Pointwise Control and Uniform Control
For a family of functions, a bound at each fixed input is not necessarily a bound that works at every input. A second distinction is equally important: bounded sets of values need not have compact closure in the target. These differences matter when asking whether a family of continuous functions is compact, or relatively compact, in a function space.
Let \(X\) be a set, let \((Y,d_Y)\) be a metric space, and let \(\mathcal{F}\) be a family of maps \(f:X\to Y\). Fix a point \(y_0\in Y\) when \(Y\) is nonempty. Boundedness of subsets of \(Y\) is measured using \(d_Y\).
Uniform boundedness does not depend on which \(y_0\) is chosen: if \(y_1\in Y\), then \(d_Y(f(x),y_1)\le d_Y(f(x),y_0)+d_Y(y_0,y_1)\). Pointwise boundedness allows the bound to depend on \(x\); uniform boundedness requires one bound for all inputs and all functions. Neither definition asserts that the family is compact in a function-space topology.
What Relative Compactness Gives
Proof. Let \(A\) be a subset of a metric space and suppose \(\overline{A}\) is compact. The open balls \(B(y,1)\), with \(y\in\overline{A}\), cover \(\overline{A}\). Compactness gives a finite subcover \(B(y_1,1),\dots,B(y_k,1)\). Choose \(y_0\in\overline{A}\) if the set is nonempty. For each \(y\in\overline{A}\), some \(i\) satisfies \(d_Y(y,y_i)<1\), so \(d_Y(y,y_0)<1+d_Y(y_i,y_0)\). The finitely many numbers \(1+d_Y(y_i,y_0)\) have a finite maximum. Thus \(\overline{A}\), and hence \(A\), is bounded. If \(A=\varnothing\), it is bounded as well.
Now suppose \(A\subseteq\mathbb{R}^m\) is bounded. Its closure is bounded and closed. By the Heine–Borel theorem, a subset of \(\mathbb{R}^m\) is compact exactly when it is closed and bounded. Therefore \(\overline{A}\) is compact, so \(A\) is relatively compact. The reverse implication follows from the first part. \(\square\)
In particular, for maps into \(\mathbb{R}^m\), pointwise relative compactness and pointwise boundedness are equivalent: apply the proposition to the values at each fixed \(x\). For a general metric target, pointwise relative compactness still implies pointwise boundedness, but the converse can fail.
Worked Example: Pointwise Relative Compactness in \(\mathbb{R}^2\)
Let \(X=[0,1]\), and let \(f_n:X\to\mathbb{R}^2\) be the constant function \(f_n(x)=(1/n,(-1)^n)\), for \(n\in\mathbb{N}\). At any \(x\), the set of values is \(\{(1/n,(-1)^n):n\in\mathbb{N}\}\), independent of \(x\).
For every \(n\), the squared Euclidean norm is \((1/n)^2+((-1)^n)^2=1/n^2+1\le2\). Hence all values lie in the closed ball of radius \(\sqrt{2}\) centred at \((0,0)\). The closure of the value set is closed and bounded, so it is compact by Heine–Borel. This verifies pointwise relative compactness at every \(x\). It also gives a uniform bound: the distance of every \(f_n(x)\) from \((0,0)\) is at most \(\sqrt{2}\). The closure contains the two limit points \((0,1)\) and \((0,-1)\), corresponding to even and odd indices; compactness does not require that the value set itself be closed.
Check: the bound uses the same radius for every \(n\) and \(x\), since \(1/n^2+1\le2\) for every \(n\ge1\). Thus the family is uniformly bounded as well as pointwise relatively compact.
Pointwise Bounds Need Not Be Uniform
Worked Example: Continuous Functions with No Uniform Bound
On \(X=[0,1]\), with its usual topology, define \(f_n(x)=n^2x(1-x)^n\) for \(n\in\mathbb{N}\). Each \(f_n\) is continuous. We show first that the family is pointwise bounded.
At \(x=0\) and \(x=1\), every value \(f_n(x)\) is zero. Fix \(x\) with \(0<x<1\), and put \(q=1-x\), so \(0<q<1\). The positive terms \(a_n=n^2q^n\) satisfy \[ \frac{a_{n+1}}{a_n}=q\left(1+\frac1n\right)^2\longrightarrow q<1. \] Choose \(r\) with \(q<r<1\). For all sufficiently large \(n\), the displayed ratio is less than \(r\). From that index onward, each term is at most \(r\) times its predecessor, so the tail is bounded by its first term. The finitely many earlier terms are bounded too. Thus \(\{f_n(x):n\in\mathbb{N}\}\) is bounded for every fixed \(x\).
But there is no common bound over all \(n\) and \(x\). Set \(x_n=1/(n+1)\), which belongs to \([0,1]\). Substitution gives \[ f_n(x_n)=\frac{n^2}{n+1}\left(\frac{n}{n+1}\right)^n. \] By the binomial formula, \((1+1/n)^n=\sum_{k=0}^n {n\choose k}n^{-k}<3\): indeed \({n\choose k}n^{-k}\le1/k!\), and \(\sum_{k=0}^\infty1/k!<3\). Consequently \((n/(n+1))^n>1/3\). Also \(n^2/(n+1)\ge n/2\), since \(2n^2\ge n(n+1)\) for \(n\ge1\). Therefore \(f_n(x_n)>n/6\), which grows without bound.
Check: the same family is bounded at each fixed \(x\), but values at the varying inputs \(x_n\) exceed \(n/6\). Hence pointwise boundedness does not imply uniform boundedness, even for continuous functions on a compact interval.
Common pitfall. A tempting but invalid argument is: for each \(x\), choose a pointwise bound \(M_x\), then take \(\max_{x\in X}M_x\). Pointwise boundedness supplies no reason that these bounds have a finite maximum, or even that \(x\mapsto M_x\) is bounded. In the example, the bound at each fixed \(x\) exists, while the values \(f_n(x_n)>n/6\) rule out every finite uniform bound.
Bounded Values Do Not Always Have Compact Closure
Worked Example: A Bounded Set That Is Not Relatively Compact
Let \(Y=\mathbb{N}\) with the discrete metric \(d_Y(j,k)=0\) if \(j=k\), and \(d_Y(j,k)=1\) if \(j\ne k\). The set \(A=Y\) is bounded: every two of its points are at distance at most \(1\). It is closed, because its complement in \(Y\) is empty, so \(\overline{A}=Y\).
The discrete open cover \(\{\{j\}:j\in\mathbb{N}\}\) has no finite subcover: any finite union of singleton sets contains only finitely many natural numbers. Thus \(Y\) is not compact and \(A\) is not relatively compact. This shows why boundedness alone cannot replace relative compactness in an arbitrary metric target.
For example, take \(X\) to be a one-point space and let the family consist of constant maps with values \(j\in\mathbb{N}\). It is pointwise bounded in this discrete metric, but it is not pointwise relatively compact.
Pointwise Relative Compactness Is Not Function-Space Compactness
Let \(K\) be a compact metric space and \(Y\) a metric space. For continuous maps \(f,g:K\to Y\), the uniform metric is \[ d_\infty(f,g)=\sup_{x\in K}d_Y(f(x),g(x)). \] This supremum is finite: each continuous image of the compact space \(K\) is compact, and compact metric sets are bounded by the proposition above. Compactness in this metric is stronger than pointwise compactness of the values.
Worked Example: Pointwise Relative Compactness Without Uniform Compactness
For \(n\in\mathbb{N}\), let \(f_n:[0,1]\to\mathbb{R}\) be \(f_n(x)=x^n\). At each fixed \(x\in[0,1]\), all the values lie in \([0,1]\), whose closure in \(\mathbb{R}\) is compact by Heine–Borel. Thus this family is pointwise relatively compact.
Nevertheless, the sequence \((f_n)\) has no uniformly convergent subsequence. For any subsequence with indices \(n_j\to\infty\), its pointwise limit is \(0\) for \(0\le x<1\), because \(x^{n_j}\to0\), and is \(1\) at \(x=1\). This pointwise limit is discontinuous at \(1\). A uniform limit of continuous real-valued functions on \([0,1]\) is continuous: given a uniform limit \(g\), a point \(x\), and \(\varepsilon>0\), choose one function \(f_{n_j}\) with \(\sup_z|f_{n_j}(z)-g(z)|<\varepsilon/3\); continuity of that function gives a neighborhood of \(x\) on which its values differ from \(f_{n_j}(x)\) by less than \(\varepsilon/3\). The triangle inequality then makes the values of \(g\) differ from \(g(x)\) by less than \(\varepsilon\). So the discontinuous pointwise limit cannot be a uniform limit.
In a metric space, compactness implies that every sequence has a convergent subsequence. Therefore, if the family \(\{f_n:n\in\mathbb{N}\}\) had compact closure in the uniform metric, the sequence would have a uniformly convergent subsequence. The preceding argument rules this out, so the family is not relatively compact in that function space.
The missing control here is equicontinuity: one choice of input tolerance must work for all functions in the family. The family \(x^n\) is not equicontinuous at \(1\). Given any \(\delta>0\), choose \(x=1-\min(\delta/2,1/2)\), so \(x\in[0,1)\) and \(|1-x|<\delta\). For sufficiently large \(n\), \(x^n<1/2\); hence \(|f_n(1)-f_n(x)|>1/2\), despite the input points being less than \(\delta\) apart.
The distinction can be summarized as follows. Pointwise relative compactness controls the set of possible outputs at each fixed input. Uniform boundedness controls all outputs at once, but does not itself supply compactness in a general target. For compactness of a family of continuous maps, one also needs control over how outputs change as inputs vary; equicontinuity provides that control.
Check Your Understanding
Use the definitions to distinguish the value sets at individual inputs from the family as a whole.
- Why does pointwise relative compactness imply pointwise boundedness for maps into any metric space?
- For maps into \(\mathbb{R}^m\), why are pointwise boundedness and pointwise relative compactness equivalent?
- In the family \(f_n(x)=n^2x(1-x)^n\), which inputs demonstrate failure of a uniform bound?
- Why is \(\mathbb{N}\) with the discrete metric bounded but not relatively compact?
- For \(f_n(x)=x^n\) on \([0,1]\), what is the pointwise limit of any subsequence whose indices tend to infinity, and why does it obstruct uniform convergence?