The Bounded Tietze Extension Theorem

Extending Bounded Continuous Functions

A continuous function defined on a closed subset need not be given by a simple formula on the whole space. The bounded Tietze extension theorem guarantees that, in a normal space, it can nevertheless be extended continuously. Its range conclusion is particularly useful: if the original function takes values in a closed interval, the extension can be chosen to stay in that same interval.

Throughout, a space \(X\) is normal if every pair of disjoint closed subsets of \(X\) has disjoint open neighborhoods. A subset \(A\subseteq X\) has the subspace topology. A function \(f:A\to\mathbb{R}\) is bounded if there is a real number \(M\ge 0\) such that \(|f(a)|\le M\) for every \(a\in A\). An extension of \(f\) is a function \(F:X\to\mathbb{R}\) satisfying \(F|_A=f\).

The theorem includes degenerate cases. If \(A=\varnothing\), the constant function \(F(x)=a\) works when \(X\ne\varnothing\); if \(X=\varnothing\), the unique empty function works. If \(a=b\), then \(f\) is constant on \(A\), and that same constant on \(X\) is an extension. For the main proof, we can therefore assume \(A\ne\varnothing\) and \(a<b\).

Proof by Successive Approximation

We use the Urysohn lemma in its standard form: if \(X\) is normal and \(C,D\subseteq X\) are disjoint closed sets, there is a continuous \(u:X\to[0,1]\) with \(u=0\) on \(C\) and \(u=1\) on \(D\). The earlier material on Urysohn's lemma establishes this result. We first prove the approximation step that converts it into an extension.

Proof. Define \(C=\{x\in A:h(x)\le-r/3\}\) and \(D=\{x\in A:h(x)\ge r/3\}\). These sets are closed in \(A\), because \(h\) is continuous and the corresponding intervals are closed in \(\mathbb{R}\). Since \(A\) is closed in \(X\), closed subsets of \(A\) are closed in \(X\). Also \(C\cap D=\varnothing\), since no real number is both at most \(-r/3\) and at least \(r/3\). Apply the Urysohn lemma to obtain \(u:X\to[0,1]\), with \(u=0\) on \(C\) and \(u=1\) on \(D\), and set \(g=(r/3)(2u-1)\). Then \(g\) is continuous and has values in \([-r/3,r/3]\).

For \(x\in C\), \(g(x)=-r/3\), so \(h(x)\in[-r,-r/3]\) gives \(|h(x)-g(x)|\le 2r/3\). For \(x\in D\), \(g(x)=r/3\), so \(h(x)\in[r/3,r]\) also gives \(|h(x)-g(x)|\le 2r/3\). If \(x\in A\setminus(C\cup D)\), then \(-r/3<h(x)<r/3\), while \(g(x)\in[-r/3,r/3]\); the distance between these two numbers is at most \(2r/3\). These cases cover \(A\), proving the bound. \(\square\)

To prove the theorem, first consider a continuous \(h:A\to[-1,1]\). The lemma gives a continuous \(g_1:X\to[-1/3,1/3]\) whose restriction approximates \(h\) with error at most \(2/3\). The residual \(h-g_1|_A\) is therefore a continuous function on \(A\) bounded in absolute value by \(2/3\). Apply the lemma again to this residual, and continue. More precisely, at stage \(n\), the residual \(h_{n-1}\) is bounded in absolute value by \(r_{n-1}=(2/3)^{n-1}\). The lemma gives \(g_n:X\to[-r_{n-1}/3,r_{n-1}/3]\), and we set \(h_n=h_{n-1}-g_n|_A\). Then \(|h_n|\le r_n=(2/3)^n\) on \(A\).

Proof of the theorem. Suppose first that \(f:A\to[-1,1]\) is continuous. Construct the functions \(g_n\) by the preceding procedure, starting with \(h_0=f\). The series \(F=\sum_{n=1}^{\infty}g_n\) converges uniformly on \(X\): for every \(N\ge0\), the absolute value of its tail after term \(N\) is bounded at every \(x\in X\) by \(\sum_{n=N+1}^{\infty}(2/3)^{n-1}/3=(2/3)^N\), which tends to zero as \(N\) tends to infinity. Each partial sum is continuous, so its uniform limit \(F\) is continuous. For completeness, at any \(x\in X\) and any \(\varepsilon>0\), choose a partial sum \(S\) with \(|F-S|<\varepsilon/3\) everywhere. Continuity of \(S\) gives a neighborhood of \(x\) on which \(|S(y)-S(x)|<\varepsilon/3\). The triangle inequality then gives \(|F(y)-F(x)|<\varepsilon\) on that neighborhood.

For every \(x\in X\), the bounds on \(g_n\) give \(|F(x)|\le\sum_{n=1}^{\infty}(2/3)^{n-1}/3=1\), so \(F:X\to[-1,1]\). On \(A\), the first \(N\) terms sum to \(f-h_N\), by the definition of each residual. Since \(|h_N|\le(2/3)^N\), these partial sums converge to \(f\) on \(A\). Thus \(F|_A=f\).

Now let \(a<b\) and \(f:A\to[a,b]\). Put \(c=(a+b)/2\) and \(r=(b-a)/2\), and define \(h=(f-c)/r\). Then \(h:A\to[-1,1]\) is continuous. The result just proved gives a continuous \(H:X\to[-1,1]\) extending \(h\). Set \(F=c+rH\). It extends \(f\), and \(H(x)\in[-1,1]\) implies \(F(x)\in[c-r,c+r]=[a,b]\) for every \(x\in X\). This proves the theorem, including its range conclusion. \(\square\)

Checking the Hypotheses in Examples

What the Hypotheses Do—and a Common Pitfall

The closedness of \(A\) is essential to the general guarantee. It ensures that the closed level sets used in the approximation proof are closed in \(X\), so the Urysohn lemma applies. Without closedness, even a bounded continuous function on \(A\) can fail to extend. For example, take \(A=(0,1)\subseteq\mathbb{R}\) and \(f(x)=\sin(1/x)\). This is bounded and continuous on \(A\), but it has no continuous extension to \(\mathbb{R}\). Indeed, the sequences \(x_n=1/(\pi/2+2\pi n)\) and \(y_n=1/(3\pi/2+2\pi n)\), for \(n\in\mathbb{N}\), lie in \((0,1)\) for all sufficiently large \(n\), and both tend to \(0\). Yet \(f(x_n)=1\) and \(f(y_n)=-1\). Continuity of an extension at \(0\) would force both sequences of values to tend to the same value there, a contradiction.

Common pitfall. Do not weaken the range conclusion to “the extension stays strictly between the endpoints.” The theorem guarantees values in the closed interval \([a,b]\), not in \((a,b)\). In the first example, the prescribed value \(f(0)=-1\) forces any extension to take the endpoint value \(-1\). More generally, an extension cannot have range inside \((a,b)\) when \(f\) takes the value \(a\) or \(b\) at a point of \(A\).

Normality is the separation property used in the proof: it provides the Urysohn functions that control the successive approximations. Boundedness supplies a finite interval in which to begin, while closedness allows the relevant level sets to be separated in \(X\). The theorem is therefore an extension result with explicit requirements, not a claim that every continuous function on every subset can be extended.