Choosing Between Hölder, Minkowski, And Jensen

Read the Shape of the Estimate First

Integral estimates often become simpler once the expression to be bounded is classified by its structure. A product of functions suggests Hölder’s inequality; a sum inside an \(L^p\) norm suggests Minkowski’s inequality; and a convex function applied to an average suggests Jensen’s inequality. The choice is not merely a matter of taste: each inequality has specific hypotheses, and applying the right one at the right stage avoids unnecessary estimates or invalid steps.

Throughout, \((X,\mathcal{M},\mu)\) is a measure space, with no assumption that \(\mu\) is finite unless stated. Functions in \(L^p(\mu)\) are measurable, real- or complex-valued functions with finite \(L^p\) norm, understood up to equality \(\mu\)-a.e. For \(1\le p<\infty\), \(\|f\|_p=(\int_X|f|^p\,d\mu)^{1/p}\); \(\|f\|_\infty\) is the essential supremum of \(|f|\). The inequalities below apply to measurable representatives, and their integral conclusions are unchanged by modifying those representatives on null sets.

In the forms used here, Hölder says that if \(1\le p,q\le\infty\) are conjugate exponents, meaning \(1/p+1/q=1\) with \(1/\infty=0\), then \(f\in L^p(\mu)\) and \(g\in L^q(\mu)\) imply \(\int_X|fg|\,d\mu\le\|f\|_p\|g\|_q\). Minkowski says that for \(1\le p\le\infty\) and \(f,g\in L^p(\mu)\), \(\|f+g\|_p\le\|f\|_p+\|g\|_p\). These inequalities, including their endpoint cases, were established in the earlier material on Hölder and Minkowski. Jensen’s integral inequality says that if \(\mathbb{P}\) is a probability measure, \(f\) is an integrable real-valued function, \(\phi\) is convex on an interval containing the values of \(f\), and \(\phi(f)\) is integrable, then \(\phi(\int f\,d\mathbb{P})\le\int\phi(f)\,d\mathbb{P}\). We use these standard results with their hypotheses checked below.

Worked Estimates by Structure

Combining the Choices

Some targets contain more than one structure. A sum of products, for example, may require first separating the sum inside a norm and then estimating each product. The order matters: Minkowski handles the sum, while Hölder handles each product. The following result records this common two-stage estimate.

Proof. For each \(j\), the definition of essential supremum gives \(|b_j|\le\|b_j\|_\infty\) \(\mu\)-a.e. Therefore \(|a_jb_j|^p\le\|b_j\|_\infty^p|a_j|^p\) a.e.; integrating shows that \(a_jb_j\in L^p(\mu)\) and \(\|a_jb_j\|_p\le\|a_j\|_p\|b_j\|_\infty\). For \(N=1\), this is the claimed inequality. For \(N>1\), repeated application of Minkowski to the finite sum gives \(\|\sum_{j=1}^Na_jb_j\|_p\le\sum_{j=1}^N\|a_jb_j\|_p\). Substituting the bound for each product proves the result. No finiteness or sigma-finiteness assumption on \(\mu\) is needed. \(\square\)

This proposition illustrates a useful order of operations. A norm of a sum is not itself a product, so first use the inequality designed for sums. Afterward, estimate each product. The same reasoning works with Hölder in place of the \(L^\infty\) product bound when the factors have suitable conjugate exponents, provided each resulting product belongs to the \(L^p\) space required by Minkowski.

Proof. Define a probability measure \(\mathbb{P}\) on \(E\) by \(\mathbb{P}(A)=\mu(A)/\mu(E)\) for measurable \(A\subseteq E\). It is a probability measure because \(\mathbb{P}(E)=1\). The integral with respect to it satisfies \(\int_E h\,d\mathbb{P}=(1/\mu(E))\int_Eh\,d\mu\) for integrable \(h\). Apply integral Jensen for \(\mathbb{P}\) to \(f\) and \(\phi\), using the stated integrability and convexity hypotheses, to obtain the displayed inequality. \(\square\)

A Common Normalization Error

Jensen’s probability-measure form cannot be applied to an unnormalized finite measure as if its total mass were one. For a concrete countercheck, take \(E=[0,2]\), Lebesgue measure, \(f(x)=x\), and \(\phi(t)=t^2\). A mistaken use of probability Jensen would assert \(\int_0^2x^2\,dm\ge(\int_0^2x\,dm)^2\). But the left side is \(8/3\), while the right side is \(2^2=4\), so this asserted inequality is false.

The correction is to divide by the mass \(m(E)=2\). The average of \(f\) is \((1/2)\int_0^2x\,dm=1\), and the average of \(f^2\) is \((1/2)\int_0^2x^2\,dm=4/3\). Jensen correctly gives \(1^2\le4/3\). Equivalently, in unnormalized form the inequality is \(\int_E f^2\,dm\ge(\int_E f\,dm)^2/m(E)\), not \(\int_E f^2\,dm\ge(\int_E f\,dm)^2\).

A different mistake is to use Minkowski on a product simply because the expression contains two functions, or to use Jensen without an average. Instead, inspect the precise target: \(\int|fg|\) has product form; \(\|f+g\|_p\) has a sum inside a norm; and \(\phi(\int f\,d\mathbb{P})\) has a convex function of an average. In a longer calculation, identify the innermost structure first, check its hypotheses, and then move outward.