Function Spaces

Francesco Beda (1840-1900) was an Italian painter.

The topics are

• Pointwise Convergence 
• Compact Open Topology and Joint Continuity
• Uniform Convergence
• Uniform Convergence on Compacta
• Compactness and Equicontinuity
• Even Continuity
• Function Spaces

\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}

Pointwise Convergence.

Let \({\cal F}\) be a family of functions that are defined from a set \(X\) into a topological space \((Y,\tau_{Y})\). Then \({\cal F}\) is contained in the product space
\[Y^{X}=\prod_{x\in X}Y.\]
Suppose that the product space \(Y^{X}\) is endowed with the product topology. By referring to Remark~\ref{top220}, since \({\cal F}\) is a subset of \(Y^{X}\), the relativized product topology to \({\cal F}\) is called the topology of pointwise convergence for \({\cal F}\). Remark \ref{top220} says that a net \(\{f_{\alpha}\}_{\alpha\in\Lambda}\) converges to \(f\) if and only if \(\{f_{\alpha}(x)\}_{\alpha\in\Lambda}\) converges to \(f(x)\) for each \(x\) in \(X\). By referring to (\ref{top215}), a subbase for the topology of pointwise convergence is the family of all subsets of the form \(\{f:f(x)\in O\}\), where \(x\in X\) and \(O\in\tau_{Y}\).

For each \(x\in X\), we can define a function \(e_{x}\) on \({\cal F}\) by \(e_{x}(f)=f(x)\) for all \(f\in {\cal F}\), which is regarded as the projection into the \(x\)-th coordinate space. Proposition \ref{top216} says that the function \(e_{x}\) is continuous and open with respect to the topology of pointwise convergence, which is the coarsest topology for \({\cal F}\) such that each function (i.e. projection) \(e_{x}\) is continuous. A function \(g\) defined on a topological space into \({\cal F}\) is continuous with respect to the topology of pointwise convergence if and only if \(e_{x}\circ g\) is continuous for each point \(x\in X\) (Proposition \ref{top217}).

Definition. A family \({\cal F}\) of functions that are defined from a set \(X\) into a topological space \((Y,\tau_{Y})\) is said to be {\bf pointwise closed} if and only if \({\cal F}\) is a closed subset of the product space \(Y^{X}\). \(\sharp\)

Let \(A\) be a subset of \(X\). We define
\[{\cal F}(A)=\left\{f(x):x\in A\mbox{ and }f\in {\cal F}\right\}.\]
If \(x\in X\), then \({\cal F}(\{x\})\) is abbreviated to \({\cal F}(x)\). Then, we have
\[e_{x}({\cal F})=\left\{e_{x}(f):f\in {\cal F}\right\}=\left\{f(x):f\in {\cal F}\right\}={\cal F}(x).\]

Proposition. Let \({\cal F}\) be a family of functions that are defined from a set \(X\) into a topological space \((Y,\tau_{Y})\). Suppose that the following conditions are satisfied:

• ${\cal F}$ is pointwise closed in \(Y^{X}\);
• for each \(x\in X\), the set \({\cal F}(x)\) has a compact closure.

Then \({\cal F}\) is compact with respect to the topology of pointwise convergence. If we assume further that \(Y\) is a Hausdorff space, then the converse also holds true. \(\sharp\)

Let \({\cal F}\) be a family of functions that are defined from a set \(X\) into a topological space \((Y,\tau_{Y})\), and let \(A\) be a subset of \(X\). We define a function \(R:{\cal F}\rightarrow Y^{A}\) defined by \(R(f)=f|A\) for each \(f\in {\cal F}\). Let \(\tau_{A}\) consist of the inverse images of the open sets in \(Y^{A}\) under \(R\). Then \(\tau_{A}\) is the coarsest topology for \({\cal F}\) such that \(R\) is continuous. A subbase for \(\tau_{A}\) is the family of sets of the form \(\{f:f(x)\in O\}\), where \(x\in A\) and \(O\in\tau_{Y}\). We also see that a net \(\{f_{\alpha}\}_{\alpha\in\Lambda}\) in \({\cal F}\) converges to \(f\) with respect to \(\tau_{A}\) if and only if \(\{f_{\alpha}(x)\}_{\alpha\in\Lambda}\) converges to \(f(x)\) for each \(x\in A\). Therefore, we may call \(\tau_{A}\) the topology of pointwise convergence on \(A\). We say that \(A\) distinguishes members of the family \({\cal F}\) if and only if, for any \(f,g\in {\cal F}\) with \(f\neq g\), there exists \(x\in A\) such that \(f(x)\neq g(x)\). Then, the map \(R\) is one-to-one if and only if \(A\) distinguishes members of the family \({\cal F}\).

Proposition. Let \({\cal F}\) be a family of functions that are defined from a set \(X\) into a Hausdorff topological space \((Y,\tau_{Y})\), and let \(A\) be a subset of \(X\).

(i) The family \({\cal F}\) endowed with the topology \(\tau_{A}\) of pointwise convergence on \(A\) is a Hausdorff space if and only if \(A\) distinguishes members of \({\cal F}\).

(ii) Suppose that \({\cal F}\) is compact with respect to the topology \(\tau_{X}\) of pointwise convergence on \(X\), and that \(A\) distinguishes members of \({\cal F}\). Then \(\tau_{X}\) and \(\tau_{A}\) are identical.

Let \({\cal F}\) be a family of functions from a set \(X\) into a uniform space \((Y,{\cal V})\). Then \({\cal F}\) is a subset of the product \(\prod_{x\in X}Y\), and the relativized product uniformity is called the  uniformity of pointwise convergence. This is sometimes abbreviated as the \(P\)-uniformity. Let \(A\) be a subset of \(X\). The uniformity of pointwise convergence on \(A\), or simply the \(P_{A}\)-uniformity, is defined to be the coarsest uniformity which makes that the restriction map \(R\) from \({\cal F}\) into the family of all functions defined on \(A\) into \(Y\) is uniformly continuous.

Proposition. Let \({\cal F}\) be a family of functions from a set \(X\) into a uniform space \((Y,{\cal V})\), and let \(A\) be a subset of \(X\).

(i) The family of all sets of the form \(\{(f,g):(f(x),g(x))\in V\}\) for \(V\in {\cal V}\) and \(x\in A\) is a subbase for the \(P_{A}\)-uniformity.

(ii) The topology of the \(P_{A}\)-uniformity is the topology of pointwise convergence on \(A\).

(iii) A net \(\{f_{\alpha}\}_{\alpha\in\Lambda}\) is a Cauchy net if and only if \(\{f_{\alpha}(x)\}_{\alpha\in\Lambda}\) is a Cauchy net for each \(x\in A\).

(iv) If \((Y,{\cal V})\) is complete and \(R({\cal F})\) is closed in \(Y^{A}\) with respect to the pointwise convergence on \(A\), then \({\cal F}\) is complete with respect to the \(P_{A}\)-uniformity. \(\sharp\)

\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}

Compact Open Topology and Joint Continuity.

Given a topology for a family \({\cal F}\) of functions defined on a topological space \(X\) into a topological space \(Y\), we may ask whether \(f(x)\) is continuous simultaneously in \(f\) and \(x\). Formally the question is: for which topologies for \({\cal F}\) is the map \({\cal F}\times X\) which carries \((f,x)\) onto \(f(x)\) continuous, where \({\cal F}\times X\) is endowed with the product topology. For each subset \(K\) of \(X\) and each subset \(U\) of \(Y\), we define \(W(K,U)\) to be the set of all members of \({\cal F}\) which carry \(K\) into \(U\); that is,
\[W(K,U)=\{f:f(K)\subseteq U\}.\]
The family of all sets of the form \(W(K,U)\) for a compact subset \(K\) of \(X\) and an open subset \(U\) of \(Y\) is a subbase for the {\bf compact open topology} for \({\cal F}\). The family of finite intersections of sets of the form \(W(K,U)\) is a base for the compact open topology; that is, each member of this base is the form
\[\bigcap_{i=1}^{n}W\left (K_{i},U_{i}\right ),\]
where each \(K_{i}\) is a compact subset of \(X\) and each \(U_{i}\) is an open subset of \(Y\).

Proposition. The compact open topology \(\tau_{C}\) contains the topology \(\tau_{P}\) of pointwise convergence.

(i) If the range space \(Y\) is Hausdorff, then \(({\cal F},\tau_{C})\) is Hasudorff. and is regular if \(Y\) is regular and the members of \({\cal F}\) are continuous.

(ii) If \(Y\) is regular and the members of \({\cal F}\) are continuous, then \(({\cal F},\tau_{C})\) is regular . \(\sharp\)

Let \(Q\) be the map from \({\cal F}\times X\) into \(Y\) defined by \((f,x)\mapsto f(x)\). A topology for \({\cal F}\) can induce the product topology \({\cal F}\times X\). We may ask whether \(Q\) is continuous with respect to this product topology \({\cal F}\times X\).

Definition. Let \({\cal F}\) be a family of functions from a topological space \(X\) into a topological space \(Y\).

• A topology for \({\cal F}\) is said to be jointly continuous when the map \(Q:{\cal F}\times X\rightarrow Y\) defined by \((f,x)\mapsto f(x)\) is continuous.
• Let \(A\) be a subset of \(X\). A topology for \({\cal F}\) is said to be jointly continuous on a set \(A\) when the restriction \(Q|_{{\cal F}\times A}\) is continuous on \({\cal F}\times A\).
• A topology for \({\cal F}\) is jointly continuous on compacta when it is jointly continuous on each compact subset of \(X\). \(\sharp\)

We have the following observations.

• The topology of pointwise convergence for \({\cal F}\) is usually not jointly continuous.
• If a topology for \({\cal F}\) is jointly continuous, then each finer topology is also jointly continuous. Therefore the natural problem is to find the coarsest jointly continuous topology when it exists.
• Assuming that \({\cal F}\) is a family of continuous functions. The discrete topology for \({\cal F}\) is jointly continuous. For if \(U\) is an open subset of \(Y\), then
  \[Q^{-1}(U)=\{(f,x):f(x)\in U\}=\bigcup_{f\in {\cal F}}\{f\}\times f^{-1}(U),\]
  which is the union of open sets.
• Suppose that \({\cal F}\) owns a topology that is jointly continuous on compacta. Then each member \(f\in {\cal F}\) is continuous on each compact set \(K\); that is, the restriction \(f|_{K}\) is continuous.

Proposition. Each topology which is jointly continuous on compacta is finer than the compact open topology \(\tau_{C}\). If \(X\) is regular or Hausdorff and each member of \({\cal F}\) is continuous on every compact subset of \(X\), then \(\tau_{C}\) is jointly continuous on compacta. \(\sharp\)

We have the following observations.

• If \(X\) is locally compact, then a topology is jointly continuous on compacta if and only if it is jointly continuous. Hence, if \(X\) is locally compact regular space, then the compact open topology for a family of continuous functions is the smallest jointly continuous topology.
• If a topology \(\tau\) for a family \({\cal F}\) is jointly continuous on compacta, then \(\tau_{P}\subseteq\tau_{C}\subseteq\tau\), where \(\tau_{C}\) is the compact open topology and \(\tau_{P}\) is the topology of pointwise convergence.
• If the topological space \(({\cal F},\tau )\) is compact and the range space \(Y\) is Hausdorff, then \(({\cal F},\tau_{P})\) is Hausdorff and \(\tau =\tau_{C}=\tau_{P}\).

Proposition. Let \(X\) be a topological space which is either regular or Hausdorff, let \(Y\) be a Hausdorff space, let \({\cal C}\) be the family of all functions from \(X\) into \(Y\) which are continuous on each compact subset of \(X\), and let \(\tau_{C}\) and \(\tau_{P}\) be respectively the compact open topology and the topology of pointwise convergence. Then a subfamily \({\cal F}\) of \({\cal C}\) is \(\tau_{C}\)-compact if and only if

• ${\cal F}$ is \(\tau_{C}\)-closed in \({\cal C}\);
• ${\cal F}(x)$ has a compact closure for each member \(x\) of \(X\);
• the topology \(\tau_{P}\) for the \(\tau_{P}\)-closure of \({\cal F}\) in \(Y^{X}\) is jointly continuous on compacta. \(\sharp\)

\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}

Uniform Convergence.

Let \({\cal F}\) be a family of functions from a set \(X\) into a uniform space \((Y,{\cal V})\). For \(V\in {\cal V}\), we define
\[{\cal W}(V)=\left\{(f,g):(f(x),g(x))\in V\mbox{ for each }x\in X\right\}.\]
Then \({\cal W}(V)(f)\) is the set of all \(g\) such that \(g(x)\in V(f(x))\) for every \(x\) in \(X\). For \(U,V\in {\cal V}\), it is easy to see that
\[{\cal W}(V^{-1})=({\cal W}(V))^{-1},\quad{\cal W}(U\cap V)={\cal W}(U)\cap {\cal W}(V)\mbox{ and }{\cal W}(U)\circ {\cal W}(V)\subseteq {\cal W}(U\circ V).\]
Therefore the family of all sets \({\cal W}(V)\) for \(V\in {\cal V}\) is a base for a uniformity \({\cal U}\) for \({\cal F}\) by Proposition~\ref{top222}. The family \({\cal U}\) is said to be the uniformity of uniform convergence, or simply the u.c. uniformity. The topology for \({\cal U}\) is said to be the topology of uniform convergence, or the u.c. topology.

Proposition. We have the following properties.

(i) The uniformity of uniform convergence is finer than the uniformity of pointwise convergence.

(ii) The uniform convergence implies pointwise convergence.

Proof. To prove part (i), for \(y\in X\) and \(V\in {\cal V}\), we have
\[\left\{(f,g):(f(x),g(x))\in V\mbox{ for all }x\in X\right\}\subseteq\left\{(f,g):(f(y),g(y))\in V\right\},\]
which says that each member of the defining base for \({\cal U}\) is a subset of a member of the defining subbase for the pointwise uniformity. It follows that the u.c. topology is finer than the pointwise.

To prove part (ii), for a net \(\{f_{\alpha}\}_{\alpha\in\Lambda}\) in \({\cal F}\) converges to \(f\) with respect to the u.c. topology if and only if the net is eventually in \(W(V)(f)\) for each \(V\in {\cal V}\), and this is true if and only if there exists \(\beta\in\Lambda\) satisfying \(f_{\alpha}(x)\in V(f(x))\) for all \(\in X\) when \(\alpha\succ\beta\). \(\blacksquare\)

Proposition. Let \({\cal F}\) be the family of all functions from a set \(X\) into a uniform space \((Y,{\cal V})\), and let \({\cal U}\) be the uniformity of uniform convergence for \({\cal F}\).

(i) The uniformity \({\cal U}\) is generated by the family of all pseudo-metrics of the form
\[d^{*}(f,g)=\sup_{x\in X}d(f(x),g(x)),\]
where \(d\) is a bounded member of the gage of \((Y,{\cal V})\).

(ii) A net \(\{f_{\alpha}\}_{\alpha\in\Lambda}\) in \({\cal F}\) converges uniformly to \(f\) if and only if it is a Cauchy net relative to \({\cal U}\) and \(\{f_{\alpha}(x)\}_{\alpha\in\Lambda}\) converges to \(f(x)\) for each \(x\) in \(X\).

(iii) If \((Y,{\cal V})\) is complete so is the uniform space \(({\cal F},{\cal U})\). \(\sharp\)

Proposition. Let \({\cal F}\) be the family of all continuous functions on a topological space \(X\) to a uniform space \((Y,{\cal V})\), and let \({\cal U}\) be the uniformity of uniform convergence. Then

(i) The family \({\cal F}\) is closed in the space of all functions on \(X\) to \(Y\), and conequently \(({\cal F},{\cal U})\) is complete if \((Y,{\cal V})\) is complete.

(ii) The topology of uniform convergence is jointly continuous. \(\sharp\)

\begin{equation}{\labe{top242}}\mbox{}\end{equation}

Definition \ref{top242}. Let \({\cal F}\) be a family of functions from a set \(X\) into a uniform space \((Y,{\cal V})\), and let \({\cal A}\) is a family of subsets of \(X\). The uniformity of uniform convergence on \({\cal A}\), denoted by \({\cal U}_{\cal A}\), owns a subbase consisting of all sets of the form
\[\left\{(f,g):(f(x),g(x))\in V\mbox{ for all }x\in A\right\}\]
for \(V\in {\cal V}\) and \(A\in {\cal A}\). \(\sharp\)

The uniformity \({\cal U}_{\cal A}\) may be described in another way as follows. For each \(A\in {\cal A}\), let \(R_{A}\) be the map which carries \(f\) into the restriction of \(f\) to \(A\); that is, \(R_{A}(f)=f|_{A}\). Then \(R_{A}\) carries \({\cal F}\) into a family of functions from \(A\) into \(Y\). This family may be assigned the uniformity of uniform convergence, and the uniformity \({\cal U}_{\cal A}\) may be described as
the coarsest which makes each \(R_{A}\) uniformly continuous.

\begin{equation}{\labe{top243}}\mbox{}\end{equation}

Proposition \ref{top243}. Let \(X\) be a topological space, let \((Y,{\cal V})\) be a uniform space, let \({\cal A}\) be a family of subsets of \(X\) which covers \(X\), let \({\cal G}\) be the family of all functions from \(X\) into \(Y\), and let \({\cal F}\) be the family of all functions which are continuous on each member of \({\cal A}\).

(i) The uniformity \({\cal U}_{\cal A}\) of uniform convergence on \({\cal A}\) is finer than the uniformity of pointwise convergence and coarser than that of uniform convergence on \(X\).

(ii) A net \(\{f_{\alpha}\}_{\alpha\in\Lambda}\) converges to \(f\) with respect to the topology for \({\cal U}_{\cal A}\) if and only if it is a Cauchy net with respect to \({\cal U}_{\cal A}\) and converges to \(f\) pointwise.

(iii) If \((Y,{\cal V})\) is complete, then \({\cal G}\) is complete with respect to \({\cal U}_{\cal A}\).

(iv) The family \({\cal F}\) is a closed subset of \({\cal G}\) with respect to the topology for \({\cal U}_{\cal A}\), and hence if \((Y,{\cal V})\) is complete so is \(({\cal F},{\cal U}_{\cal A})\).

(v) The topology of \({\cal U}_{\cal A}\) for \({\cal F}\) is jointly continuous on each member of \({\cal A}\). \(\sharp\)

\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}

Uniform Convergence on Compacta.

Suppose that \({\cal F}\) is a family of continuous functions on a topological space \(X\) to a uniform space \((Y,{\cal V})\). The uniformity of uniform convergence on compacta is the uniformity \({\cal U}_{\cal C}\), where \({\cal C}\) is the family of all compact subsets of \(X\). The topology of \({\cal U}_{\cal C}\) is sometimes called the topology of compact convergence. It will be proved that this topology is identical with the compact open topology which is constructed from the topology of \(X\) and the topology of the uniformity \({\cal V}\). Thus the uniformity \({\cal U}_{\cal C}\) depends on the uniformity \({\cal V}\) for \(Y\), but the topology of \({\cal U}_{\cal C}\) depends only on the topology of \({\cal V}\).

Proposition. Let \({\cal F}\) be a family of continuous functions on a topological space \(X\) to a uniform space \((Y,{\cal V})\). Then, the topology of uniform convergence on compacta is the compact open topology. \(\sharp\)

If the uniform space \((Y,{\cal V})\) is complete and \({\cal A}\) is a family of subsets of the topological space \(X\) then the family of all functions on \(X\) to \(Y\) which are continuous on each member of \({\cal A}\) is \({\cal U}_{\cal A}\)-complete, according to Proposition \ref{top243}. In order that the family of all continuous functions be complete it is then sufficient that \({\cal A}\) satisfy the condition: a function is continuous whenever it is continuous on each member of \({\cal A}\). If \(f\) is a function on \(X\) to \(Y\) and \(B\) is a subset of \(Y\), then this condition would be implied by: if \(A\cap f^{-1}(B)\) is closed for each
member \(A\) of \({\cal A}\), then \(f^{-1}(B)\) is closed. In particular, the space of all continuous functions on \(X\) to \(Y\) is complete relative to uniform convergence on compacta if \(X\) satisfies the condition: if a subset \(A\) of \(X\) intersects each closed compact set in a closed set, then \(A\) is closed. Such a topological space is called a \(k\)-space. It is clear that the family \({\cal C}\) of closed compact sets determines the topology of a \(k\)-space entirely, for \(A\) is closed if and only if \(A\cap C\in {\cal C}\) for each \(C\) in \({\cal C}\). By complementation it follows that a subset \(U\) of a \(k\)-space is open if and only if \(U\cap C\) is open in \(C\) for each closed compact set \(C\).

Proposition. The family of all continuous functions on a \(k\)-space to a complete uniform space is complete relative to the uniform convergence on compacta. \(\sharp\)

Proposition. If \(X\) is a Hausdorff space which is either locally compact or satisfies the first axiom of countability, then \(X\) is a \(k\)-space. \(\sharp\)

\begin{equation}{\label{e}}\tag{E}\mbox{}\end{equation}

Compactness and Equicontinuity.

Let \({\cal F}\) be a family of maps of a topological space \(X\) into a uniform space \((Y,{\cal V})\). The family \({\cal F}\) is equicontinuous at a point $x$ when for each member \(V\) of \({\cal V}\) there is a neighborhood \(U\) of \(x\) such that \(f(U)\subseteq V(f(x))\) for every member \(f\) of \({\cal F}\). Equivalently, \({\cal F}\) is equicontinuous at \(x\) if and only if \(\bigcap\{f^{-1}(V(f(x))):f\in {\cal F}\}\) is a neighborhood of \(x\) for each \(V\) in \({\cal V}\). Roughly speaking, \({\cal F}\) is equicontinuous at \(x\) if and only if there is a neighborhood of \(x\) whose image under every member of \({\cal F}\) is small.

Proposition. If \({\cal F}\) is equicontinuous at \(x\), then the closure of \({\cal F}\) relative to the topology \({\cal P}\) of pointwise convergence is also equicontinuous at \(x\). \(\sharp\)

A family \({\cal F}\) of functions is equicontinuous when it is equicontinuous at every point. In view of the preceding proposition the closure of an equicontinuous family relative to the topology of pointwise convergence is also equicontinuous; in particular the members of the closure are continuous functions.

Proposition. If \({\cal F}\) is an equicontinuous family, then the topology of pointwise convergence is jointly continuous and hence concides with the topology of uniform convergence on compacta. \(\sharp\)

The preceding proposition implies that an equicontinuous family of functions is compact relative to the topology of uniform convergence on compacta if it is compact relative to the pointwise topology \({\cal P}\), and the Tychonoff product theorem gives the sufficient conditions for \({\cal P}\)-compactness. In this way, equicontinuous together with certain other conditions implies compactness of a family of functions.

Proposition. If a family \({\cal F}\) of functions on a topological space \(X\) to a uniform space \((Y,{\cal V})\) is compact relative to a jointly continuous topology, then \({\cal F}\) is equicontinuous. \(\sharp\)

Theorem (Ascoli Theorem). Let \(C\) be the family of all continuous functions on a regular locally compact topological space to a Hausdorff uniform space, and let \(C\) have the topology of uniform convergence on compacta. Then, the subfamily \({\cal F}\) of \(C\) is compact if and only if

• ${\cal F}$ is closed in \(C\);
• ${\cal F}(x)$ has a compact closure for each member \(x\) of \(X\);
• the family \({\cal F}\) is equicontinuous. \(\sharp\)

A form of the Ascoli theorem is true for families of functions on a \(k\)-space (a space such that a set is closed whenever its intersection with every closed compact set is closed). A family \({\cal F}\) of functions is equicontinuous on a set \(A\) if and only if the family of all restrictions of members of \({\cal F}\) to \(A\) is an equicontinuous family. A family which is equicontinuous at every point of \(A\) is equicontinuous
on \(A\), but the converse is false. However, a family which is equicontinuous on \(A\) is equicontinuous at each point of the interior of \(A\).

Theorem (Ascoli Theorem). Let \(C\) be the family of all continuous functions on a \(k\)-space \(X\) which is either Hausdorff or regular to a Hausdorff uniform space \(Y\), and let \(C\) have the topology of uniform convergence on compacta. Then a subfamily \({\cal F}\) of \(C\) is compact if and only if

• ${\cal F}$ is closed in \(C\);
• the closure of \({\cal F}(x)\) is compact for each \(x\) in \(X\);
• ${\cal F}$ is equicontinuous on every compact subset of \(X\). \(\sharp\)

\begin{equation}{\label{f}}\tag{F}\mbox{}\end{equation}

Even Continuity.

Let \({\cal F}\) be a family of functions, each on a topological space \(X\) to a topological space \(Y\). The concept of even continuity can be described intuitively by the statement: for each \(x\) in \(X\), \(y\) in \(Y\), and \(f\) in \({\cal F}\), if \(f(x)\) is near \(y\), then \(f\) maps points near \(x\) into points near \(y\). Explicitly, the family \({\cal F}\) is evenly continuous when for each \(x\) in \(X\), \(y\) in \(Y\), and each neighborhood \(U\) of \(y\) there is a neighborhood \(V\) of \(x\) and a neighborhood \(W\) of \(y\) such that \(f(V)\subseteq U\) whenever \(f(x)\in W\). The close connection between this definition and joint continuity may be empahasized by the re-statement: \({\cal F}\) is evenly continuous if and only if for each \(x\) in \(X\) and each \(y\) in \(Y\) and for each neighborhood \(U\) of \(y\) there are neighborhoods \(V\) of \(x\) and \(W\) of \(y\) such that \(\{f:f\in {\cal F}\mbox{ and }f(x)\in W\}\times V\) is carried into \(U\) by the natural map.

Proposition. Let \({\cal F}\) be an evenly continuous family of functions on a topological space \(X\) to a regular space \(Y\) and let \({\cal P}\) be the topology of pointwise convergence. Then, the \({\cal P}\)-closure \(cl({\cal F})\) of \({\cal F}\) is evenly continuous and \({\cal P}\) is jointly continuous on \(cl({\cal F})\). \(\sharp\)

Proposition. If a family \({\cal F}\) of continuous functions on a topological space \(X\) to a regular Hausdorff space \(Y\) is compact relative to a jointly continuous topology, then \({\cal F}\) is evenly continuous. \(\sharp\)

\begin{equation}{\label{kelp56}}\tag{1}\mbox{}\end{equation}

Theorem \ref{kelp56} (Ascoli Theorem). Let \(C\) be the family of all continuous functions on a regular locally compact space \(X\) to a regular Hausdorff space \(Y\), and let \(C\) have the compact open topology. Then a subset \({\cal F}\) of \(C\) is compact if and only if

• ${\cal F}$ is closed in \(C\);
• the closure \({\cal F}(x)\) is compact for each \(x\) in \(X\);
• ${\cal F}$ is evenly continuous. \(\sharp\)

A family \({\cal F}\) of functions if evenly continuous on a set \(A\) when the family of all restrictions of members of \({\cal F}\) to \(A\) is evenly continuous. With this definition the Ascoli Theorem~\ref{kelp56} can be proved for \(k\)-space \(X\) if condition (iii) is replaced by “${\cal F}$ is evenly continuous on each compact subset of \(X\)”.

Proposition. An equicontinuous family of functions on a topological space to a uniform space is evenly continuous. \(\sharp\)

Proposition. If \({\cal F}\) is evenly continuous family of functions on a topological space \(X\) to a uniform space \(Y\), and \(x\) is a point of \(X\) such that \({\cal F}(x)\) has a compact closure, then \({\cal F}\) is equicontinuous at \(x\). \(\sharp\)

\begin{equation}{\label{g}}\tag{G}\mbox{}\end{equation}

Function Spaces.

Let \(S\) be any set, and let \(E\) be a topological vector space. The set \(F(S,E)\) of all functions on \(S\) to \(E\), with addition and scalar multiplication defined pointwise, is a vector space. For each subset \(A\) of \(S\), and for each neighborhood \(U\) of \(0\) in \(E\), let \(N(A,U)\) be the family of all members \(f\) of \(F(S,E)\) with the property that \(f(A)\subseteq U\). Observe that \(N(A,U)\) is circled if \(U\) is circled. A subset \(G\) of \(F(S,E)\) is open relative to the topology of uniform convergence on \(A\) when for each \(f\in G\) there is a neighborhood \(U\) of \(0\) in \(E\) satisfying \(f+N(A,U)\subseteq G\). It is easy to verify that this definition gives a topology for \(F(S,E)\) such that the family of sets of the form \(f+N(A,U)\) is a base for the neighborhood system of \(f\). However, this topology need not be a vector topology for \(F(S,E)\) since the neighborhoods of \(0\) need not be radial at \(0\). In fact, if \(U\) is circled, then \(N(A,U)\) is radial at \(0\) if and only if for each \(f\) there is a scalar \(t\) such that \(f\in tN(A,U)\), that is, \(f(A)\subseteq tU\). It follows that if \(G\) is a subspace of \(F(S,E)\) such that \(G\) with the relativized topology of uniform convergence on \(A\) is a topological vector space, then \(f(A)\) is a bounded subset of \(E\) for each member \(f\) of \(G\). We shall say that a function \(f\) is bounded on a set \(A\) when \(f(A)\) is a bounded subset of \(E\).

A net \(\{f_{\alpha}:\alpha\in A\}\) in \(F(S,E)\) converges to \(f\) uniformly of a set \(A\) when the net converges to \(f\) relative to the topology of uniform convergence on \(A\). Clearly this is the case if and only if \(f_{\alpha}-f\) is eventually in \(N(A,U)\) for each neighborhood \(U\) of \(0\) in \(E\). If \({\cal A}\) is an arbitrary nonempty family of subsets of \(S\), then the topology \(\tau_{\cal A}\) of uniform convergence on members of \({\cal A}\) is defined to be the weakest topology which is stronger than that of uniform convergence on \(A\) for every \(A\in {\cal A}\). Convergence relative to \(\tau_{\cal A}\) can be described in a somewhat less esoteric fashion: a net converges to \(f\) relative to \(\tau_{\cal A}\) if and only if the net converges to \(f\) uniformly on each member of \({\cal A}\). A base for the neighborhood system of \(0\) relative to \(\tau_{\cal A}\) is the family of all finite intersections of sets of the form \(N(A,U)\), where \(A\) is a member of \({\cal A}\) and \(U\) is a neighborhood of \(0\) in \(E\). Since \(N(A,U)\cap N(B,U)=N(A\cup B,U)\), the topology of uniform convergence on members of \({\cal A}\) is identical with the topology of uniform convergence on members of \({\cal B}\), where \({\cal B}\) is the family of all finite unions of members of \({\cal A}\). If the union of two members of \({\cal A}\) is always contained in some member of \({\cal A}\) (that is, if \({\cal A}\) is directed by \(\supseteq\)), then the intersection of two sets of the form \(N(A,U)\) contains a set of the same form, and consequently the family of such sets is a base for the neighborhood system of \(0\). The pointwise topology, or the topology of pointwise convergence, is the topology \(\tau_{\cal A}\) where \({\cal A}\) is the family of all sets \(\{t\}\) for all \(t\in S\). The topology of uniform convergence on \(S\) is often called the uniform topology. It is identical with the topology \(\tau_{\cal A}\) where \({\cal A}=\{S\}\).

Proposition. The family \(B(S,E)\) of all functions from \(S\) to \(E\) which are bounded on \(S\) is a subspace of \(F(S,E)\) which is closed relative to the uniform topology, and the uniform topology is a vector topology for \(B(S,E)\). \(\sharp\)

It follows from the foregoing proposition that if a net of functions, each of which is bounded on a subset \(A\) of \(S\), converges uniformly on \(A\) to a function \(f\), then \(f\) is also bounded on \(A\) (using the closeness).

Corollary. The class \(B_{\cal A}(S,E)\) of all functions \(f\) on \(S\) to a linear topological space \(E\) such that \(f\) is bounded on each member of \({\cal A}\) is closed in \(F(S,E)\) relative to the topology \(\tau_{\cal A}\) of uniform convergence on members of \({\cal A}\). Moreover, \(\tau_{\cal A}\) is a vector topology for \(B_{\cal A}(S,E)\). \(\sharp\)

A subset \(B\) of a topological vector space \(E\) is totally bounded when for each neighborhood \(U\) of \(0\) there exists a finite set \(N\) satisfying \(B\subseteq N+U\). It is clear that each compact set is totally bounded, and each totally bounded set is bounded. If \(A\) is totally bounded, then so are \(\mbox{cl}(A)\), the smallest closed circled set containing \(A\), and the product \(aA\) for each scalar \(a\). A function \(f\) on \(S\) to a linear topological space \(E\) is totally bounded on a subset \(A\) of \(S\) when \(f(A)\) is totally bounded; that is, a function \(f\) is totally bounded on \(S\) if and only if the range of \(f\) is a totally bounded subset of \(E\).

Proposition. The class \(T(S,E)\) of all functions from \(S\) to \(E\), each of which has a totally bounded range, is a subspace of \(B(S,E)\) which is closed in the uniform topology. \(\sharp\)

Corollary. The class \(T_{\cal A}(S,E)\) of all functions on \(S\) to \(E\), each of which is totally bounded on each member of \({\cal A}\), is closed relative to \(\tau_{\cal A}\) in the space \(B_{\cal A}(S,E)\) of all functions bounded on members of \({\cal A}\). \(\sharp\)

Lemma. If \(\{f_{\alpha}:\alpha\in D\}\) is a net of functions on \(S\) to \(E\) such that \(\{f_{\alpha}(t):\alpha\in D\}\) converges to \(f(t)\) for each \(t\) in a subset \(A\) of \(S\), and if \(\{f_{\alpha}:\alpha\in D\}\) is a Cauchy net relative to the topology of uniform convergence on \(A\), then \(\{f_{\alpha}:\alpha\in D\}\) converges to \(f\) uniformly on \(A\). \(\sharp\)

If \(\{f_{\alpha}:\alpha\in D\}\) is a net which is a Cauchy net relative to the uniform topology, then \(\{f_{\alpha}(t):\alpha\in D\}\) is a Cauchy net in \(E\) for each \(t\in S\), and if \(E\) is complete, it is then possible to choose a limit \(f(t)\) of \(\{f_{\alpha}(t):\alpha\in D\}\) for each \(t\). Then \(\{f_{\alpha}:\alpha\in D\}\) converges to \(f\) uniformly by the preceding lemma. Of course, the same argument applies to sequences, if \(E\) is sequentially complete.

Proposition. If \(E\) is complete or sequentially complete, then so is the space \(B(S,E)\) of bounded functions on \(S\) to \(E\) with the uniform topology. \(\sharp\)

A particular consequence of this proposition is that \(T(S,E)\) is complete or sequentially complete if \(E\) is complete or sequentially complete, since \(T(S,E)\) is closed in \(B(S,E)\). If \(\{f_{\alpha}(t):\alpha\in D\}\) converges to \(f(t)\) for each \(t\) in \(\bigcup_{A\in {\cal A}}A\), and if \(\{f_{\alpha}:\alpha\in D\}\) is a Cauchy net relative to \(\tau_{\cal A}\), then this net converges to \(f\) uniformly on members of \({\cal A}\). In view of this fact it is easy to demonstrate the following proposition.

\begin{equation}{\label{top244}}\tag{2}\mbox{}\end{equation}

Proposition \ref{top244}. If \(E\) is complete or sequentially complete, then \(B_{\cal A}(S,E)\) has the same property relative to the topology \(\tau_{\cal A}\). \(\sharp\)

If \(t\) belongs to a member \(A\) of \({\cal A}\), then the function which carries \(f\) in \(B_{\cal A}(S,E)\) into \(f(t)\) is a continuous linear on \(B_{\cal A}(S,E)\), with the topology \(\tau_{\cal A}\), into \(E\). (That is, uniform convergence on \(A\) implies convergence at each point of \(A\)). Consequently, if \(G\) is a subset of \(B_{\cal A}(S,E)\) which is bounded (resp. totally bounded) relative to \(\tau_{\cal A}\), then the set of all \(f(t)\) for \(f\in G\) is also bounded (resp. totally bounded). It follows that if \(\{f_{\alpha}:\alpha\in D\}\) is a Cauchy net in a bounded set \(G\), and if \(E\) has the property that closed bounded sets are complete, then \(\{f_{\alpha}:\alpha\in D\}\) converges to a limit \(f(t)\) for each \(t\) in \(\bigcup_{A\in {\cal A}}A\). The argument demonstrates the following proposition.

Proposition. If \(E\) has the property that each closed bounded (resp. totally bounded) set is complete, then \(B_{\cal A}(S,E)\), with the topology \(\tau_{\cal A}\), has the same property. \(\sharp\)

Of course \(T_{\cal A}(S,E)\), being a closed subspace of \(B_{\cal A}(S,E)\), shares the property stated in the foregoing proposition. If \(S\) has a topology, it is possible to consider continuous functions from \(S\) to \(E\). Although the topology of uniform convergence is independent of any topology for \(S\), the class of continuous functions on \(S\) has special properties relative to uniform convergence (Kelley and Namioka \cite [p.72]{kel61}).

Proposition. The class \(C(S,E)\) of continuous functions from a topological space \(S\) to a topological vector space \(E\) is a closed linear subspace of \(F(S,E)\) relative to the uniform topology. \(\sharp\)

Corollary. the class of bounded continuous functions on \(S\) to \(E\) is a closed linear subspace of \(B(S,E)\) with respect to the uniform topology. \(\sharp\)

A function \(f\) is continuous on a set \(A\) when the function \(f\), restricted to \(A\), is continuous with respect to the relativized topology. (It is reminded that a function may be continuous on \(A\) but fail to be continuous at any point of \(A\).) The foregoing proposition implies that if a net of functions continuous on \(A\) converges uniformly on \(A\) to a function \(f\), then \(f\) is continuous on \(A\).

Proposition. The family \(C_{\cal A}(S,E)\) of all functions which are continuous on each member of a family \({\cal A}\) of subsets of a topological space \(S\) is closed relative to \(\tau_{\cal A}\) in the space \(F(S,E)\) of all functions on \(S\) to \(E\). \(\sharp\)

A consequence of the preceding proposition is the fact that the family of all functions which are bounded and continuous on each member of \({\cal A}\) is closed relative to \({\cal A}\) in \(B_{\cal A}(S,E)\), and is complete if \(E\) is complete. A family \(M\) of functions on \(S\) to \(E\) is equicontinuous at a point $s$ of \(S\) when for each neighborhood \(U\) of \(0\) in \(E\) there exists a neighborhood \(V\) of \(s\) in \(S\) such that \(f(t)-f(s)\in U\) for all \(t\in V\) and for all \(f\in M\); that is, \(f(V)\subseteq f(s)+U\) for all \(f\in M\). The family \(M\) of functions is equicontinuous when it is equicontinuous at each point of \(S\). If the functions in a net in \(F(S,E)\) form an equicontinuous family, the net is called an equicontinuous net.

Proposition. the closure relative to the pointwise topology of a family \(M\) which is equicontinuous at a point is equicontinuous at \(t\). \(\sharp\)

The foregoing proposition implies that the pointwise closure of an equicontinuous family of functions consists of continuous functions.

Proposition. If \(M\) is an equicontinuous family of functions on \(S\) to a linear topological space \(E\), then the topology for \(M\) of pointwise convergence on a subset \(A\) of \(S\) is identical with the topology of pointwise convergence on the closure \(\mbox{cl}(A)\) of \(A\). \(\sharp\)

Proposition. Let \(M\) be an equicontinuous family of functions on \(S\) to \(E\), and let \(M\) have the topology of pointwise convergence on \(S\). Then \(f(s)\) is jointly continuous in \(f\) and in \(s\), in the sense that the function of \(M\times S\) into \(E\) defined by \((f,s)\mapsto f(s)\) is continuous relative to the product topology. \(\sharp\)

When \(E\) and \(F\) are vector spaces, the vector space of all linear functions on \(E\) into \(F\) is denoted by \(L(E,F)\).

Theorem. Let \(E\) and \(F\) be topological vector spaces, let \(F\) be complete, and let \({\cal A}\) be a family of bounded subsets of \(E\). Then with the topology \(\tau_{\cal A}\), the space \(L(E,F)\cap B_{\cal A}(E,F)\) \((\)the family of linear functions on \(E\) to \(F\) which are bounded on each member of \({\cal A})\) is a complete topological vector space. \(\sharp\)

The notion of an equicontinuous family \(M\) is particular simple in case the functions are linear. In this case, the definition can be phrased: \(M\) is equicontinuous at a point \(s\) of \(E\) if and only if for each neighborhood \(U\) of \(0\) in \(F\) there is a neighborhood \(V\) of \(0\) in \(E\) such that, if \(t\in s+V\), then \(f(t)\in f(s)+U\) for all \(f\in M\). Rephrased, for \(U\) there is \(V\) such that \(f(r)\in U\) whenever \(r\in V\) and \(f\in M\).

\begin{equation}{\label{kel61t2816}}\tag{3}\mbox{}\end{equation}

Theorem \ref{kel61t2816}. Let \(M\) be a family of linear functions on a topological vector space \(E\) to a topological vector space \(F\). Then, the following conditions are equivalent:

(i) \(M\) is an equicontinuous family;

(ii) \(M\) is equicontinuosu at some points;

(iii) for each neighborhood \(U\) of \(0\) in \(F\) there is a neighborhood \(V\) of \(0\) in \(E\) such that \(f(V)\subseteq U\) for all \(f\in M\);

(iv) if \(U\) is a neighborhood of \(0\) in \(F\), then \(\bigcap_{f\in M}f^{-1}(U)\) is a neighborhood of \(0\) in \(E\). \(\sharp\)

Theorem. If \(M\) is an equicontinuous family of linear functions from a linear topological space \(E\) to a topological vector space \(F\), then the pointwise topology and the topology of uniform convergence on totally bounded sets coincide on \(M\). \(\sharp\)

Proposition. Let \(S\) be a compact topological space, let \(S_{0}\) be a dense subset of \(S\), and let \(G\) be the space of all continuous functions on \(S\) to a compact metric space \((Z,d)\) with the topology of pointwise convergence. Then for each subset \(F\) of \(G\), the following conditions are equivalent.

(i} Each sequence in \(F\) has a cluster point in \(G\).

(ii) For all sequences \(\{s_{m}\}\) in \(S_{0}\) and \(\{f_{n}\}\) in \(F\), it is true that \(lim_{m}\lim_{n}f_{n}(s_{m})=\lim_{n}\lim_{m}f_{n}(s_{m})\) whenever each of the limits exists.

(iii) The closure of \(F\) in \(G\) is compact \((\)equivalently, each limit, relative to the topology of pointwise convergence, of members of \(F\) is continuous$)$. \(\sharp\)

Lemma. Let \(F\) be a family of continuous functions on a compact topological space \(S\) to a metric space \(Z\), and let \(f\) be a continuous function on \(S\) to \(Z\) which belongs to the closure of \(F\) relative to the topology of pointwise convergence. Then \(f\) is a cluster point of some sequence in \(F\). \(\sharp\)

As a preliminary to the next theorem we recall that if \(F\) is a countable family of functions on a set \(S\) to a metric space \(Z\), then the weakest topology \(\tau\) for \(S\) which makes each member of \(F\) continuous is pseudo-metrizable. (Convergence of a net relative to \(\tau\) is equivalent to convergence of its natural image in the countable product \(\prod_{f\in F}Z\) of metric spaces.) If each member of \(F\) is continuous relative to some compact topology \(\eta\) for \(S\) then \(\tau\) is weaker than \(\eta\) and hence \(S\), with \(\tau\), is compact.

Theorem. Let \(G\) be the family of all continuous functions on a compact space \(S\) to a compact metric space \(Z\), let \(G\) have the topology of pointwise convergence, and let \(F\) be a subfamily of \(G\) such that each sequence in \(F\) has a cluster point in \(G\). Then each function \(f\) which belongs to the closure of \(F\) is a cluster point of a sequence in \(F\), and each cluster point of a sequence in \(F\) is the limit of a subsequence. \(\sharp\)

Theorem. Let \(G\) be the family of all continuous functions on a compact space \(S\) to a compact metric space \(Z\), let \(G\) have the topology of pointwise convergence, and let \(F\) be a subfamily of \(G\) such that each sequence in \(F\) has a cluster point in \(G\). Then, the closure of \(F\) in \(G\) is compact, and each member of this closure is the limit of a sequence in \(F\). In particular, each countably compact family of continuous functions on \(S\) to \(Z\) is compact and sequentially compact. \(\sharp\)