty\), the equality criterion from the earlier material on equality cases tells whether equality can occur: the normalized \(p\)-th and \(q\)-th powers of their absolute values must agree \(\mu\)-almost everywhere.

These calculations distinguish a valid estimate from a sharp one. On finite sets, the exact integral is a sum of coordinatewise products; for indicators, it is the measure of an intersection; and for powers, direct integration can expose the exponent choice that achieves equality. In every case, the norm calculation and the equality test are separate checks: getting the conjugate exponents right ensures the inequality applies, while matching the normalized powers determines sharpness.