The First Norm Axioms for Lp

Three Axioms That Must Respect Almost-Everywhere Equality

A norm must be nonnegative, vanish only at the zero element, and scale by the absolute value of a scalar. For \(L^p\), these properties are tied to the way its elements are defined: functions that agree almost everywhere represent the same element. Thus, the zero element is not a particular function that vanishes at every point; it is the class of functions that vanish almost everywhere. We prove these first three norm axioms for \(1\le p\le\infty\), without proving the triangle inequality.

Unless an example specifies otherwise, let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite, sigma-finite, or complete. Functions are measurable and real-valued or complex-valued. Two functions \(f\) and \(g\) are equivalent if \(f=g\) \(\mu\)-a.e., meaning that \(\mu(\{x:f(x)\ne g(x)\})=0\). The elements of \(L^p(\mu)\) are these equivalence classes, not individual representatives.

In particular, the strict tail set \(\{|f|>K\}\) is the set used to test whether \(K\) is an almost-everywhere upper bound. It is not necessary that \(|f(x)|\le K\) at every point. Also, the infimum in the essential-supremum definition need not itself be an almost-everywhere bound; the infimum describes the least possible bound, whether or not it is attained as a pointwise value.

Finite-\(p\) Norms

First check that the formula does not depend on the representative. If \(f=g\) \(\mu\)-a.e., then \(|f|^p=|g|^p\) \(\mu\)-a.e. The Lebesgue integral of nonnegative measurable functions equal almost everywhere is the same, so \(\int_X|f|^p\,d\mu=\int_X|g|^p\,d\mu\). Thus both membership in \(L^p(\mu)\) and the value of \(\|f\|_p\) are unchanged when the representative is changed. The same reasoning for tail sets shows that \(\mu(\{|f|>K\})=0\) if and only if \(\mu(\{|g|>K\})=0\), so the essential supremum is also independent of the representative.

Proof. Since \(|f|^p\ge0\), its integral is nonnegative, and taking the nonnegative \(p\)-th root gives \(\|f\|_p\ge0\). If \(f=0\) \(\mu\)-a.e., then \(|f|^p=0\) \(\mu\)-a.e.; its integral and hence its norm are zero.

Conversely, suppose \(\|f\|_p=0\). Then \(\int_X|f|^p\,d\mu=0\). For each positive integer \(n\), set \(E_n=\{x:|f(x)|>1/n\}\). On \(E_n\), \(|f|^p\ge n^{-p}\), so monotonicity of the integral gives \(0\le n^{-p}\mu(E_n)\le\int_X|f|^p\,d\mu=0\). Hence \(\mu(E_n)=0\). Every point where \(|f|>0\) belongs to some \(E_n\), and therefore \(\{|f|>0\}=\bigcup_{n=1}^\infty E_n\) is null. Thus \(f=0\) \(\mu\)-a.e.

For homogeneity, \(|cf|^p=|c|^p|f|^p\), so \(\int_X|cf|^p\,d\mu=|c|^p\int_X|f|^p\,d\mu<\infty\). If \(c\ne0\), taking the \(p\)-th root gives \(\|cf\|_p=|c|\|f\|_p\). If \(c=0\), then \(cf\) is the zero function, so \(\|cf\|_p=0=|c|\|f\|_p\). This also proves \(cf\in L^p(\mu)\) in every case. \(\square\)

The Essential-Supremum Norm

For \(L^\infty(\mu)\), nonnegativity follows from the fact that the essential supremum of \(|f|\) is an infimum over \(K\ge0\). To prove definiteness and homogeneity carefully, use the tail-set definition. For a measurable \(f\), write \(B_f=\{K\ge0:\mu(\{|f|>K\})=0\}\). Membership in \(B_f\) means exactly that \(K\) is an almost-everywhere upper bound for \(|f|\).

Proof. Nonnegativity holds because \(\|f\|_\infty=\inf B_f\) and \(B_f\subseteq[0,\infty)\). If \(f=0\) \(\mu\)-a.e., then for every \(K\ge0\), the set \(\{|f|>K\}\) is null; in particular \(0\in B_f\), and \(\|f\|_\infty=0\).

Now suppose \(\|f\|_\infty=0\). The set \(B_f\) is nonempty because \(f\in L^\infty(\mu)\), and its infimum is zero. For every \(n\ge1\), the definition of infimum gives a \(K_n\in B_f\) with \(0\le K_n<1/n\). Since \(K_n<1/n\), \(\{|f|>1/n\}\subseteq\{|f|>K_n\}\), which is null. The union over \(n\) of \(\{|f|>1/n\}\) is \(\{|f|>0\}\), so \(f=0\) \(\mu\)-a.e.

If \(c=0\), then \(\|cf\|_\infty=0=|c|\|f\|_\infty\). If \(c\ne0\), then for every \(K\ge0\), \(\{|cf|>|c|K\}=\{|f|>K\}\). Consequently, \(K\in B_f\) if and only if \(|c|K\in B_{cf}\); equivalently, \(B_{cf}=\{|c|K:K\in B_f\}\). Taking infima gives \(\|cf\|_\infty=|c|\|f\|_\infty\). In particular, \(cf\) has finite essential supremum, so it belongs to \(L^\infty(\mu)\). \(\square\)

Null Sets, Equivalence Classes, and a Common Pitfall

A common error is to replace the essential supremum by the pointwise supremum. On \([0,1]\) with Lebesgue measure, let \(h(0)=7\) and \(h(x)=0\) for \(x>0\). The pointwise supremum of \(|h|\) is \(7\), but for every \(K\ge0\), \(\{|h|>K\}\) is either empty or contained in \(\{0\}\), so it has measure zero. Hence \(\|h\|_\infty=0\), not \(7\); also, \(\|h\|_p=0\) for every finite \(p\ge1\). The correction is to compute norms on the almost-everywhere class and, for \(L^\infty\), test the measures of strict tail sets rather than inspecting exceptional point values.

The results here establish nonnegativity, definiteness on equivalence classes, and absolute homogeneity for \(L^p(\mu)\), including \(p=\infty\). They do not establish the triangle inequality, which is a separate property requiring a separate argument. In particular, calling these quantities norms does not make that remaining axiom automatic.