From a Normed Space to a Complete Space
The previous tutorial established that \(C(K)\), the continuous real-valued functions on a nonempty compact set \(K\subseteq\mathbb{R}\), is a normed space under the supremum norm. Completeness asks whether a Cauchy sequence in this norm always converges to an element of the same space. Here the norm has a concrete interpretation: Cauchy in norm means that the functions become uniformly close to one another across all of \(K\).
There are two points to check. A sequence of real numbers that is Cauchy at each point has a pointwise limit, but pointwise convergence alone does not guarantee continuity. The Uniform Cauchy Criterion from earlier in the course addresses this issue by turning uniform Cauchy behavior into uniform convergence. We will also give a direct proof using a subsequence whose successive norm differences have a summable bound. This method makes the uniform error estimate explicit. The theorem that uniform limits of continuous functions are continuous then ensures that the limit remains in \(C(K)\).
Throughout, \(K\) is fixed, nonempty, and compact, and the norm on \(C(K)\) is \(\|f\|_\infty=\sup_{x\in K}|f(x)|\). Compactness ensures, by the Extreme-Value Theorem, that every continuous function on \(K\) is bounded, so this norm is finite for every member of \(C(K)\).
Extracting a Subsequence with Summable Differences
The Cauchy property gives control over every pair of sufficiently late terms. To make a limit estimate, it is useful to select particular terms so that the distance from one selected term to the next is at most a prescribed summable sequence. The selection must ensure that each selected index is late enough for the corresponding Cauchy bound.
Proof. Since \((f_n)\) is Cauchy in the supremum norm, for each positive integer \(j\) there is an integer \(N_j\) such that $$ m,n\geq N_j\quad\Longrightarrow\quad \|f_m-f_n\|_\infty<2^{-j}. $$ Choose \(n_1\geq N_1\). Having chosen \(n_j\), choose \(n_{j+1}\) so that $$ n_{j+1}\geq\max\{N_{j+1},n_j+1\}. $$ This recursive choice gives \(n_{j+1}>n_j\), and it ensures \(n_j\geq N_j\) for every \(j\): this holds for \(j=1\) by choice, and the choice of \(n_{j+1}\) ensures it at the next index. Thus both \(n_j\) and \(n_{j+1}\) are at least \(N_j\). Applying the Cauchy bound for \(j\) gives $$ \|f_{n_{j+1}}-f_{n_j}\|_\infty<2^{-j}. $$ For \(p>q\), write $$ f_{n_p}-f_{n_q} =\sum_{j=q}^{p-1}(f_{n_{j+1}}-f_{n_j}). $$ The triangle inequality for the norm now yields $$ \|f_{n_p}-f_{n_q}\|_\infty \leq\sum_{j=q}^{p-1}\|f_{n_{j+1}}-f_{n_j}\|_\infty <\sum_{j=q}^{p-1}2^{-j} <\sum_{j=q}^{\infty}2^{-j}=2^{1-q}. $$ This proves both claims. \(\square\)
The condition \(n_j\geq N_j\) at every stage is essential. It is not enough to require only that the next index \(n_{j+1}\) be at least \(N_j\): the Cauchy estimate applies to a pair only when both indices meet its threshold. The recursive choice above handles both requirements while keeping the indices strictly increasing.
Completeness of \(C(K)\)
Proof. Let \((f_n)\) be Cauchy in \(C(K)\), and select the subsequence \((f_{n_j})\) from the Summable-Subsequence Estimate. Fix \(x\in K\). For \(p>q\), the pointwise difference is bounded by the supremum norm, so $$ |f_{n_p}(x)-f_{n_q}(x)| \leq\|f_{n_p}-f_{n_q}\|_\infty <2^{1-q}. $$ As \(q\) tends to infinity, this bound tends to zero. Hence \((f_{n_j}(x))\) is a Cauchy sequence of real numbers. Completeness of \(\mathbb{R}\) gives a real limit at each \(x\in K\). Define \(f:K\to\mathbb{R}\) by $$ f(x)=\lim_{j\to\infty}f_{n_j}(x). $$
We next show that the convergence to \(f\) is uniform. Fix \(q\geq1\) and \(x\in K\). For every \(p>q\), the estimate already proved gives $$ |f_{n_p}(x)-f_{n_q}(x)|<\sum_{j=q}^{p-1}2^{-j}. $$ Let \(p\) tend to infinity. The left side converges to \(|f(x)-f_{n_q}(x)|\), and the finite sums on the right are bounded above by the infinite tail. Therefore $$ |f(x)-f_{n_q}(x)|\leq\sum_{j=q}^{\infty}2^{-j}=2^{1-q}. $$ This holds for every \(x\in K\). Taking the supremum gives $$ \|f-f_{n_q}\|_\infty\leq2^{1-q}, $$ so \(f_{n_q}\) converges uniformly to \(f\).
Every \(f_{n_q}\) is continuous on \(K\). By the Theorem on Uniform Limits of Continuous Functions, \(f\) is continuous on \(K\). Since \(K\) is compact, the Extreme-Value Theorem implies that \(f\) is bounded. Thus \(f\in C(K)\), rather than merely being a pointwise-defined function on \(K\).
It remains to show that the full sequence converges to \(f\), not only the selected subsequence. Let \(\varepsilon>0\). Since \((f_n)\) is Cauchy, choose \(N\) such that $$ m,n\geq N\quad\Longrightarrow\quad\|f_m-f_n\|_\infty<\frac{\varepsilon}{2}. $$ Choose \(j\) large enough that \(n_j\geq N\) and \(2^{1-j}<\varepsilon/2\). For every \(n\geq N\), both \(n\) and \(n_j\) are at least \(N\). The triangle inequality and the uniform estimate for the subsequence give $$ \|f_n-f\|_\infty \leq\|f_n-f_{n_j}\|_\infty+\|f_{n_j}-f\|_\infty <\frac{\varepsilon}{2}+2^{1-j} <\varepsilon. $$ Therefore \(f_n\) converges to \(f\) in the supremum norm. We have shown that every Cauchy sequence in \(C(K)\) converges to a member of \(C(K)\), which is completeness. \(\square\)
Worked Examples
Worked Example: A Geometric Polynomial Sequence
Let \(K=[-1,1]\), and for each positive integer \(n\) define $$ f_n(x)=\sum_{k=0}^{n}\left(\frac{x}{3}\right)^k. $$ Each \(f_n\) is a polynomial, hence belongs to \(C(K)\). If \(m>n\), then for every \(x\in[-1,1]\), $$ |f_m(x)-f_n(x)| =\left|\sum_{k=n+1}^{m}\left(\frac{x}{3}\right)^k\right| \leq\sum_{k=n+1}^{m}\left(\frac{1}{3}\right)^k \leq\sum_{k=n+1}^{\infty}\left(\frac{1}{3}\right)^k =\frac{1}{2}\,3^{-n}. $$ The last quantity tends to zero as \(n\) tends to infinity. The same bound, with the smaller of two indices in place of \(n\), proves that \((f_n)\) is Cauchy in the supremum norm.
The finite geometric-sum identity gives $$ \left(1-\frac{x}{3}\right)f_n(x) =1-\left(\frac{x}{3}\right)^{n+1}. $$ Since \(1-x/3\neq0\) for \(x\in[-1,1]\), the pointwise limit is \(f(x)=1/(1-x/3)=3/(3-x)\). The error can also be bounded uniformly: $$ |f(x)-f_n(x)| =\left|\frac{(x/3)^{n+1}}{1-x/3}\right| \leq\frac{3^{-(n+1)}}{2/3} =\frac{1}{2}\,3^{-n}. $$ Thus \(f_n\) converges uniformly to the continuous function \(3/(3-x)\) on \(K\), in agreement with completeness.
Worked Example: Cauchy Sequences on a Finite Compact Set
Take \(K=\{-1,2,4\}\). A function on this finite set is continuous, and its supremum norm is the maximum of the absolute values at the three points. Define \(f_n\) by $$ f_n(-1)=1+\frac{1}{n},\qquad f_n(2)=-2+\frac{2}{n},\qquad f_n(4)=3-\frac{1}{n}. $$ For positive integers \(m,n\), the difference at the three points has absolute values $$ \left|\frac{1}{n}-\frac{1}{m}\right|,\qquad 2\left|\frac{1}{n}-\frac{1}{m}\right|,\qquad \left|\frac{1}{n}-\frac{1}{m}\right|. $$ Consequently, $$ \|f_n-f_m\|_\infty =2\left|\frac{1}{n}-\frac{1}{m}\right|, $$ which tends to zero as \(m,n\) tend to infinity. The limit \(f\) has values \(f(-1)=1\), \(f(2)=-2\), and \(f(4)=3\). Directly, $$ \|f_n-f\|_\infty =\max\left\{\frac{1}{n},\frac{2}{n},\frac{1}{n}\right\} =\frac{2}{n}\longrightarrow0. $$ Here completeness reduces to convergence at finitely many coordinates, with the maximum norm measuring the largest coordinate error.
Worked Example: Uniform Convergence to an Absolute-Value Function
Let \(K=[-2,2]\) and define $$ g_n(x)=\sqrt{x^2+\frac{1}{n}}. $$ Each \(g_n\) is continuous. Its candidate limit is \(g(x)=|x|\), which is also continuous. For every \(x\in K\), $$ 0\leq g_n(x)-|x| =\frac{1/n}{\sqrt{x^2+1/n}+|x|} \leq\frac{1}{\sqrt{n}}. $$ The final inequality also holds at \(x=0\), where the difference is exactly \(1/\sqrt n\); when \(x\neq0\), the denominator is at least \(1/\sqrt n\). Therefore $$ \|g_n-g\|_\infty\leq\frac{1}{\sqrt n}\longrightarrow0. $$ In particular, for any positive integers \(m,n\), the triangle inequality gives $$ \|g_n-g_m\|_\infty \leq\|g_n-g\|_\infty+\|g-g_m\|_\infty \leq\frac{1}{\sqrt n}+\frac{1}{\sqrt m}. $$ This proves directly that \((g_n)\) is Cauchy in \(C(K)\) and illustrates how an explicit uniform error estimate controls both convergence and the Cauchy property.
Why Compactness and Uniformity Matter
The completeness proof combines two different facts. The Cauchy property in the supremum norm gives uniform control over the domain; this is what permits the subsequence estimates to hold simultaneously for every \(x\in K\). Uniform convergence then preserves continuity, by the earlier theorem on uniform limits. Compactness has a distinct role: it guarantees that every continuous function on \(K\), including the limit, is bounded and therefore belongs to the space on which the supremum norm is defined.
A common pitfall is to construct a pointwise limit and stop there. Pointwise convergence alone does not establish convergence in the supremum norm, nor does it by itself ensure that the limit is continuous. In the proof, the summable subsequence estimate gives a bound independent of \(x\), which is precisely what establishes uniform convergence. Another pitfall is to check only that the subsequence converges. The final Cauchy estimate is needed to show that the entire original sequence converges to the same limit.
The result strengthens the normed-space conclusion of the previous tutorial: \(C(K)\) is not only a space where norms and distances are available; its Cauchy sequences have limits inside the space. By the definition of a Banach space, \(C(K)\) with the supremum norm is therefore a Banach space. This makes uniform approximation arguments stable: a sequence of continuous functions that becomes uniformly Cauchy cannot converge to a discontinuous or unbounded object outside \(C(K)\).
Check Your Understanding
Use the subsequence estimate and the completeness proof to answer the following questions.
- Why must both selected indices in a successive pair meet the Cauchy threshold for that pair’s bound?
- How does summability of the successive norm differences give a uniform bound on the distance between two selected terms?
- Which earlier theorem ensures that the uniform limit constructed in the proof is continuous?
- Where is compactness used to ensure that the limit belongs to \(C(K)\)?
- Why does convergence of the selected subsequence alone not finish the proof of completeness for the original sequence?