Function Space Topologies and the Exponential Law: Examples and Counterexamples

Let $I:=[0,1]$ carry the metric $d(s,t)=\mathrm{\mid }s-t\mathrm{\mid }$ inherited from $\mathbb{R}$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset) and let $\mathbb{R}$ carry the same metric. Write $C(I,\mathbb{R})$ for the continuous real functions on $I$, $\bar{\rho }$ for the uniform metric of For a nonempty set $X$ and a metric space $(Y,d)$ the uniform metric $\bar{\rho }(f,g)={\sup }_x\min \{d(f(x),g(x)),1\}$ is a metric on $Y^X$ and $d_{\mathrm{\infty }}$ for the supremum metric of The supremum metric $d_{\mathrm{\infty }}(f,g)={\sup }_x\mathrm{\mid }f(x)-g(x)\mathrm{\mid }$ is a metric on the bounded real-valued functions on a nonempty set. Then:

1. every $f\in C(I,\mathbb{R})$ is bounded, so $d_{\mathrm{\infty }}$ is defined on $C(I,\mathbb{R})$;
2. $\bar{\rho }$ and $d_{\mathrm{\infty }}$ are uniformly equivalent on $C(I,\mathbb{R})$, hence induce the same topology there (Topologically, uniformly and Lipschitz equivalent metrics on a set, Lipschitz equivalence implies uniform equivalence implies topological equivalence);
3. $(C(I,\mathbb{R}),\bar{\rho })$ is a complete metric space (Complete metric space: every Cauchy sequence converges in the space).

Claim 2 is this page's guarantee that no second notion of convergence has been created. For a nonempty set $X$ and a metric space $(Y,d)$ the uniform metric $\bar{\rho }(f,g)={\sup }_x\min \{d(f(x),g(x)),1\}$ is a metric on $Y^X$ mints a metric on $Y^X$ that is not the published supremum metric — it truncates distances at $1$ and needs no boundedness hypothesis — and a reader who has met $d_{\mathrm{\infty }}$ first is entitled to ask whether "uniform convergence" now means two things. On the set where both are defined it does not: the two metrics take different values but have the same open sets, so they have the same convergent sequences, the same continuous functions and the same closed sets.

• The two metrics really are different functions. Take $f$ constant $0$ and $g$ constant $2$: then $d_{\mathrm{\infty }}(f,g)=2$ while $\bar{\rho }(f,g)=1$. What claim 2 says is that this difference is invisible to the topology, not that it does not exist. In particular an assertion about the value of the distance — a diameter, a Lipschitz constant, a radius — must name which metric it means.

• Completeness is inherited from $\mathbb{R}$ and from closedness, in that order. $\mathbb{R}$ is complete, so all the real functions on $I$ are complete in the uniform metric; the continuous ones form a closed subset by the uniform limit theorem (A uniform limit of continuous functions is continuous, so $C(X,Y)$ is closed in $Y^X$ under the uniform metric); and a closed subset of a complete space is complete. Completeness in $d_{\mathrm{\infty }}$ follows as well, since uniformly equivalent metrics have the same Cauchy sequences and the same convergent sequences, both conditions being expressed with $\epsilon$ and $\delta$ alone.

• Compactness of $I$ is used only for claim 1. Boundedness of every continuous function is what makes $d_{\mathrm{\infty }}$ defined at all, and that is the extreme value theorem. On a non-compact domain the supremum metric is unavailable on all of $C(X,\mathbb{R})$, while the uniform metric $\bar{\rho }$ remains defined; that is the whole reason this page mints the truncated metric.

Let $I:=[0,1]$ with the metric inherited from $\mathbb{R}$, let $a_k:=1\mathrm{/}\iota (k+2)$ for $k\in \mathbb{N}$ (The canonical natural $\iota (n)=n\cdot 1_F$ of a field), and let $f_k:I\to \mathbb{R}$ be the moving spike

$f_k(t)=\frac{t}{a_k}(0\le t\le a_k),f_k(t)=2-\frac{t}{a_k}(a_k\le t\le 2a_k),f_k(t)=0(2a_k\le t\le 1),$

which is exactly the family built in FALSE: a pointwise convergent sequence of continuous functions converges uniformly on every compact set: each $f_k$ is continuous, and $(f_k)$ converges pointwise to the constant function $\mathbf{0}$, which is continuous. Write $\bar{\rho }$ for the uniform metric on ${\mathbb{R}}^I$ (For a nonempty set $X$ and a metric space $(Y,d)$ the uniform metric $\bar{\rho }(f,g)={\sup }_x\min \{d(f(x),g(x)),1\}$ is a metric on $Y^X$).

This example traces the one family through all three topologies of the A page:

1. $f_k\to \mathbf{0}$ in the topology of pointwise convergence (The topology of pointwise convergence on $Y^X$, which is the product topology, and its restriction to $C(X,Y)$);
2. $\bar{\rho }(f_k,\mathbf{0})=1$ for every $k$, so $(f_k)$ does not converge to $\mathbf{0}$ in the topology of uniform convergence (Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^X$ and on $C(X,Y)$);
3. $(f_k)$ does not converge to $\mathbf{0}$ in the topology of compact convergence either (The topology of compact convergence on $C(X,Y)$ for metric $X$ and $Y$: uniform convergence on each compact subset of $X$).

So the two inclusions of On $C(X,Y)$ with $X$ and $Y$ metric, uniform convergence is finer than compact convergence, which is finer than pointwise convergence are strict on $C(I,\mathbb{R})$ at the leftmost step: pointwise convergence is strictly weaker than convergence on compact sets. The two rightmost topologies coincide here, because $I$ is itself compact; separating those two needs a domain that is not compact, and the next counterexample on this page does it on $\mathbb{R}$.

• Nothing is lost and nothing is gained by the truncation. The uniform metric truncates distances at $1$, and here the spikes never exceed $1$, so $\bar{\rho }(f_k,\mathbf{0})$ is the honest supremum of $\mathrm{\mid }{f}_k\mathrm{\mid }$. A family of spikes of height $5$ would have $\bar{\rho }=1$ as well, which is exactly the sense in which the uniform metric records "not close" without recording how far.

• The failure is at a moving point. For each fixed $t$ the values $f_k(t)$ are eventually $0$; what prevents a single index from serving every $t$ is that the place where $f_k$ equals $1$ depends on $k$ and never disappears. That is the quantifier order of Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^X$ and on $C(X,Y)$, seen in one family.

• On this domain the two right-hand topologies cannot be separated. Since $I=[0,1]$ is compact, $K=I$ is an admissible compact set and convergence on every compact subset of $I$ is convergence on $I$ itself, that is uniform convergence. Separating them needs a domain that is not compact, and the next counterexample on this page uses $\mathbb{R}$.

Refuted claim: if a sequence $(g_k)$ in $C(X,\mathbb{R})$ converges to $g$ uniformly on every compact subset of $X$, that is in the topology of compact convergence (The topology of compact convergence on $C(X,Y)$ for metric $X$ and $Y$: uniform convergence on each compact subset of $X$), then it converges to $g$ uniformly (Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^X$ and on $C(X,Y)$).

The witness is $X=\mathbb{R}$ with its usual metric and

$g_k(x):=\frac{x}{\iota (k+1)}(k\in \mathbb{N},x\in \mathbb{R}),$

$\iota$ being the canonical natural of $\mathbb{R}$ (The canonical natural $\iota (n)=n\cdot 1_F$ of a field), so that $\iota (k+1)\ge 1>0$ and the quotient is defined. These converge to the constant function $\mathbf{0}$ uniformly on every compact subset of $\mathbb{R}$ and satisfy $\bar{\rho }(g_k,\mathbf{0})=1$ for every $k$, so they do not converge uniformly.

This is the strictness of the right-hand inclusion of On $C(X,Y)$ with $X$ and $Y$ metric, uniform convergence is finer than compact convergence, which is finer than pointwise convergence; the left-hand one is separated on the companion example of this page. Note that the domain has to be non-compact for such a witness to exist, since on a compact domain the two topologies coincide.

• What goes wrong is arbitrarily far out. On any fixed bounded region the maps $g_k$ do flatten to $0$, and a compact subset of $\mathbb{R}$ is bounded; the discrepancy $\mathrm{\mid }{g}_k(x)\mathrm{\mid }=\mathrm{\mid }x\mathrm{\mid }\mathrm{/}(k+1)$ reaches $1$ only at $x=k+1$, which escapes every compact set as $k$ grows. Uniform convergence asks for control at every point at once, including those.

• The same family shows the two topologies are different as topologies, not merely that one sequence behaves differently in them: the difference is already visible in a basic neighbourhood, since $B_{\bar{\rho }}(\mathbf{0},1\mathrm{/}2)$ contains no $g_k$ while every compact-convergence neighbourhood of $\mathbf{0}$ contains a tail of them.

• The index shift is the usual one. $\mathbb{N}$ contains $0$, so the family is written with $\iota (k+1)$; at $k=0$ this is $1$ and $g_0$ is the identity, which is exactly the intended first term. Writing $x\mathrm{/}\iota (k)$ would divide by $0$ at $k=0$.

Let $\mathbb{R}$ carry its usual metric $d(s,t)=\mathrm{\mid }s-t\mathrm{\mid }$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded) and let $C(\mathbb{R},\mathbb{R})$ carry the compact-open topology (The compact-open topology on $C(X,Y)$ for a metric domain $X$, with subbasis $S(K,V)=\{f:f[K]\subseteq V\}$). For a natural $m\ge 1$ write $[-m,m]:=\{t\in \mathbb{R}:-\iota (m)\le t\le \iota (m)\}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, The canonical natural $\iota (n)=n\cdot 1_F$ of a field). Then:

1. every $[-m,m]$ is a compact subset of $\mathbb{R}$, and every compact $K\subseteq \mathbb{R}$ is contained in some $[-m,m]$;
2. for each $f\in C(\mathbb{R},\mathbb{R})$ the sets $B_{[-m,m]}(f,\epsilon )=\{g\in C(\mathbb{R},\mathbb{R}):\mathrm{\mid }f(t)-g(t)\mathrm{\mid }<\epsilon \text{ for every }t\in [-m,m]\}(m\ge 1,\epsilon >0)$ form a neighbourhood base at $f$ in the compact-open topology (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open);
3. $\mathbb{R}$ is a locally compact metric space (Locally compact metric space: every point has a compact neighbourhood), so the evaluation map $e:C(\mathbb{R},\mathbb{R})\times \mathbb{R}\to \mathbb{R}$ is continuous (If $X$ is a locally compact metric space then the evaluation map is continuous for the compact-open topology, The evaluation map $e:C(X,Y)\times X\to Y$, $e(f,x)=f(x)$).

The quantity ${\sup }_{t\in [-m,m]}\mathrm{\mid }f(t)-g(t)\mathrm{\mid }$ of the title exists and is a maximum, by fact (U3) of The topology of compact convergence on $C(X,Y)$ for metric $X$ and $Y$: uniform convergence on each compact subset of $X$; the formulation in claim 2 avoids writing it, which is what keeps the empty compact set harmless elsewhere on this page.

• Claim 2 is what makes the compact-open topology on $C(\mathbb{R},\mathbb{R})$ concrete. A general neighbourhood in it involves an arbitrary compact set and an arbitrary open subset of the target; claim 2 replaces both by a bound on a symmetric interval and a single $\epsilon$, and the intervals may be indexed by the naturals. That is the shape a metrization proof would exploit, and this library does not carry out that proof.

• Local compactness of $\mathbb{R}$ is where Heine-Borel is spent. In a general metric space a closed ball need not be compact, and then nothing above survives; what makes $\mathbb{R}$ work is that closed bounded sets are compact (Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line). The contrast is $\mathbb{Q}$, where the evaluation map is not continuous at all.

• The intervals $[-m,m]$ exhaust $\mathbb{R}$, and that is claim 1's real content. Every compact subset sits inside one of countably many of them, so the compact sets, of which there are very many, are controlled by a countable family. Nothing about metrizability follows from this alone, and none is claimed.

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $\mathbb{R}$ carry its usual metric (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded). Put

$\mathcal{L}:=\{f:X\to \mathbb{R}:f\text{ is Lipschitz with constant }1\}\subseteq {\mathbb{R}}^X$

(Lipschitz map, $\alpha$-Hölder map for rational $0<\alpha \le 1$, and contraction, The topology of pointwise convergence on $Y^X$, which is the product topology, and its restriction to $C(X,Y)$). Then:

1. $\mathcal{L}$ is uniformly equicontinuous (Equicontinuity at a point, uniform equicontinuity, and pointwise boundedness of a family of maps between metric spaces), with $\delta :=\epsilon$ serving at every $\epsilon$;
2. for every nonempty $A\subseteq X$ the distance function ${\phi }_A(x):=d(x,A)$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) belongs to $\mathcal{L}$;
3. if $X\mathrm{\ne }\mathrm{\varnothing }$ then $\mathcal{L}$ is not pointwise bounded, since it contains every constant function.

So equicontinuity and pointwise boundedness are genuinely independent hypotheses: this family has the first and not the second, and the next counterexample on this page has the second and not the first.

• A common constant is what makes the family equicontinuous, not Lipschitzness of each member. Every member of $\mathcal{L}$ is Lipschitz, but so is every member of $\{x\mapsto \iota (k)x:k\in \mathbb{N}\}$ on $\mathbb{R}$, and that family is not equicontinuous at any point: the constants grow without bound. Fixing the constant at $1$ is the hypothesis doing the work, exactly as the last remark of Equicontinuity at a point, uniform equicontinuity, and pointwise boundedness of a family of maps between metric spaces records.

• Claim 2 is why equicontinuity is worth defining at all here. The distance functions ${\phi }_A$ are the standard supply of Lipschitz maps in a metric space, and they are what an Ascoli-type argument on a later page will use; that they all sit in one uniformly equicontinuous family is the reason such arguments do not need any hypothesis on $A$ beyond nonemptiness.

• Claim 3 is a warning about reading the two hypotheses as one. Pointwise boundedness is a condition on the values and equicontinuity a condition on the variation; a family may satisfy either without the other, and the theorem that uses both needs both.

Refuted claim: a pointwise bounded family of continuous maps between metric spaces is equicontinuous (Equicontinuity at a point, uniform equicontinuity, and pointwise boundedness of a family of maps between metric spaces).

The witness is the family of moving spikes on $I:=[0,1]$ already built in FALSE: a pointwise convergent sequence of continuous functions converges uniformly on every compact set: with $a_k:=1\mathrm{/}\iota (k+2)$ (The canonical natural $\iota (n)=n\cdot 1_F$ of a field),

$f_k(t)=\frac{t}{a_k}(0\le t\le a_k),f_k(t)=2-\frac{t}{a_k}(a_k\le t\le 2a_k),f_k(t)=0(2a_k\le t\le 1)\mathrm{.}$