When Supports Move Outward
On a compact domain, controlling a function on compact subsets can amount to controlling it on the entire domain. The real line is different: a compact subset can never contain points arbitrarily far to the right. This lets us move a fixed-shaped function outward, so that every compact set eventually misses it, even though its height never decreases. The resulting sequence converges to zero in the compact-open topology but not uniformly.
Throughout, \(\mathbb{R}\) has its standard topology, and \(C(\mathbb{R},\mathbb{R})\) denotes the set of continuous functions from \(\mathbb{R}\) to \(\mathbb{R}\). For \(f\in C(\mathbb{R},\mathbb{R})\), its support is the closed set \(\operatorname{supp}(f)=\overline{\{x\in\mathbb{R}:f(x)\ne 0\}}\), where the closure is taken in \(\mathbb{R}\).
A basic neighborhood in a topology generated by a subbasis is a finite intersection of subbasic sets. Thus, to prove convergence in the compact-open topology, it suffices to show that every such finite intersection containing the proposed limit eventually contains every term of the sequence.
A Triangular Bump with Fixed Height
Worked Example: Computing One Moving Bump
For each \(n\in\mathbb{N}\), define \(f_n:\mathbb{R}\to\mathbb{R}\) by \(f_n(x)=\max\{1-2|x-n|,0\}\). We compute its continuity, support, and supremum norm.
The absolute-value function is continuous. Also, \(\max\{t,0\}=(t+|t|)/2\), a continuous function of \(t\). Therefore \(f_n\) is continuous as a composition of continuous functions. It is nonzero exactly when \(1-2|x-n|>0\), equivalently when \(|x-n|<1/2\). This is the interval \((n-1/2,n+1/2)\), so its closure, and hence \(\operatorname{supp}(f_n)\), is \([n-1/2,n+1/2]\).
For every \(x\), the maximum defining \(f_n(x)\) is at least \(0\), and \(1-2|x-n|\le 1\). Thus \(0\le f_n(x)\le 1\). At \(x=n\), \(f_n(n)=\max\{1-2|n-n|,0\}=\max\{1,0\}=1\). Consequently \(\sup_{x\in\mathbb{R}}|f_n(x)|=1\), and the supremum is attained.
Check: At either support endpoint, for example \(x=n+1/2\), the formula gives \(f_n(n+1/2)=\max\{1-2(1/2),0\}=0\). The endpoints belong to the support, which is the closure of the nonzero set, but the function is zero there.
We will use the fact that compact subsets of \(\mathbb{R}\) are bounded. Here is the short argument needed for this construction. If \(K\) is compact, the open sets \((-m,m)\), for \(m\in\mathbb{N}\), cover \(K\). A finite subcover exists by compactness. Since these intervals are nested, their largest one in that finite subcover contains all of \(K\). Thus \(K\subseteq(-M,M)\) for some \(M\in\mathbb{N}\). If \(K=\varnothing\), every assertion that it is bounded holds as well.
Worked Example: A Compact Interval Misses the Supports
Take \(K=[-3,4]\). For \(n\ge 5\), the left endpoint of \(\operatorname{supp}(f_n)\) satisfies \(n-1/2\ge 5-1/2=4.5>4\). Therefore \([n-1/2,n+1/2]\cap[-3,4]=\varnothing\), and \(f_n(x)=0\) for every \(x\in K\). In particular \(f_n(K)=\{0\}\) for \(n\ge5\).
The threshold matters: for \(n=4\), the support is \([4-1/2,4+1/2]=[3.5,4.5]\), which intersects \(K\). At the intersection point \(x=3.5\), \(f_4(3.5)=\max\{1-2|3.5-4|,0\}=\max\{0,0\}=0\). This illustrates why the claim about eventual disjointness is made for \(n\ge5\), rather than \(n\ge4\).
Check: For \(n=5\), the support is \([4.5,5.5]\), entirely to the right of \(K\), and \(f_5(4)=\max\{1-2|4-5|,0\}=\max\{-1,0\}=0\). This agrees with the conclusion that \(f_5\) vanishes on \(K\).
The Convergence Result
Proof. First let \(K\subseteq\mathbb{R}\) be compact. If \(K\) is empty, \(\operatorname{supp}(f_n)\cap K=\varnothing\) for every \(n\). Otherwise the boundedness argument above gives an \(M\in\mathbb{N}\) with \(K\subseteq(-M,M)\). If \(n\ge M+1\), then every point of \(\operatorname{supp}(f_n)=[n-1/2,n+1/2]\) is at least \(n-1/2\ge M+1/2>M\), so \(\operatorname{supp}(f_n)\cap K=\varnothing\). In either case, the supports eventually miss \(K\).
Now consider a subbasic neighborhood \([K,U]\) of the zero function, with \(K\) compact and \(U\) open. Being a neighborhood of zero means that the zero function maps \(K\) into \(U\). If \(K\ne\varnothing\), this implies \(0\in U\). Once \(\operatorname{supp}(f_n)\cap K=\varnothing\), every \(x\in K\) has \(f_n(x)=0\), so \(f_n(K)=\{0\}\subseteq U\). If \(K=\varnothing\), then \(f_n(K)=\varnothing\subseteq U\) for every \(n\). Thus every subbasic neighborhood of zero eventually contains \(f_n\). Given a finite intersection of such neighborhoods, take the largest of their finitely many thresholds; all \(f_n\) beyond that threshold lie in the intersection. This proves compact-open convergence to zero.
For every \(n\), the computation in the first example gives \(\sup_{x\in\mathbb{R}}|f_n(x)|=1\). Uniform convergence to zero would require that, for every \(\varepsilon>0\), there be \(N\) such that \(|f_n(x)|<\varepsilon\) for every \(x\in\mathbb{R}\) whenever \(n\ge N\). Choose \(\varepsilon=1/2\). For every \(n\), at \(x=n\) we have \(|f_n(n)|=1\), which is not less than \(1/2\). No such \(N\) exists, so convergence is not uniform. \(\square\)
Worked Example: Testing a Specific Compact-Open Neighborhood
Let \(K=[-2,4]\) and \(U=(-1/10,1/10)\). The set \([K,U]\) is a subbasic open set in the compact-open topology. For \(n\ge5\), the support begins at \(n-1/2\ge4.5>4\). Hence \(f_n(x)=0\) for every \(x\in K\), so \(f_n(K)=\{0\}\subseteq(-1/10,1/10)\). Therefore \(f_n\in[K,U]\) for every \(n\ge5\).
For comparison, at the point \(x=5\), \(f_5(5)=\max\{1-2|5-5|,0\}=1\), which is not in \(U\). This does not contradict \(f_5\in[K,U]\), because \(5\notin K=[-2,4]\). The subbasic condition tests the image of \(K\), not the image of all of \(\mathbb{R}\).
Check: The zero function belongs to \([K,U]\) because it maps every \(x\in K\) to \(0\), and \(-1/10<0<1/10\). The computed eventual membership is therefore membership in a neighborhood of the proposed limit.
The Construction Is Not Tied to One Formula
Worked Example: Wider Bumps of Height Three
Define \(g_n(x)=3\max\{1-|x-(2n+1)|,0\}\). The expression inside the maximum is positive exactly when \(|x-(2n+1)|<1\), or \(2n<x<2n+2\). Thus \(\operatorname{supp}(g_n)=[2n,2n+2]\). The function is continuous by the same absolute-value and maximum calculation used for \(f_n\).
For all \(x\), \(0\le\max\{1-|x-(2n+1)|,0\}\le1\), so \(0\le g_n(x)\le3\). At the center \(x=2n+1\), \(g_n(2n+1)=3\max\{1-|0|,0\}=3\). Therefore \(\sup_{x\in\mathbb{R}}|g_n(x)|=3\) for every \(n\).
If \(K\) is a nonempty compact set, choose \(M\in\mathbb{N}\) with \(K\subseteq(-M,M)\). For \(n\ge M\), the left endpoint of \(\operatorname{supp}(g_n)\) is \(2n\ge2M>M\), so this support misses \(K\). For empty \(K\), disjointness holds for every \(n\). Hence the supports escape every compact set. The same subbasic-neighborhood argument in the theorem proves that \(g_n\) converges compact-open to zero. It cannot converge uniformly to zero: with \(\varepsilon=1\), the value \(g_n(2n+1)=3\) violates the required bound \(|g_n(x)|<1\) for every \(n\).
Check: At a support endpoint, \(g_n(2n)=3\max\{1-|2n-(2n+1)|,0\}=3\max\{0,0\}=0\). At the center it equals \(3\), as calculated above. The endpoint and center values match the claimed support and height.
Why the Distinction Matters
Compact-open convergence tests functions on one compact set at a time. The compact set may be chosen freely, but after it is fixed, the functions in this construction eventually vanish throughout it. Uniform convergence imposes a different requirement: one index must control the values at every point of the domain, including points that move farther out as the index increases.
Common pitfall. A tempting but incorrect calculation is: “For each fixed \(x\), \(f_n(x)=0\) for all sufficiently large \(n\); therefore \(\sup_x|f_n(x)|\to0\).” The first statement is true: once \(n>x+1/2\), the point \(x\) lies to the left of the support, so \(f_n(x)=0\). But the supremum takes all \(x\) at the same index \(n\). Substituting \(x=n\) gives \(f_n(n)=1\), and thus \(\sup_x|f_n(x)|=1\) for every \(n\). The error is allowing the point used to test the supremum to vary with \(n\) while treating a pointwise statement as if it controlled all points at once.
Moving supports therefore give a direct way to distinguish control on every fixed compact subset from control on the whole real line. The construction needs no decrease in height: it is the location of the nonzero values, not their size, that changes.
Check Your Understanding
Use the definitions of support, compact-open convergence, and uniform convergence to answer these questions.
- For \(h_n(x)=\max\{2-4|x-n|,0\}\), compute \(\operatorname{supp}(h_n)\) and \(\sup_{x\in\mathbb{R}}|h_n(x)|\).
- For \(K=[-1,6]\), find an integer \(N\) such that \(\operatorname{supp}(f_n)\cap K=\varnothing\) for all \(n\ge N\).
- Explain why eventual vanishing on each fixed compact set does not imply uniform convergence on \(\mathbb{R}\).
- For the wider bumps \(g_n\), which value of \(x\) gives the supremum \(3\)?
- In the definition of a subbasic set \([K,U]\), what is tested: values on \(K\), or values on all of \(\mathbb{R}\)?