Function Spaces

In standard terminology,

map = mapping = continuous function.

Given spaces X and Y, denote by Y^{X} the set of all maps X → Y. (If X is a discrete space with m points and Y has n points, there are n^{m} maps.) We wish to topologize this set in some reasonable manner. When this is done, it is commonly known as a function space (although the term mapping space is often used too, and is more accurate).

Example If X is compact and Y is a metric space, it is reasonable to define the distance between two maps g: X → Y and h: X → Y as

D(g, h) = max_{x∈X} D(g(x), h(x)).  (1)

Note that D(g(x), h(x)) is a continuous real-valued function of x, and because X is compact, the maximum exists. It is readily verified that this defines a metric on Y^{X}, known as the metric of uniform convergence.

Generally, suppose that f: W × X → Y is a map, where we impose no conditions on the spaces. For each w ∈ W, we can define a map f_{w}: X → Y by f_{w}(x) = f(w, x); then we construct a function f̂: W → Y^{X} by f̂(w) = f_{w}. This suggests the following axiom:

f̂: W → Y^{X} is continuous if and only if f: W × X → Y is continuous. (2)

An important special case is when W = Y^{X} and f̂ is the identity map; then f becomes the evaluation function e: Y^{X} × X → Y, given by e(g, x) = g(x). It is easy to recover f from f̂ as

f(w, x) = e(f̂(w), x).  (3)

Uniqueness of such a topology on Y^{X} is not difficult to see. (If we have two candidates, we take W = Y^{X} with either topology and f̂ as the identity map. Then (2) shows that e is continous with either topology on Y^{X}, also that f̂ is continuous even when the two copies of Y^{X} have different topologies, in either order. It follows that the two candidate topologies on Y^{X} coincide.)

The following result is far beyond the scope of this note.

Theorem 4 In general, no topology on Y^{X} satisfies axiom (2) for all W.

Moreover, even when such a topology exists, it may not be the most appropriate one for a specific context. Still, we can look for situations where there is a topology that satisfies the axiom.

Theorem 5 If X is compact and Y is a metric space, the topology on Y^{X} defined by the metric (1) does satisfy axiom (2).

Proof First, assume that f: W × X → Y is continuous. We have to show that f̂ is continuous. Take w_{0} ∈ W and a disk neighborhood N(f_{w_{0}}, p) of f_{w_{0}} = f̂(w_{0}) in Y^{X}; we seek a neighborhood M of w_{0} in W such that f̂(M) ⊂ N(f_{w_{0}}, p). Let P be the open set consisting of all points (w, x) ∈ W × X such that D(f(w_{0}, x), f(w, x)) < p; evidently, w_{0} × X ⊂ P. By the tube lemma, since X is assumed compact, there is a neighborhood M of w_{0} such that M × X ⊂ P. Then for any w ∈ M,

D(f_{w_{0}}(x), f_{w}(x)) = D(f(w_{0}, x), f(w, x)) < p for all x ∈ X,

as required.

For the converse, we see from (3) that we need only show that e is continuous. Take a point (g, x_{0}) ∈ Y^{X} × X and a disk neighborhood N(y_{0}, p) of y_{0} = g(x_{0}) in Y. We need neighborhoods N(g, q) of g in Y^{X} and V of x_{0} in X such that e(N(g, q) × V) ⊂ N(y_{0}, p). We take q = p/2 and choose V such that g(V) ⊂ N(y_{0}, p/2). Then D(g(x), h(x)) < p/2 for any h ∈ N(g, q) and any x; also, D(y_{0}, g(x)) < p/2 if x ∈ V. By the triangle inequality, D(y_{0}, e(h, x)) = D(y_{0}, h(x)) < p.

In spite of the previous caution, there is a topology on Y^{X} that works much of the time in practice (and necessarily includes the above example as a special case).

Definition 6 Given a compact subspace K ⊂ X and an open set U ⊂ Y, denote by N(K, U) the set of all maps g: X → Y that satisfy g(K) ⊂ U. The compact-open topology on Y^{X} is defined as having the collection of all such subsets N(K, U) of Y^{X} as a subbasis.

Lemma 7 If f: W × X → Y is continuous and we give Y^{X} the compact-open topology, then f̂: W → Y^{X} is continuous.

Proof We have only to prove that f̂^{−1}(N(K, U)) is open. Take any w_{0} ∈ W such that f̂(w_{0}) ∈ N(K, U), i. e. w_{0} × K ⊂ f^{−1}(U). By the tube lemma, there is a neighborhood M of w_{0} in W such that M × K ⊂ f^{−1}(U), i. e. f̂(M) ⊂ N(K, U).

Lemma 8 The evaluation function e: Y^{X} × X → Y is continuous if we give Y^{X} the compact-open topology and X is locally compact.

Proof To show that e is continuous at (g, x_{0}) ∈ Y^{X} × X, take a neighborhood U of e(g, x_{0}) = g(x_{0}). We seek neighborhoods N of g and V of x_{0} such that e(N ×V) ⊂ U. Now g^{−1}(U) is a neighborhood of x_{0}. Since X is locally compact, there is a compact subspace K ⊂ g^{−1}(U) that contains a neighborhood V of x_{0}. We take N = N(K, U). Since g(K) ⊂ U, g ∈ N. For any h ∈ N, h(V) ⊂ h(K) ⊂ U, i. e. e(N ×V) ⊂ U.

Combining these two lemmas with equation (3), we have the main theorem.

Theorem 9 Let X be locally compact and Y and W any spaces. If we give Y^{X} the compact-open topology, then f: W × X → Y is continuous if and only if f̂: W → Y^{X} is continuous.