From the \(L^2\) Norm to Essential Boundedness
The \(L^2\) norm measures the integral of a function’s squared size. Another useful measure asks instead for the smallest bound that holds outside a set of measure zero. This leads to \(L^\infty\), the space of essentially bounded measurable functions. As in \(L^1\) and \(L^2\), functions equal almost everywhere represent the same element; values on a null set do not affect the element or its norm.
The symbol \(\infty\) does not mean that the norm is an integral with an infinite exponent. The \(L^\infty\) norm is an essential supremum. It measures the largest size that persists on sets of positive measure, disregarding exceptional null sets. The Least Almost-Everywhere Bound theorem established earlier in the course will be useful: for an essentially bounded \(f\), its essential supremum is itself an almost-everywhere bound.
This definition is unchanged if \(f\) is replaced by an almost-everywhere equal representative: an almost-everywhere bound for one is also such a bound for the other. The Least Almost-Everywhere Bound theorem says that \(|f|\leq\|f\|_\infty\) almost everywhere. In particular, \(\|f\|_\infty=0\) exactly when \(f=0\) almost everywhere, so the norm is definite on equivalence classes.
The norm is homogeneous. For \(c\neq0\), the almost-everywhere bounds for \(|cf|\) are precisely \(|c|\) times the almost-everywhere bounds for \(|f|\), and therefore \(\|cf\|_\infty=|c|\|f\|_\infty\). For \(c=0\), the function \(cf\) is zero everywhere, so \(\|cf\|_\infty=0=|c|\|f\|_\infty\) directly. Minkowski’s inequality, established earlier for \(p=\infty\), gives the triangle inequality.
Worked Example: A Bound That Is Not Attained
On \((0,1)\) with Lebesgue measure, let \(f(x)=x\). Since \(0<x<1\), the number \(1\) is an almost-everywhere bound for \(|f|\). If \(0\leq M<1\), then $$ \{x\in(0,1):|f(x)|>M\}=(M,1), $$ which has measure \(1-M>0\). Thus no \(M<1\) is an almost-everywhere bound, and \(\|f\|_\infty=1\). The function never takes the value \(1\) on its domain, but its essential supremum is still \(1\). Attainment of the bound is not required.
Worked Example: Changing a Value on a Null Set
On \([0,1]\) with Lebesgue measure, define \(g(x)=x\) for \(x>0\) and \(g(0)=10^6\). The function is pointwise bounded, and for every \(M<1\), the set \(\{x:|g(x)|>M\}\) contains \((M,1]\), which has positive measure. On the other hand, \(|g(x)|\leq1\) except at \(x=0\), a null set. Hence \(\|g\|_\infty=1\). The exceptional value does not change the equivalence class. More generally, a finite change at a single point on a Lebesgue measure space does not affect the \(L^\infty\) element.
Worked Example: Essential Boundedness Does Not Require Finite Measure
On \(\mathbb{R}\) with Lebesgue measure, the constant function \(h(x)=1\) has \(\|h\|_\infty=1\), so it belongs to \(L^\infty(\mathbb{R},\lambda)\). But $$ \int_{\mathbb{R}}|h|\,d\lambda =\lambda(\mathbb{R}) =\infty. $$ Thus \(h\notin L^1(\mathbb{R},\lambda)\). Essential boundedness concerns the size of a function, not the total measure of the region on which it is nonzero. In particular, membership in \(L^\infty\) alone does not imply membership in \(L^1\) on a space of infinite measure.
Bounded Functions on Finite-Measure Spaces
On a finite-measure space, essential boundedness does imply membership in every \(L^p\) space with finite exponent \(p\geq1\). The finite measure controls how much area contributes to the integral, while the essential bound controls the size of the integrand. This gives an inclusion in the opposite direction from the finite-measure inclusion of \(L^q\) in \(L^p\) discussed earlier.
Proof. By the Least Almost-Everywhere Bound theorem, \(|f|\leq\|f\|_\infty\) almost everywhere. Raising this nonnegative inequality to the power \(p\) gives \(|f|^p\leq\|f\|_\infty^p\) almost everywhere. Monotonicity of the Lebesgue integral under almost-everywhere order therefore yields $$ \int_X |f|^p\,d\mu \leq \int_X\|f\|_\infty^p\,d\mu =\|f\|_\infty^p\mu(X)<\infty. $$ Thus \(f\in L^p(X,\mu)\). Taking the \(p\)th root of the inequality proves the stated norm bound. This argument also covers \(\|f\|_\infty=0\): then \(f=0\) almost everywhere and both sides of the bound are zero.
Worked Example: Comparing \(L^\infty\) and \(L^2\) on a Finite Interval
On \((0,3)\), define \(f(x)=4\) for \(0<x<1\) and \(f(x)=-2\) for \(1\leq x<3\). The first interval has positive measure, so any almost-everywhere bound for \(|f|\) must be at least \(4\); also, \(|f|\leq4\) everywhere. Therefore \(\|f\|_\infty=4\). Direct integration gives $$ \|f\|_2^2 =\int_0^1 16\,dx+\int_1^3 4\,dx =16+8=24, \qquad \|f\|_2=\sqrt{24}. $$ The theorem gives the bound \(\|f\|_2\leq 3^{1/2}\cdot4=4\sqrt{3}\), and \(\sqrt{24}=2\sqrt{6}\leq4\sqrt{3}\), since \(24\leq48\). The general bound need not be an equality.
Completeness of \(L^\infty\)
Completeness means that every Cauchy sequence in the norm converges to an element of the space. For \(L^\infty\), the key idea is to select a subsequence whose successive differences have summable norms. Outside the union of countably many null sets, those differences are bounded pointwise by a summable numerical series. The subsequence then has a pointwise limit there, and the tail bounds give convergence in the \(L^\infty\) norm.
Proof. Let \((F_n)\) be a Cauchy sequence in \(L^\infty(X,\mu)\), with each \(F_n\) an equivalence class. For each positive integer \(j\), choose \(N_j\) so that $$ \|F_r-F_s\|_\infty<2^{-j} \quad\text{whenever }r,s\geq N_j. $$ Choose strictly increasing indices \(n_j\) with \(n_j\geq N_j\). Then $$ \|F_{n_{j+1}}-F_{n_j}\|_\infty<2^{-j}. $$ Choose measurable representatives \(f_{n_j}\) for these classes and set \(u_j=f_{n_{j+1}}-f_{n_j}\). By the Least Almost-Everywhere Bound theorem, for each \(j\) there is a measurable null set \(N'_j\) outside which \(|u_j|\leq2^{-j}\). Let $$ E=X\setminus\bigcup_{j=1}^{\infty}N'_j. $$ The set \(E\) is measurable, and its complement is null. For every \(x\in E\) and every \(m>k\), $$ \left|\sum_{j=k}^{m-1}u_j(x)\right| \leq\sum_{j=k}^{m-1}2^{-j} \leq 2^{1-k}. $$ Hence the partial sums of \(\sum_j u_j(x)\) are Cauchy for each \(x\in E\), so that series converges there.
Define \(f(x)=f_{n_1}(x)+\sum_{j=1}^{\infty}u_j(x)\) on \(E\), and set \(f(x)=0\) outside \(E\). On \(E\), this is the pointwise limit of the measurable partial sums \(f_{n_1}+\sum_{j=1}^{m-1}u_j\). Since \(E\) is measurable, defining the function to be zero on its complement preserves measurability. Also, \(f\) is essentially bounded: \(f_{n_1}\) is essentially bounded, and on \(E\) the absolute value of the added series is at most \(\sum_{j=1}^{\infty}2^{-j}=1\). Thus \(f\) represents an element \(F\in L^\infty(X,\mu)\).
The partial sum \(f_{n_1}+\sum_{j=1}^{m-1}u_j\) equals \(f_{n_m}\) pointwise, by telescoping. On \(E\), the remaining tail has absolute value at most $$ \sum_{j=m}^{\infty}2^{-j}=2^{1-m}. $$ Consequently \(\|F_{n_m}-F\|_\infty\leq2^{1-m}\), so the subsequence converges to \(F\) in \(L^\infty\).
Finally, let \(\varepsilon>0\). Since \((F_n)\) is Cauchy, choose \(N\) such that \(\|F_r-F_s\|_\infty<\varepsilon/2\) whenever \(r,s\geq N\). Choose \(m\) with \(n_m\geq N\) and \(\|F_{n_m}-F\|_\infty<\varepsilon/2\). For every \(n\geq N\), the triangle inequality gives $$ \|F_n-F\|_\infty \leq\|F_n-F_{n_m}\|_\infty+\|F_{n_m}-F\|_\infty <\varepsilon. $$ Therefore the full sequence converges to \(F\) in \(L^\infty\), proving completeness.
What the Essential Supremum Does—and Does Not—Measure
The \(L^\infty\) norm records a size bound that holds almost everywhere. It ignores how small a set is if that set is null, but it does not ignore a set merely because its measure is small: if a function takes a large value on any set of positive measure, that value can affect its essential supremum. For instance, an indicator of a positive-measure set has \(L^\infty\) norm \(1\), regardless of how small that positive measure is. An indicator of a null set, by contrast, represents the zero element and has norm \(0\).
On a finite-measure space, the inclusion theorem explains one connection to \(L^p\): a bounded function has finite \(p\)th-power integral. The converse does not generally hold. For example, \(x^{-1/4}\) on \((0,1)\) belongs to \(L^2\), as seen in the \(L^2\) tutorial, but it is not essentially bounded. For any finite \(M\), the set where \(x^{-1/4}>M\) contains a positive-length interval next to zero (when \(M>0\)); hence no finite \(M\) is an almost-everywhere bound.
The finite-measure hypothesis also matters. The constant function \(1\) on \(\mathbb{R}\) belongs to \(L^\infty\), but it does not belong to \(L^p\) for any finite \(p\), because its \(p\)th-power integral over \(\mathbb{R}\) is infinite. Thus \(L^\infty\) membership alone guarantees neither integrability nor finite-exponent \(L^p\) membership on an infinite-measure space.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why can a function have essential supremum \(1\) even if it never takes the value \(1\)?
- What is the \(L^\infty\) norm of an indicator of a null set? What if the set has positive measure?
- Why does the inclusion of \(L^\infty\) in \(L^p\) require the finite-measure hypothesis in the theorem?
- In the completeness proof, why can the chosen subsequence’s successive differences be bounded pointwise outside a single null set?
- Why must homogeneity of the \(L^\infty\) norm be checked separately when the scalar is zero?