What Completeness Adds to the \(L^p\) Metric
The triangle inequality in \(L^p\) makes \(d_p(f,g)=\|f-g\|_p\) a metric on equivalence classes of functions equal almost everywhere. A metric tells us when two functions are close, and therefore what it means for a sequence to converge. Completeness asks a further question: if the terms of a sequence become arbitrarily close to one another, must there be a limit in the same space?
This is a question about the space, not just about a particular sequence. In a complete space, a sequence that is internally Cauchy cannot “run out” of the space. That property matters when functions are built by approximation: estimates may show that successive approximations are close without yet exhibiting their limit. Completeness ensures that such a Cauchy process has a legitimate limit in \(L^p\).
The definition compares pairs of terms and does not require a limit to be specified. By contrast, convergence in \(L^p\) to \(f\) requires that \(\|f_n-f\|_p\) tend to zero. Every convergent sequence in a metric space is Cauchy, by the triangle inequality. Completeness is the converse at the level of the whole space.
The \(L^p\) distance is the norm of a difference, so completeness of \(L^p\) means that every \(L^p\)-Cauchy sequence has an \(L^p\) limit that is itself an element of \(L^p\). The limit is an equivalence class modulo equality almost everywhere, just like every other element of the space.
The Completeness Theorem for \(L^p\)
Earlier in this course, completeness was established for \(L^1\), \(L^2\), and \(L^\infty\). The theorem states the full result, including those cases. Thus the additional finite-exponent cases not already covered are \(1<p<\infty\) with \(p\ne2\). No assumption that \(\mu(X)\) is finite is needed.
The assertion that the limit belongs to \(L^p\) is essential. Pointwise limits of measurable functions are measurable, but measurability alone does not guarantee a finite \(L^p\) norm. The completeness theorem gives more: when the sequence is Cauchy in the \(L^p\) norm, there is a measurable limit whose equivalence class belongs to \(L^p\), and the sequence converges to it in that norm. The proof of the general theorem uses a carefully chosen subsequence with summably small successive differences; the next tutorial develops that proof.
Basic Facts About Cauchy Sequences
Two useful consequences of the definitions require only the metric properties already established. First, a Cauchy sequence cannot have terms with arbitrarily large norm. Second, a Cauchy sequence is determined by any one of its convergent subsequences: if a subsequence converges, then the entire Cauchy sequence converges to the same limit.
Proof. Apply the Cauchy condition with \(\varepsilon=1\). There is an integer \(N\) such that \(\|f_n-f_m\|_p<1\) whenever \(n,m\geq N\). Fix \(m=N\). The reverse triangle inequality gives $$ \|f_n\|_p\leq\|f_n-f_N\|_p+\|f_N\|_p<1+\|f_N\|_p $$ for every \(n\geq N\). The finitely many numbers \(\|f_1\|_p,\ldots,\|f_{N-1}\|_p\) are finite. Taking \(C\) to be the maximum of those numbers and \(1+\|f_N\|_p\) bounds every term. If \(N=1\), the same choice \(C=1+\|f_1\|_p\) suffices. Thus the sequence is bounded.
Proof. Let \(\varepsilon>0\). Since \((x_n)\) is Cauchy, choose \(N\) such that \(d(x_n,x_m)<\varepsilon/2\) whenever \(n,m\geq N\). Since \(x_{n_k}\to x\), choose \(k\) large enough that \(n_k\geq N\) and \(d(x_{n_k},x)<\varepsilon/2\). For every \(n\geq N\), the triangle inequality gives $$ d(x_n,x)\leq d(x_n,x_{n_k})+d(x_{n_k},x)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. $$ This is precisely convergence of \((x_n)\) to \(x\). Conversely, if a sequence converges, each of its subsequences converges to the same limit. Thus, for a Cauchy sequence, finding a convergent subsequence is enough to identify convergence of the full sequence.
In \(L^p\), these statements use \(d_p(f,g)=\|f-g\|_p\). Boundedness is often a useful preliminary check, but it is not a substitute for completeness: a bounded sequence need not be Cauchy or convergent. The subsequence proposition, on the other hand, explains why many completeness arguments first select a convenient subsequence and then return to the original sequence.
Worked Examples
Worked Example: A Cauchy Sequence on a Finite Space
Let \(X=\{a,b,c\}\) have counting measure, and fix \(1\leq p\leq\infty\). Define $$ f_n(a)=1+\frac1n,\qquad f_n(b)=2-\frac1n,\qquad f_n(c)=3. $$ For finite \(p\), the \(L^p\) norm is the \(p\)-th root of the sum of the \(p\)-th powers over the three points. Hence, for any positive integers \(n,m\), $$ \|f_n-f_m\|_p =\left(2\left|\frac1n-\frac1m\right|^p\right)^{1/p} =2^{1/p}\left|\frac1n-\frac1m\right|. $$ For \(p=\infty\), the norm is the largest absolute coordinate difference, so $$ \|f_n-f_m\|_\infty=\left|\frac1n-\frac1m\right|. $$ Since \(1/n\to0\), both expressions tend to zero as \(n,m\to\infty\), and \((f_n)\) is Cauchy for every exponent under consideration.
The coordinatewise candidate limit is \(f(a)=1,\ f(b)=2,\ f(c)=3\). For finite \(p\), $$ \|f_n-f\|_p =\left(\left|\frac1n\right|^p+\left|-\frac1n\right|^p\right)^{1/p} =\frac{2^{1/p}}{n}\longrightarrow0. $$ For \(p=\infty\), \(\|f_n-f\|_\infty=1/n\to0\). In this finite-space example the limit is directly visible, and it belongs to \(L^p\).
Worked Example: Cauchy Constant Functions on a Finite-Measure Set
Let \(E\in\mathcal{F}\) satisfy \(0<\mu(E)<\infty\), and fix \(1\leq p<\infty\). Set \(c_n=4+(-1)^n/n\) and \(f_n=c_n\mathbf{1}_E\). For any \(n,m\), $$ \|f_n-f_m\|_p =|c_n-c_m|\,\mu(E)^{1/p}. $$ The scalar sequence \(c_n\) is Cauchy because \(|c_n-4|=1/n\to0\). Therefore the displayed norm tends to zero as \(n,m\to\infty\), so \((f_n)\) is Cauchy in \(L^p\). Its candidate limit is \(f=4\mathbf{1}_E\), and $$ \|f_n-f\|_p =\frac{\mu(E)^{1/p}}{n}\longrightarrow0. $$ The finite, positive measure of \(E\) ensures that these constant multiples of its indicator have finite \(L^p\) norm.
For \(p=\infty\), the same sequence also converges: because \(\mu(E)>0\), the essential supremum of \(|(c_n-4)\mathbf{1}_E|\) is \(1/n\). Thus \(\|f_n-f\|_\infty=1/n\to0\). In this example, convergence follows from convergence of a single scalar coefficient, with the norm scaling determined by the measure of \(E\).
Worked Example: Pointwise Convergence Need Not Give \(L^\infty\) Convergence
On \((0,1)\) with Lebesgue measure, let \(f_n=\mathbf{1}_{(0,1/n)}\). For each \(x\in(0,1)\), eventually \(1/n\leq x\), so \(f_n(x)=0\) eventually. Thus \(f_n\to0\) pointwise on \((0,1)\). For any finite \(p\geq1\), $$ \|f_n\|_p =\left(\int_0^1|f_n(x)|^p\,dx\right)^{1/p} =\left(\frac1n\right)^{1/p}\longrightarrow0. $$ So this sequence converges to zero in every finite-\(p\) norm.
For the essential-supremum norm, however, \(|f_n|=1\) on a set of measure \(1/n>0\). Consequently \(\|f_n\|_\infty=1\) for every \(n\), and the sequence does not converge to zero in \(L^\infty\). More strongly, it is not Cauchy there: if \(m>n\), then \(f_n-f_m=1\) on \((1/m,1/n)\), a set of positive measure, so \(\|f_n-f_m\|_\infty=1\). This example shows why pointwise convergence alone cannot replace the norm estimates in a completeness argument.
Completeness on Finite Counting Spaces
The first example reflects a general elementary fact: \(L^p\) on a finite set with counting measure is complete. This can be verified directly from completeness of the real numbers and gives a concrete model for what a limit in \(L^p\) means.
Proof. Write \(X=\{x_1,\ldots,x_r\}\), and let \((f_n)\) be Cauchy in \(L^p(X)\). For each coordinate \(j\), if \(p<\infty\), then $$ |f_n(x_j)-f_m(x_j)|^p \leq \sum_{i=1}^r|f_n(x_i)-f_m(x_i)|^p =\|f_n-f_m\|_p^p. $$ Taking \(p\)-th roots gives \(|f_n(x_j)-f_m(x_j)|\leq\|f_n-f_m\|_p\). If \(p=\infty\), the definition of the maximum norm gives the same inequality directly. Thus, for each \(j\), the real sequence \((f_n(x_j))\) is Cauchy. Completeness of \(\mathbb{R}\) provides a real limit \(a_j=\lim_{n\to\infty}f_n(x_j)\). Define \(f(x_j)=a_j\).
For finite \(p\), there are only \(r\) coordinates, so $$ \|f_n-f\|_p^p=\sum_{j=1}^r|f_n(x_j)-a_j|^p\longrightarrow0, $$ because every summand tends to zero and the sum is finite. Hence \(\|f_n-f\|_p\to0\). For \(p=\infty\), each of the finitely many coordinate differences tends to zero, so their maximum tends to zero: $$ \|f_n-f\|_\infty=\max_{1\leq j\leq r}|f_n(x_j)-a_j|\longrightarrow0. $$ Every real-valued function on a finite counting space has finite \(L^p\) norm, so \(f\in L^p(X)\). Therefore \((f_n)\) converges in \(L^p(X)\), proving completeness.
Why Completeness Matters
Completeness is the bridge between estimates and existence. A construction may produce a sequence of functions whose pairwise \(L^p\) distances become small, without giving a formula for a limit at the outset. In a complete \(L^p\) space, the Cauchy condition alone guarantees that there is a limit in the same space. The norm then controls how accurately the sequence approximates that limit.
A common pitfall is to infer \(L^p\) convergence from pointwise convergence, or to assume that boundedness implies convergence. Neither inference is valid in general. The indicator example shows that pointwise convergence can coexist with failure of \(L^\infty\) convergence, while boundedness by itself says nothing about whether the terms approach one another. Completeness applies specifically to Cauchy sequences in the norm metric.
Check Your Understanding
Use the definitions and results above to answer the following questions.
- What distinguishes the definition of a Cauchy sequence from the definition of convergence in \(L^p\)?
- Where is the reverse triangle inequality used to show that a Cauchy sequence in \(L^p\) is bounded?
- Why does a convergent subsequence of a Cauchy sequence force the full sequence to converge to the same limit?
- In the finite counting-space proof, why does coordinatewise convergence imply convergence in the \(L^p\) norm?
- For the indicators of \((0,1/n)\), why does the \(L^p\) norm tend to zero for finite \(p\) while the \(L^\infty\) norm does not?