From \(L^p\) Membership to Size
In the previous tutorial, \(L^p(X,\mu)\) was defined as a space of equivalence classes: functions that agree almost everywhere represent the same element. Membership tells us that the relevant integral is finite, but it does not yet assign a single measure of size to each element. The \(L^p\) norm supplies that measurement.
For finite \(p\), the norm is built from the integral of \(|f|^p\). At \(p=\infty\), an integral is not used; instead, size is measured by the smallest bound that holds outside a null set. Both definitions must respect almost-everywhere equality, because elements of \(L^p\) are equivalence classes rather than particular pointwise representatives.
Definition of the \(L^p\) Norm
Fix a measure space \((X,\mathcal{F},\mu)\). When \(1\leq p<\infty\), every representative \(f\) of an element of \(L^p(X,\mu)\) has a finite integral of \(|f|^p\). Taking the \(p\)-th root gives a nonnegative quantity with the same scaling as the function itself.
At the endpoint \(p=\infty\), define the essential supremum by taking the infimum of all nonnegative almost-everywhere bounds. This is different from the ordinary supremum: values on a null set do not affect it.
The restriction \(p\geq1\) in the finite-\(p\) definition matters when the expression is called a norm. For \(0<p<1\), the same integral expression can be defined for suitable functions, but it does not in general satisfy the triangle inequality. Here, \(L^p\) and \(\|\cdot\|_p\) refer to \(1\leq p<\infty\), unless the endpoint \(p=\infty\) is specified.
Well-Definedness and Basic Norm Facts
The first issue is whether choosing a different representative could change the formula. It cannot: if \(f=g\) almost everywhere, then \(|f|^p=|g|^p\) almost everywhere, and the integral is invariant under almost-everywhere equality. The essential-supremum definition also depends only on almost-everywhere bounds, so it is unchanged when a representative is modified on a null set.
Proof. If two representatives agree almost everywhere, their absolute values raised to the power \(p\) agree almost everywhere. Invariance of the integral under almost-everywhere equality therefore makes their \(p\)-th power integrals, and hence their norms, equal. Nonnegativity follows because \(|f|^p\geq0\).
If \(\|f\|_p=0\), then \(\int_X|f|^p\,d\mu=0\). The Zero Integral Criterion for nonnegative measurable functions implies \(|f|^p=0\) almost everywhere, so \(f=0\) almost everywhere. Thus \(f\) represents the zero element of \(L^p\). Conversely, if \(f=0\) almost everywhere, then \(|f|^p=0\) almost everywhere and its integral, and therefore its norm, is zero.
For a scalar \(a\), the pointwise identity \(|af|^p=|a|^p|f|^p\) and nonnegative homogeneity of the integral give $$ \|af\|_p =\left(|a|^p\int_X|f|^p\,d\mu\right)^{1/p} =|a|\left(\int_X|f|^p\,d\mu\right)^{1/p} =|a|\|f\|_p. $$ When \(a=0\), both sides equal zero, so the identity holds for every real scalar. This proves the claims.
For \(p=\infty\), the same scaling rule follows from the definition by almost-everywhere bounds. If \(a\neq0\), a bound \(|f|\leq M\) almost everywhere is equivalent to \(|af|\leq |a|M\) almost everywhere. Taking the infimum of such bounds gives \(\|af\|_\infty=|a|\|f\|_\infty\). If \(a=0\), both norms are zero. Also, \(\|f\|_\infty=0\) implies \(f=0\) almost everywhere: for every positive integer \(n\), the definition of the infimum gives an almost-everywhere bound smaller than \(1/n\), so \(|f|\leq1/n\) almost everywhere. Outside the union of the corresponding countably many exceptional null sets, these inequalities hold for every \(n\), forcing \(f=0\).
Worked Examples of \(L^p\) Norms
Worked Example: The Norm of an Indicator Function
Let \(A\) be a measurable set with \(\mu(A)=3/8\), and take \(f=\mathbf{1}_A\). Since \(\mathbf{1}_A^p=\mathbf{1}_A\) for every finite \(p\geq1\), $$ \|f\|_p =\left(\int_X\mathbf{1}_A\,d\mu\right)^{1/p} =\left(\frac{3}{8}\right)^{1/p}. $$ The set \(A\) has positive measure, and \(|f|=1\) on \(A\) while \(|f|=0\) off \(A\). Thus no number less than \(1\) is an almost-everywhere bound, and \(1\) is a bound everywhere. It follows that \(\|f\|_\infty=1\).
For example, the finite norms at \(p=1\) and \(p=2\) are \(3/8\) and \(\sqrt{3/8}\), respectively. This calculation also shows why the positive-measure condition matters for the \(L^\infty\) value: if \(\mu(A)=0\), then the indicator represents the zero element and its \(L^\infty\) norm is zero.
Worked Example: The Norm of the Coordinate Function
On \(X=(0,1)\) with Lebesgue measure, let \(f(x)=x\). For finite \(p\geq1\), $$ \|f\|_p =\left(\int_0^1x^p\,dx\right)^{1/p} =\left(\frac{1}{p+1}\right)^{1/p}. $$ In particular, \(\|f\|_1=1/2\) and \(\|f\|_2=1/\sqrt{3}\).
The ordinary supremum of \(f\) on \((0,1)\) is not attained, since \(x<1\) at every point of the domain. Nevertheless, \(\|f\|_\infty=1\). The value \(1\) is an almost-everywhere bound because \(x\leq1\) on \((0,1)\). If \(0\leq M<1\), then $$ \{x\in(0,1):x>M\}=(M,1), $$ which has measure \(1-M>0\). Thus \(M\) is not an almost-everywhere bound. No smaller nonnegative bound works, proving that the essential supremum is \(1\).
Worked Example: A Step Function with Two Values
On \(X=(0,3)\) with Lebesgue measure, define \(f(x)=2\) for \(0<x<1\) and \(f(x)=-1\) for \(1\leq x<3\). The first region has measure \(1\), and the second has measure \(2\). For every finite \(p\geq1\), $$ \int_0^3|f(x)|^p\,dx =1\cdot 2^p+2\cdot1^p =2^p+2, $$ so $$ \|f\|_p=(2^p+2)^{1/p}. $$ In particular, \(\|f\|_1=4\), and \(\|f\|_2=\sqrt{6}\).
The absolute value equals \(2\) on a set of measure \(1\) and equals \(1\) on the rest of the domain. Hence \(|f|\leq2\) everywhere, while every \(M<2\) fails to bound \(|f|\) on the positive-measure interval \((0,1)\). Therefore \(\|f\|_\infty=2\). The sign of \(f\) does not affect any of these calculations because the definitions use \(|f|\).
The Essential Supremum Really Is a Least Bound
An infimum need not, in general, belong to the set whose infimum is being taken. For the essential supremum, however, the infimum itself is an almost-everywhere bound. This fact makes the definition especially useful: it is not merely a limiting value of bounds but the smallest bound that works outside a null set.
Proof. Let \(S=\{M\geq0:\ |f|\leq M\text{ almost everywhere}\}\). Essential boundedness means \(S\) is nonempty, and \(M_*=\inf S\) is finite. For each positive integer \(n\), the definition of infimum allows a choice \(M_n\in S\) such that $$ M_*\leq M_n<M_*+\frac{1}{n}. $$ For each \(n\), the set \(N_n=\{x:|f(x)|>M_n\}\) is null. The union \(N=\bigcup_{n=1}^{\infty}N_n\) is null by countable subadditivity. If \(x\notin N\), then \(|f(x)|\leq M_n\) for every \(n\). Since \(M_n<M_*+1/n\), it follows that \(|f(x)|\leq M_*+1/n\) for every \(n\). Letting \(n\) increase gives \(|f(x)|\leq M_*\). Thus this inequality holds outside the null set \(N\). As \(M_*\) is itself an almost-everywhere bound and is the infimum of all such bounds, it is the least one.
This proof handles a detail that can be missed if one treats the essential supremum as an ordinary maximum. A function may never attain its essential supremum, and the exceptional sets associated with a sequence of approximate bounds need not be the same. Taking their countable union ensures that all the bounds hold together outside one null set.
Order and Norm Size
The integral definition makes a basic comparison principle immediate: if one function is no larger in absolute value than another almost everywhere, its norm cannot be larger. This is a useful way to estimate a norm without calculating its integral exactly.
Proof. For finite \(p\), the function \(t\mapsto t^p\) is nondecreasing on \([0,\infty)\). Thus \(|f|^p\leq|g|^p\) almost everywhere. Monotonicity of the Lebesgue integral under almost-everywhere order gives $$ \int_X|f|^p\,d\mu\leq\int_X|g|^p\,d\mu. $$ Both sides are finite and nonnegative. Taking their \(p\)-th roots, which preserves order, proves \(\|f\|_p\leq\|g\|_p\).
For \(p=\infty\), every almost-everywhere bound for \(|g|\) is also an almost-everywhere bound for \(|f|\). Therefore the set of admissible bounds for \(g\) is contained in the set of admissible bounds for \(f\). Taking infima gives \(\|f\|_\infty\leq\|g\|_\infty\), as required.
Worked Example: Estimating a Norm by Comparison
On \((0,1)\), let \(f(x)=x(1-x)\) and \(g(x)=x\), with Lebesgue measure. Since \(0<1-x<1\) on this interval, $$ |f(x)|=x(1-x)\leq x=|g(x)|. $$ Both functions belong to \(L^p(0,1)\) for every finite \(p\geq1\), since they are bounded on a set of finite measure. Monotonicity of the norm therefore gives $$ \|f\|_p\leq\|g\|_p=(p+1)^{-1/p}. $$ For \(p=\infty\), the same comparison yields \(\|f\|_\infty\leq\|g\|_\infty=1\). This estimate does not require evaluating \(\int_0^1[x(1-x)]^p\,dx\).
Interpreting the Norm
For finite \(p\), the norm combines the size of a function with how much of the space carries that size. The indicator example makes this visible: a function with value \(1\) on a set of measure \(m\) has norm \(m^{1/p}\). The essential-supremum norm behaves differently: it records the least bound outside null sets, without weighting that bound by the measure of the region where it occurs.
Neither formula depends on values at individual points when those points form a null set. Likewise, the essential supremum need not equal the pointwise supremum, as the coordinate function on \((0,1)\) demonstrates. These distinctions are not technical distractions; they follow directly from defining \(L^p\) elements as almost-everywhere equivalence classes.
The results here establish representative independence, definiteness, scaling, and order comparison. They do not yet prove the triangle inequality, the remaining central property needed to verify fully that these expressions are norms. The next tutorial develops further properties of the \(L^p\) norm.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why does changing a representative on a null set leave its finite-\(p\) norm unchanged?
- If \(\mu(A)=5\), what is the \(L^p\) norm of \(\mathbf{1}_A\) for finite \(p\), and what is its \(L^\infty\) norm?
- Why is the \(L^\infty\) norm of \(f(x)=x\) on \((0,1)\) equal to \(1\), even though \(f\) never takes the value \(1\)?
- What does the zero-norm criterion imply about a function whose \(L^p\) norm is zero?
- If \(|f|\leq|g|\) almost everywhere and both functions belong to \(L^p\), what comparison follows for their norms?