Why an Extension May Need Range Control
Suppose a continuous function on a closed subset takes values only between two real numbers \(a\) and \(b\). An extension supplied by a Tietze theorem is continuous on the whole space, but its values away from the original subset need not remain between those bounds. A simple operation fixes this: values below \(a\) are raised to \(a\), values above \(b\) are lowered to \(b\), and values already in the interval are left unchanged.
Throughout, \((X,\mathcal{T})\) is a normal topological space, \(A\subseteq X\) is closed with the subspace topology, and \(a,b\in\mathbb{R}\) satisfy \(a\le b\). A function is bounded if its absolute value is bounded above by some real constant. We use the bounded Tietze extension theorem in the following form: if \(f:A\to\mathbb{R}\) is bounded and continuous, then there is a continuous \(G:X\to\mathbb{R}\) with \(G|_A=f\). This existence result is established in the earlier material on Tietze extension; the theorem in this form does not require the extension to have its values in the same interval as \(f\).
The formula has the following equivalent piecewise form, including the endpoints:
Why Clipping Preserves Continuity
Proof. For real numbers \(u,v\), the identities \(\max(u,v)=(u+v+|u-v|)/2\) and \(\min(u,v)=(u+v-|u-v|)/2\) express maximum and minimum using addition, subtraction, and absolute value. Each of these operations is continuous on \(\mathbb{R}\), so both the maximum and minimum of two continuous real-valued functions are continuous. Applying this fact first to \(\min(t,b)\), then to \(\max(a,\min(t,b))\), proves continuity of \(c_{a,b}\). By its piecewise formula, every output is either \(a\), \(t\in[a,b]\), or \(b\), so it lies in \([a,b]\). If \(a\le t\le b\), then \(\min(t,b)=t\) and \(\max(a,t)=t\); hence \(c_{a,b}(t)=t\). This argument also covers \(a=b\): the map is then constantly \(a\). \(\square\)
The last property is what makes clipping useful for extension problems. It changes values outside the target interval, but leaves every value already in that interval untouched.
Worked Example: Clipping an Extension on the Real Line
Take \(X=\mathbb{R}\) with its standard topology, \(A=[-1,1]\), and \(f:A\to[1,3]\) given by \(f(x)=2+x\). The formula \(G(x)=2+x\) defines a continuous real-valued function on all of \(\mathbb{R}\) and satisfies \(G|_A=f\), but \(G\) does not take all its values in \([1,3]\).
Clip \(G\) to the interval \([1,3]\). If \(x\le-1\), then \(G(x)=2+x\le1\), so \(c_{1,3}(G(x))=1\). If \(-1\le x\le1\), then \(1\le2+x\le3\), so \(c_{1,3}(G(x))=2+x\). If \(x\ge1\), then \(G(x)=2+x\ge3\), so \(c_{1,3}(G(x))=3\). Thus the clipped function is
At \(x=-1\), the first and middle formulas both give \(1\); at \(x=1\), the middle and last formulas both give \(3\). The lemma gives continuity without needing a separate piecewise continuity argument. On \(A\), \(G(x)=f(x)\in[1,3]\), so clipping does not change \(G(x)\). For instance, at \(x=-2\), \(G(-2)=2+(-2)=0\) and \(H(-2)=1\); at \(x=0\), \(G(0)=2+0=2\) and \(H(0)=2\); at \(x=2\), \(G(2)=2+2=4\) and \(H(2)=3\). Therefore \(H\) takes values in \([1,3]\) and extends \(f\).
The Range-Preserving Extension Theorem
Proof. Since \([a,b]\) is bounded, \(f\), viewed as a real-valued function, is bounded. By the bounded Tietze extension theorem stated above, there is a continuous \(G:X\to\mathbb{R}\) with \(G|_A=f\). Define \(h=c_{a,b}\circ G\). The clipping lemma gives continuity of \(c_{a,b}\), so \(h\) is continuous as a composition of continuous functions. The same lemma gives \(h(x)\in[a,b]\) for every \(x\in X\). Finally, if \(x\in A\), then \(G(x)=f(x)\in[a,b]\), and clipping is the identity on \([a,b]\). Therefore \(h(x)=c_{a,b}(G(x))=f(x)\), which proves \(h|_A=f\).
This proof does not require \(a<b\). If \(a=b\), then \([a,b]=\{a\}\), so \(f\) has the constant value \(a\) wherever it is defined, and clipping sends every real number to \(a\). If \(A=\varnothing\), the extension condition is vacuous; the proof still applies, or one may take the constant function \(h(x)=a\) on \(X\). When \(X=\varnothing\), the unique function with domain \(X\) is continuous and the range condition is vacuous. \(\square\)
The argument separates the extension problem into two tasks. The Tietze theorem supplies a continuous real-valued extension; clipping then imposes the desired range without changing any prescribed values. In particular, the range-control step does not depend on how the original extension was constructed.
Worked Example: Clipping on a Closed Line in the Plane
Give \(X=\mathbb{R}^2\) its Euclidean topology, and let \(A=\mathbb{R}\times\{0\}\), which is closed. Define \(f:A\to[0,1]\) by \(f(x,0)=(1+\sin x)/2\). The inequality \(-1\le\sin x\le1\) gives \(0\le(1+\sin x)/2\le1\). A continuous real-valued extension is \(G(x,y)=(1+\sin x)/2+2y\): it is continuous, and \(G(x,0)=f(x,0)\).
Set \(H(x,y)=c_{0,1}(G(x,y))\). The clipping lemma shows that \(H\) is continuous and has values in \([0,1]\). For every \((x,0)\in A\), \(G(x,0)=f(x,0)\in[0,1]\), so \(H(x,0)=f(x,0)\). To see what clipping changes off \(A\), evaluate at two points. Since \(\sin(\pi/2)=1\),
Since \(\sin(3\pi/2)=-1\),
These calculations show why the unrestricted extension \(G\) itself need not be range-preserving, while its clipped version is. On the line \(y=0\), clipping leaves the prescribed values unchanged.
Worked Example: The Degenerate Interval
Let \(X=\mathbb{R}\) with its standard topology, let \(A=[0,1]\), and take the target interval \([4,4]=\{4\}\). Any function \(f:A\to\{4\}\) has \(f(x)=4\) for every \(x\in A\). Define \(h(x)=4\) for every \(x\in\mathbb{R}\). For any open set \(V\subseteq\mathbb{R}\), the preimage \(h^{-1}(V)\) is \(\mathbb{R}\) if \(4\in V\), and is \(\varnothing\) if \(4\notin V\); both are open. Thus \(h\) is continuous, its values lie in \([4,4]\), and \(h|_A=f\).
The clipping formula gives the same result: for every \(t\in\mathbb{R}\), \(c_{4,4}(t)=\max(4,\min(t,4))=4\). This checks that the theorem includes the case of a one-point interval, rather than requiring a division by \(b-a\).
Common Pitfall: Reversing the Maximum and Minimum
For clipping to \([a,b]\), the order of the operations matters: first cap the value above at \(b\), then raise values below \(a\) to \(a\). For example, on \([1,3]\) the incorrect expression \(\min(1,\max(t,3))\) sends \(t=2\) to \(\min(1,3)=1\), even though \(2\) is already in the target interval. The correct computation is \(\max(1,\min(2,3))=\max(1,2)=2\). For an outside value such as \(t=4\), the correct expression gives \(\max(1,\min(4,3))=\max(1,3)=3\); for \(t=0\), it gives \(\max(1,\min(0,3))=\max(1,0)=1\).
Another important check is that range control must not alter the data on \(A\). In the theorem, this follows from \(f(A)\subseteq[a,b]\) and the identity \(c_{a,b}(t)=t\) on that interval. If the original function takes even one value outside the proposed target interval, clipping will change that value; it cannot then be an extension of the original function.
Check Your Understanding
For each question, pay attention to the interval endpoints and to whether the original values are changed.
- Write the piecewise formula for \(c_{-2,5}(t)\), including the cases \(t=-2\) and \(t=5\).
- If \(G:X\to\mathbb{R}\) is continuous and \(f=G|_A\) takes values in \([a,b]\), explain why \(c_{a,b}\circ G\) still agrees with \(f\) on \(A\).
- What does the clipping map become when \(a=b\), and why does the extension proof still work?
- For clipping to \([0,2]\), compute \(c_{0,2}(-3)\), \(c_{0,2}(1)\), and \(c_{0,2}(4)\).
- Why is continuity of an arbitrary extension not, by itself, enough to ensure that its values lie in the original function’s target interval?