The Geometric-Series Proof of Tietze Extension

From Approximation to Extension

Suppose \(A\) is a closed subset of a normal space \(X\), and \(f:A\to\mathbb{R}\) is bounded and continuous. A useful way to construct an extension is to approximate \(f\) on \(A\) by a continuous function on all of \(X\), then approximate the remaining error, and repeat. If each new error is a fixed fraction of the previous one, the continuous functions produced at the successive stages form a geometrically controlled series.

Here, normal means that \(X\) is \(T_1\) and any two disjoint closed subsets of \(X\) have disjoint open neighborhoods. We use the bounded approximation lemma from the earlier material on Tietze extension. We state its quantitative form explicitly, since the constants determine the convergence argument.

The lemma controls two different quantities: the size of each new function on the whole space, and the error it leaves on \(A\). The error factor \(2/3\) is less than \(1\), so repeated applications make the residuals tend uniformly to zero. Meanwhile the bounds \(L/3\) on the terms make their sum converge.

The Geometric-Series Construction

Proof. If \(A=\varnothing\), take \(F=0\). If \(A\ne\varnothing\) and \(M=0\), then \(|f(a)|\leq 0\) implies \(f(a)=0\) for every \(a\in A\), so again \(F=0\) works. Assume now \(A\ne\varnothing\) and \(M>0\). Put \(q=2/3\), and start with the residual \(r_0=f\), which satisfies \(|r_0(a)|\leq M\).

We define continuous functions \(g_n:X\to\mathbb{R}\) and residuals \(r_n:A\to\mathbb{R}\) successively. If \(r_n\) is continuous and \(|r_n(a)|\leq Mq^n\) on \(A\), apply the bounded approximation lemma with \(L=Mq^n\). It gives \(g_n\) with

Define \(r_{n+1}=r_n-g_n|_A\). This is continuous on \(A\), and the displayed estimate gives \(|r_{n+1}(a)|\leq Mq^{n+1}\). Since this bound holds for \(r_0\) and the construction preserves it at every step, it holds for all \(n\geq 0\).

Consider the partial sums \(S_N=\sum_{n=0}^{N}g_n\). For integers \(m>N\) and every \(x\in X\), the triangle inequality and the bounds on the terms give

For the last inequality, the infinite geometric tail is \(\sum_{n=N+1}^{\infty}q^n=q^{N+1}/(1-q)=3q^{N+1}\), because \(q=2/3\). As \(Mq^{N+1}\to 0\), the partial sums are uniformly Cauchy on \(X\). For each \(x\), \((S_N(x))\) is therefore a Cauchy sequence of real numbers and has a limit. Define \(F(x)=\lim_{N\to\infty}S_N(x)\). The same tail bound, passed to the limit in \(m\), gives \(|F(x)-S_N(x)|\leq Mq^{N+1}\) for every \(x\). Thus \(S_N\to F\) uniformly on \(X\).

Each \(S_N\) is continuous because it is a finite sum of continuous functions. To verify continuity of \(F\), fix \(x_0\in X\) and \(\varepsilon>0\). Choose \(N\) so that \(|F(x)-S_N(x)|<\varepsilon/3\) for every \(x\in X\). By continuity of \(S_N\) at \(x_0\), there is an open neighborhood \(U\) of \(x_0\) such that \(|S_N(x)-S_N(x_0)|<\varepsilon/3\) for \(x\in U\). For each such \(x\),

Therefore \(F\) is continuous. Finally, the definition of the residuals gives \(f-S_N|_A=r_{N+1}\). Hence, for every \(a\in A\),

Since \(S_N(a)\to F(a)\), it follows that \(F(a)=f(a)\). Thus \(F|_A=f\), as required. \(\square\)

Following the Construction in Examples

Why the Geometric Bounds Matter

The construction works because a single bound controls every point of \(A\) at every stage. The residual estimate is uniform: \(|r_n(a)|\leq M(2/3)^n\) for all \(a\in A\). It follows both that the series of functions converges uniformly on \(X\) and that its sum agrees with \(f\) uniformly on \(A\). These are separate uses of geometric decay: one controls the tails of the series on \(X\), and the other controls the error on the subset.

Common pitfall. Do not confuse the error ratio \(2/3\) with the term bound \(1/3\). When \(M=3\), the successive uniform bounds on the terms are \(1\), \(2/3\), \(4/9\), and so on. Their total bound is \(1/(1-2/3)=3\), not \(1/(1-1/3)\). The latter incorrectly treats \(1/3\) as the ratio between successive terms. The factor \(1/3\) is the initial term-size factor; the ratio is \(2/3\).

Another error is to assert continuity from pointwise convergence alone. For example, on \([0,1]\) with the subspace topology from \(\mathbb{R}\), each function \(u_n(x)=x^n\) is continuous. Its pointwise limit is \(0\) for \(0\leq x<1\) and \(1\) at \(x=1\), which is not continuous. The extension construction avoids this failure by proving a single tail bound valid at every \(x\in X\), and hence uniform convergence.