When Do Cauchy Sequences Have Limits in the Space?
The Cauchy condition compares sufficiently late terms with one another, without specifying a limit. As discussed in Cauchy Sequences in a Normed Space, norm convergence always implies the Cauchy property, but the reverse implication is not automatic: a sequence may approach an object that does not belong to the space. Completeness is the property that rules out this failure.
There are two parts to this definition. The sequence must satisfy the Cauchy condition using the norm on \(V\), and its limit must lie in \(V\). Completeness is therefore a property of the normed space, not merely of the vector space considered without a norm. The same set of functions can behave differently when equipped with different norms.
Completeness of Continuous Functions in the Supremum Norm
A central example is \(C[a,b]\), the vector space of continuous real-valued functions on a closed interval \([a,b]\), where \(a<b\). Equip it with the supremum norm \(\|f\|_\infty=\sup_{x\in[a,b]}|f(x)|\). Every continuous function on a closed bounded interval is bounded, so this norm is finite. The key is that a Cauchy sequence in this norm is uniformly Cauchy, and its pointwise limit is also its uniform limit.
Proof. Let \((f_n)\) be a Cauchy sequence in \(C[a,b]\). Fix \(x\in[a,b]\). For every \(m,n\), $$ |f_n(x)-f_m(x)|\leq \|f_n-f_m\|_\infty. $$ Since \((f_n)\) is Cauchy in the supremum norm, the right-hand side can be made arbitrarily small for all sufficiently large \(m,n\). Thus \((f_n(x))\) is a Cauchy sequence of real numbers. Completeness of \(\mathbb{R}\) gives a real limit at each \(x\); define \(f(x)=\lim_{n\to\infty}f_n(x)\).
It remains to show that the convergence to \(f\) is uniform. Let \(\varepsilon>0\). By the Cauchy property, there is an index \(N\) such that $$ m,n\geq N\quad\Longrightarrow\quad \|f_n-f_m\|_\infty<\frac{\varepsilon}{2}. $$ Fix \(n\geq N\) and \(x\in[a,b]\). For every \(m\geq N\), we have $$ |f_n(x)-f_m(x)|\leq\|f_n-f_m\|_\infty<\frac{\varepsilon}{2}. $$ Let \(m\) tend to infinity. Since \(f_m(x)\to f(x)\), continuity of the absolute value gives $$ |f_n(x)-f(x)|\leq\frac{\varepsilon}{2}<\varepsilon. $$ This estimate holds for every \(x\in[a,b]\), so \(\|f_n-f\|_\infty\leq\varepsilon/2<\varepsilon\) for every \(n\geq N\). Hence \(f_n\to f\) uniformly. By the Theorem on Uniform Limits of Continuous Functions, \(f\) is continuous on \([a,b]\), so \(f\in C[a,b]\). We have found a limit in the space, proving completeness. \(\square\)
The proof uses two distinct facts. Completeness of the real numbers supplies the pointwise values \(f(x)\). Uniform control then shows that these pointwise limits fit together into a uniform limit, and the theorem on uniform limits ensures that this limit remains continuous. A pointwise limit alone would not guarantee continuity or convergence in the supremum norm.
Worked Example: A Cauchy Sequence of Continuous Functions
On \([0,1]\), define $$ f_n(x)=\frac{x}{1+nx},\qquad n\geq1. $$ Each \(f_n\) is continuous, since its denominator is positive throughout \([0,1]\). For \(m,n\geq1\), direct subtraction gives $$ f_n(x)-f_m(x) =\frac{x}{1+nx}-\frac{x}{1+mx} =\frac{(m-n)x^2}{(1+nx)(1+mx)}. $$ For \(x\in[0,1]\), the denominator is at least \(1\) and \(x^2\leq1\). Therefore $$ |f_n(x)-f_m(x)|\leq |m-n|. $$ This bound is not useful for large indices, so instead estimate each term for \(x\in[0,1]\): $$ 0\leq f_n(x)=\frac{x}{1+nx}\leq\frac{1}{n}. $$ Indeed, \(nx\leq 1+nx\) implies \(x/(1+nx)\leq1/n\). It follows that $$ \|f_n-f_m\|_\infty \leq\|f_n\|_\infty+\|f_m\|_\infty \leq\frac{1}{n}+\frac{1}{m}. $$ Given \(\varepsilon>0\), choose \(N>2/\varepsilon\). If \(m,n\geq N\), then $$ \|f_n-f_m\|_\infty\leq\frac{1}{n}+\frac{1}{m} \leq\frac{2}{N}<\varepsilon. $$ Thus \((f_n)\) is Cauchy in \(C[0,1]\). Completeness guarantees a continuous uniform limit in that space. In fact, for every fixed \(x>0\), \(f_n(x)\to0\), and \(f_n(0)=0\), so the limit is the zero function.
Finite-Dimensional Spaces as Complete Spaces
Completeness is not limited to function spaces. Every norm on a finite-dimensional real vector space gives a complete normed space. For \(\mathbb{R}^d\), this follows from the Equivalence of Norms in Finite Dimensions, established earlier in the course, together with completeness of \(\mathbb{R}\).
Worked Example: Completeness of \(\mathbb{R}^2\) with an Arbitrary Norm
Let \(N\) be any norm on \(\mathbb{R}^2\), and let \((z_n)\) be Cauchy in \(N\), writing \(z_n=(u_n,v_n)\). By equivalence of norms in finite dimensions, there are constants \(c,C>0\) such that $$ c\|z\|_2\leq N(z)\leq C\|z\|_2 \qquad\text{for every }z\in\mathbb{R}^2. $$ For any \(m,n\), this gives $$ \|z_n-z_m\|_2\leq \frac{1}{c}N(z_n-z_m). $$ Since \((z_n)\) is Cauchy in \(N\), it is Cauchy in the Euclidean norm. In particular, $$ |u_n-u_m|\leq\|z_n-z_m\|_2,\qquad |v_n-v_m|\leq\|z_n-z_m\|_2, $$ so \((u_n)\) and \((v_n)\) are Cauchy sequences in \(\mathbb{R}\). Let \(u_n\to u\) and \(v_n\to v\), and put \(z=(u,v)\). Then $$ \|z_n-z\|_2 =\sqrt{(u_n-u)^2+(v_n-v)^2}\longrightarrow0. $$ Using the other norm comparison, $$ N(z_n-z)\leq C\|z_n-z\|_2\longrightarrow0. $$ Thus \(z_n\to z\) in \(N\), and \(z\in\mathbb{R}^2\). The argument works for every norm \(N\) on \(\mathbb{R}^2\), so each such normed space is complete.
Closed Subspaces Preserve Completeness
A closed subspace of a complete normed space is itself complete, using the norm inherited from the larger space. Closedness prevents a Cauchy sequence in the subspace from converging to a point that lies outside it.
Proof. Let \((w_n)\) be a Cauchy sequence in \(W\). The norm on \(W\) is the restriction of the norm on \(V\), so the same Cauchy estimates show that \((w_n)\) is Cauchy in \(V\). Since \(V\) is complete, there exists \(v\in V\) such that \(w_n\to v\) in \(V\). Because \(W\) is closed, this limit belongs to \(W\). To see why closedness applies, if \(v\notin W\), the open complement \(V\setminus W\) would contain a ball centered at \(v\); convergence would force all sufficiently late \(w_n\) into that ball, contradicting \(w_n\in W\). Thus \(v\in W\), and the convergence in \(V\) is also convergence in the restricted norm on \(W\). Every Cauchy sequence in \(W\) therefore converges in \(W\), as required. \(\square\)
Worked Example: Continuous Functions Vanishing at Both Endpoints
For \(a<b\), consider the subspace $$ W=\{f\in C[a,b]: f(a)=0\text{ and }f(b)=0\}. $$ It is a linear subspace: sums and scalar multiples of functions that vanish at both endpoints still vanish there. It is also closed in the supremum norm. Indeed, suppose \(f_n\in W\) and \(f_n\to f\) in that norm. Since \(f_n(a)=0\) for every \(n\), $$ |f(a)|=|f(a)-f_n(a)|\leq\|f-f_n\|_\infty\longrightarrow0, $$ so \(f(a)=0\). The same argument at \(b\) gives \(f(b)=0\), hence \(f\in W\). The space \(C[a,b]\) is complete in the supremum norm, so the theorem on closed subspaces shows that \(W\) is complete as well.
An Incomplete Normed Space of Smooth Functions
A normed space need not be complete, even when its elements are continuous or differentiable functions. The next example uses the supremum norm on \(C^1[-1,1]\), the space of continuously differentiable functions on \([-1,1]\). The sequence will converge uniformly to a function that is not differentiable, so its limit in the larger space \(C[-1,1]\) does not belong to \(C^1[-1,1]\).
Worked Example: A Cauchy Sequence with No Limit in \(C^1[-1,1]\)
For each positive integer \(n\), define $$ g_n(x)=\sqrt{x^2+\frac{1}{n}},\qquad -1\leq x\leq1. $$ The quantity under the square root is positive, and $$ g_n'(x)=\frac{x}{\sqrt{x^2+1/n}}, $$ which is continuous on \([-1,1]\). Thus \(g_n\in C^1[-1,1]\). For every real \(x\), $$ 0\leq g_n(x)-|x| =\sqrt{x^2+\frac1n}-|x| \leq\frac{1}{\sqrt n}. $$ For the upper bound, \(\sqrt{x^2+1/n}\leq |x|+1/\sqrt n\), since both sides are nonnegative and $$ \left(|x|+\frac1{\sqrt n}\right)^2 =x^2+\frac{2|x|}{\sqrt n}+\frac1n \geq x^2+\frac1n. $$ Consequently, \(\|g_n-|\cdot|\|_\infty\leq1/\sqrt n\), so \(g_n\) converges uniformly to \(|x|\). It follows directly from the triangle inequality that \((g_n)\) is Cauchy in the supremum norm: for \(m,n\geq N\), $$ \|g_n-g_m\|_\infty \leq\|g_n-|\cdot|\|_\infty+\|g_m-|\cdot|\|_\infty \leq\frac1{\sqrt n}+\frac1{\sqrt m} \leq\frac{2}{\sqrt N}. $$ Given \(\varepsilon>0\), choose \(N\) so large that \(2/\sqrt N<\varepsilon\).
If \((g_n)\) converged in \(C^1[-1,1]\) with the supremum norm to some \(h\in C^1[-1,1]\), it would also converge uniformly to \(h\). But it already converges uniformly to \(|x|\); by uniqueness of uniform limits, \(h(x)=|x|\) for every \(x\in[-1,1]\). The function \(|x|\) is not differentiable at \(0\): its left-hand difference quotients equal \(-1\), while its right-hand difference quotients equal \(1\). This contradicts \(h\in C^1[-1,1]\). Therefore this Cauchy sequence has no limit in \(C^1[-1,1]\) under the supremum norm, and that normed space is not complete.
Why Completeness Matters
Completeness lets us turn estimates between approximations into the existence of an actual object in the space. In \(C[a,b]\) with the supremum norm, a uniformly Cauchy sequence of continuous functions has a continuous uniform limit. In an incomplete space, the same kind of Cauchy estimates may produce a limit only after enlarging the space, as the \(C^1[-1,1]\) example demonstrates.
A common pitfall is to show that a sequence is Cauchy and then conclude it converges without checking completeness. Another is to say that a space is complete without naming its norm: the example shows that the supremum norm on \(C^1[-1,1]\) does not ensure limits remain differentiable. In function spaces, the norm determines which notion of closeness is being used, while the underlying space determines whether the resulting limits are allowed members.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What two requirements are contained in the definition of a complete normed space?
- In the proof that \(C[a,b]\) is complete, why does the Cauchy condition imply that \(f_n(x)\) is Cauchy for each fixed \(x\)?
- Where does the proof of completeness of \(C[a,b]\) use the fact that the interval is closed and bounded?
- Why does a closed subspace of a complete normed space contain the limit of each of its Cauchy sequences?
- What is the uniform limit of \(g_n(x)=\sqrt{x^2+1/n}\), and why does it show that \(C^1[-1,1]\) is incomplete in the supremum norm?