1 Introduction↩︎

At the foot of pages 129 in [1], in our discussion of locally convex spaces in constructive analysis, we wrote:

The proofs of the next five results are similar to those of their counterparts in metric space theory ... and are left as an exercise.

It turns out that some of those proofs are not quite so straightforward as our quote suggests, and one theorem appears not to be constructively provable.¹ In the present article² we provide proofs of those results (corrected as necessary), as well as correcting some others in [1]. To do so, we give an amplified presentation of some fundamental elements of the theory of locally convex spaces within the framework of Bishop’s constructive analysis (a good introduction to which is given by the articles [2], [3] in [4]).

2 Locally convex topologies↩︎

A locally convex space is a pair \(\left( X,\left( p_{i}\right) _{i\in I}\right)\), where

• \(X\) is a real or complex linear space, with inequality relation \(\neq\);

• for each \(i\) in the index set \(I\), \(p_{i}\) is a seminorm on \(X\);

• \(x\neq0\) if and only if \(p_{i}(x)>0\) for some \(i\in I\).³

We call the functions \(p_{i}\) the defining seminorms of the locally convex space. If it is clear what the defining seminorms are, we refer to \(X\) itself as a locally convex space. Given an inhabited⁴ finitely enumerable subset of \(I\), a point \(x_{0}\in X\), and \(r>0\), we define the open and closed \(F\)-balls with centre \(x_{0}\) and radius \(r\) to be, respectively,\[\begin{align} B^{F}(x_{0},r) & \equiv\left\{ x\in X:\sum_{i\in F}p_{i}(x-x_{0})<r\right\} \text{ and}\\ \overline{B}^{F}(x_{0},r) & \equiv\left\{ x\in X:\sum_{i\in F}p_{i}(x-x_{0})\leq r\right\} . \end{align}\] The locally convex topology \(\tau_{X}\) on \(X\) consists of all finite unions of open \(F\)-balls. We also denote \(B^{F}(0,r)\) by \(B^{F}(r)\), and \(\overline{B}^{F}(0,r)\) by \(\overline{B}^{F}(r)\).

Let \(S\subset X\). The closure of \(S\) in \(X\) is the set \(S^{c}\) of all \(x\in X\) such that \(S\cap B^{F}(x,r)\) is inhabited for each finitely enumerable \(F\subset I\) and each \(\varepsilon>0\). We say that \(S\) is closed in \(X\) if \(S=S^{c}\); dense in \(X\) if \(S^{c}=X\); and separable if it has a countable dense subset. On the other hand, \(S\) is bounded if there exist \(c>0\), a finitely enumerable \(F\subset I\), and \(r>0\) such that \(S\subset cB^{F}(r)\).

For example, a normed linear space \(\left( X,\left\Vert \;\right\Vert \right)\) is a locally convex space in which the family of defining seminorms comprises the single norm \(\left\Vert \;\right\Vert\), and the locally convex topology is just the standard metric topology associated with that norm.⁵ If \(Y\) is also a normed linear space, then the linear space \(\mathcal{B}(X,Y)\) of bounded linear mappings of \(X\) into \(Y\) has a locally convex structure with defining seminorms \(\left\Vert \;\right\Vert _{x}:T\rightsquigarrow\left\Vert Tx\right\Vert\),\(\;\)indexed by the vectors \(x\in X\) with \(\left\Vert x\right\Vert \leq1\). In the special case where \(Y\) is the groundfield (\(\mathbb{R}\) or \(\mathbb{C}\)) the locally convex structure gives us the weak\(^{\ast}\)topology on the dual space \(X^{\ast}\) of all bounded linear functionals on \(X\). If \(X=Y=H\), where \(H\) is a Hilbert space, then the seminorms \(\left\Vert \;\right\Vert _{x}\) with \(x\in H\) and \(\left\Vert x\right\Vert \leq1\) give us the strong operator topology on the space \(\mathcal{B}(H)\) of bounded linear operators on \(H\). Another important example⁶ of a locally convex structure on \(\mathcal{B}(H)\) has defining seminorms \(T\rightsquigarrow\left\vert \left\langle Tx,y\right\rangle \right\vert\) indexed by the ordered pairs \(\left( x,y\right)\) of vectors in \(H\) with \(\left\Vert x\right\Vert ,\left\Vert y\right\Vert \leq1\); this structure gives rise to the weak operator topology on \(\mathcal{B}(H)\).

Returning to the general locally convex space \(\left( X,\left( p_{i}\right) _{i\in I}\right)\), let \(f\) be a mapping of a subset \(S\) of \(X\) into a locally convex space \(\left( Y,\left( q_{j}\right) _{j\in J}\right)\). Then \(f\) is (pointwise) continuous, in the usual topological sense, at \(a\in S\;\)if for each finitely enumerable subset \(G\) of \(J\) and each \(\varepsilon >0\), there exist a finitely enumerable subset \(F\) of \(I\) and \(\delta>0\) such that if \(x\in S\) and \(\sum_{i\in F}p_{i}(x-a)<\delta\), then \(\sum_{j\in J}q_{j}(f(x)-f(a))<\varepsilon\). We say that \(f\) is uniformly continuous on \(S\) if for each finitely enumerable subset \(G\) of \(J\) and each \(\varepsilon>0\), there exist a finitely enumerable subset \(F\) of \(I\) and \(\delta>0\) such that if \(x,y\in S\) and \(\sum_{i\in F}p_{i}(x-y)<\delta\), then \(\sum_{j\in J}q_{j}(f(x)-f(y))<\varepsilon\).

Proof. For each \(\varepsilon>0\), if \(x,y\in X\) and \(\sum_{i\in F}p_{i}(x-y)<\varepsilon\), then (since the \(p_{i}\) are seminorms)\[\begin{align} \left\vert \sum_{i\in F}p_{i}(x-s)-\sum_{i\in F}p_{i}(y-s)\right\vert & \leq\sum_{i\in F}\left\vert p_{i}(x-s)-p_{i}(y-s)\right\vert \\ & \leq\sum_{i\in F}p_{i}((x-s)-(y-s))=\sum_{i\in F}p_{i}(x-y)<\varepsilon. \end{align}\] The final conclusion of the proposition is just the case \(s=0\).  

In Proposition 5.4.1 of [1] and its proof, every instance of linear mapping should be replaced by linear functional. Here is the correct statement and proof of the more general proposition.

Proof. It is routine to show that (d)\(\;\Rightarrow\;\)(c)\(\Rightarrow \;\)(b)\(\Rightarrow\;\)(a). To complete the proof, suppose that \(u\) is continuous at \(0\), and let \(G\) be a finitely enumerable subset of \(J\). There exist \(C>0\) and a finitely enumerable set \(F\subset I\) such that if \(\sum_{i\in F}p_{i}(x)\leq C^{-1}\), then \(\sum_{j\in J}q_{j}(x)<1\). For each \(x\in X\) and each \(\varepsilon>0\) we have\[\sum_{i\in F}p_{i}\left( \frac{C^{-1}x}{\varepsilon+\sum_{i\in F}p_{i}(x)}\right) \leq C^{-1},\] so\[\sum_{j\in J}q_{j}\left( u\left( \frac{C^{-1}x}{\sum_{i\in F}p_{i}(x)+\varepsilon}\right) \right) <1\text{ }\] and therefore\[\sum_{j\in J}q_{j}(u(x))<C\left( \sum_{i\in F}p_{i}(x)+\varepsilon\right) .\] Since \(x\in X\) and \(\varepsilon>0\) are arbitrary, (?? ) now follows.  

Let \(S\) be a subset of the locally convex space \(X\). If \(F\) is a finitely enumerable subset of \(I\) and\[\rho^{F}(x,S)\equiv\inf\left\{ \sum_{i\in F}p_{i}(x-y):y\in S\right\}\] exists, then \(F\) is \(F\)-located in \(X\). If this holds for all finitely enumerable \(F\subset I\), then \(S\) is located in \(X\).

Proof. For each \(x\in X\) and each \(t>0\),\[\begin{align} \exists y\in\ker u\left( {\textstyle\sum_{i\in F}} p_{i}(x-y)\leq t\right) & \Leftrightarrow\exists z\in X\left( \sum_{i\in F}p_{i}(z)\leq1\text{ and }u(x-tz)=0\right) \\ & \Leftrightarrow u(x)\in tu(\overline{B}^{F}(1)). \end{align}\] Thus\[\left\{ t>0:\exists y\in\ker u\left( {\textstyle\sum_{i\in F}} p_{i}(x-y)\leq t\right) \right\} =\left\{ t>0:u(x)\in tu(\overline{B}^{F}(1))\right\} .\] Since \(\rho^{F}(x,\ker u)\), if it exists, equals the infimum of the left-hand set, and \(s_{x}\), if it exists, equals the infimum of the right-hand set, the first part of the lemma now follows.

Supposing that \(s_{x}\) exists and that \(u(x)\neq0\), let \(0<r<C^{-1}\left\vert u(x)\right\vert\). If \(s_{x}<r\), then there exist a positive \(t<r\) and \(z\in\overline{B}^{F}(1)\) such that \(u(x)=tu(z)=u(tz)\). But \(\sum_{i\in F}p_{i}(tz)=t\sum_{i\in F}p_{i}(z)\leq t<r\), so \(\left\vert u(tz)\right\vert \leq C\sum_{i\in F}p_{i}(tz)<Cr<\left\vert u(x)\right\vert\), a contradiction. Hence \(s_{x}\geq r>0\).  

Now let \(u\) be a continuous linear functional on our locally convex space \(X\), and let \(F\subset I\) be finitely enumerable. If\[\left\Vert u\right\Vert _{F}\equiv\sup\left\{ \left\vert u(x)\right\vert :x\in X,\,\sum_{i\in F}p_{i}(x)\leq1\right\}\] exists, we say that \(u\) is \(F\)-normed, or \(F\)-normable, and we call \(\left\Vert u\right\Vert _{F}\) the \(F\)-norm of \(u\). In that case, \[\left\Vert u\right\Vert _{F}=\sup\left\{ \left\vert u(x)\right\vert :x\in X,\,\sum_{i\in F}p_{i}(x)=1\right\} \label{zz1}\tag{1}\] provided the set on the right of (1 ) is inhabited. If this holds for all finitely enumerable \(F\subset I\), then \(u\) is normed, or normable.

Note that if \(u\) is \(F\)-normed, then \[\left\vert u(x)\right\vert \leq\left\Vert u\right\Vert _{F}\sum_{i\in F}p_{i}(x)\;\;\;(x\in X).\] Moreover, if \(C>0\) and \(\left\vert u(x)\right\vert \leq C\sum_{i\in F}p_{i}(x)\) for all \(x\in X\), then \(c\geq\left\Vert u\right\Vert _{F}\). For if \(C<\left\Vert u\right\Vert _{F}\), then there exists \(x\in X\) such that \(\sum_{i\in F}p_{i}(x)\leq1\) and \(C<\left\vert u(x)\right\vert \leq C\), which is absurd. Hence\[\left\Vert u\right\Vert _{F}=\inf\left\{ C>0:\left\vert u(x)\right\vert \leq C\sum_{i\in F}p_{i}(x)\;\;(x\in X)\right\} .\]

The following is the locally convex space analogue of the standard criterion for normability for linear functionals on normed spaces in [6] Proposition 8, page 258,[1].

Proof. Suppose first that \(u\) is \(F\)-normed; then since \(u\) is nonzero, \(\left\Vert u\right\Vert _{F}>0\). Consider any \(a\in X\). For each \(y\in\ker u\) we have\[\sum_{i\in F}p_{i}(a-y)\geq\frac{\left\vert u(a-y)\right\vert }{\left\Vert u\right\Vert _{F}}=\frac{\left\vert u(a)\right\vert }{\left\Vert u\right\Vert _{F}}.\] On the other hand, if \(0<\varepsilon<\left\Vert u\right\Vert _{F}\) and we choose \(x\in X\) such that \(\sum_{i\in F}p_{i}(x)=1\) and \(u(x)>\left\Vert u\right\Vert _{F}-\varepsilon\), then\[z\equiv a-\frac{u(a)}{u(x)}x\in\ker u\] and\[\sum_{i\in F}p_{i}(a-z)=\sum_{i\in F}p_{i}\left( \frac{u(a)}{u(x)}x\right) =\frac{u(a)}{u(x)}\sum_{i\in F}p_{i}\left( x\right) =\frac{u(a)}{u(x)}<\frac{\left\vert u(a)\right\vert }{\left\Vert u\right\Vert _{F}-\varepsilon }.\] Since \(\varepsilon\) is arbitrary, it follows that \(\inf_{y\in\ker u}\sum_{i\in F}p_{i}(a-y)\) exists and equals \(\left\vert u(a)\right\vert /\left\Vert u\right\Vert _{F}\). Since \(a\) is arbitrary, we see that \(\ker u\) is \(F\)-located.

Conversely, suppose that \(\ker u\) is \(F\)-located. Since \(u\) is nonzero, there exists \(x_{0}\) with \(u(x_{0})=1\). Thus, by Lemma 1,\[s\equiv\inf\left\{ t>0:1\in tu(\overline{B}^{F}(1))\right\}\] exists and is positive. We show that \(\left\Vert u\right\Vert _{F}\) equals \(1/s\). For each \(x\in\overline{B}^{F}(1)\) we have either \(u(x)<1/s\) or \(u(x)\neq0\). In the latter case, \[\sum_{i\in F}p_{i}\left( \frac{\left\vert u(x)\right\vert }{u(x)}x\right) \leq1\] and\[1=\frac{1}{\left\vert u(x)\right\vert }u\left( \frac{\left\vert u(x)\right\vert }{u(x)}x\right) ,\] so \(1/\left\vert u(x)\right\vert \geq s\) and therefore \(\left\vert u(x)\right\vert \leq1/s\). On the other hand, by definition of \(s\), if \(0<\varepsilon<1/s\), then there exist \(t\) with \(s\leq t<s/\left( 1-\varepsilon s\right)\), and \(z\in\overline{B}^{F}(1)\), such that \(tu(z)=1\). Then \(u(z)=1/t>1/s>1/s-\varepsilon\). Since \(\varepsilon\) is arbitrary, it follows that \(\left\Vert u\right\Vert _{F}\) exists and equals \(1/s\).  

3 Total boundedness in locally convex spaces↩︎

Let \(S\) be an inhabited subset of our locally convex space⁷ \(X\), let \(F\subset I\) be finitely enumerable, and let \(\varepsilon>0\). By an \(\left( F,\varepsilon\right)\)-approximation to \(S\) we mean a set \(T\subset S\) such that for each \(x\in S\) there exists \(y\in T\) with \(\sum_{i\in F}p_{i}(x-y)<\varepsilon\). We say that \(S\) is \(F\)-totally bounded if for each \(\varepsilon>0\) there exists a finitely enumerable \(\left( F,\varepsilon\right)\)-approximation to \(S\). If \(S\) is \(F\)-totally bounded for each finitely enumerable set \(F\subset I\), then we say that \(S\) is totally bounded in \(X\). If \(\left( X,\left\Vert \;\right\Vert \right)\) is a normed linear space, then the total boundedness of a subset of the locally convex space \(X\) is just total boundedness in the usual sense relative to the metric induced on \(X\) by \(\left\Vert \;\right\Vert\).

Proof. Choose a finitely enumerable \(\left( F,1\right)\)-approximation \(\left\{ x_{1},\ldots,x_{N}\right\}\) to \(S\), and let\[c=1+\sum_{i\in F}\sum_{j=1}^{N}p_{i}(x_{j}).\] Given \(x\in S\), choose \(k\) such that \(\sum_{i\in F}p_{i}(x-x_{k})<1\). Then \[\begin{align} \sum_{i\in F}p_{i}(x) & \leq\sum_{i\in F}(p_{i}(x-x_{k})+p_{i}(x_{k}))\\ & \leq\sum_{i\in F}p_{i}(x-x_{k})+\sum_{i\in F}p_{i}(x_{k})\leq1+\sum_{i\in F}\sum_{j=1}^{N}p_{i}(x_{j})=c \end{align}\]

Proof. Let \(G\subset J\) be finitely enumerable and \(\varepsilon>0\). There exist a finitely enumerable subset \(F\) of \(I\) and \(\delta>0\) such that if \(x,y\in S\) and \(\sum_{i\in F}p_{i}(x-y)<\delta\), then \(\sum_{j\in J}q_{j}(f(x)-f(y))<\varepsilon\). Let \(T\) be a finitely enumerable \((F,\delta )\)-approximation to \(S\). Then for each \(x\in S\) there exists \(x^{\prime}\in T\) such that \(\sum_{i\in F}p_{i}(x-x^{\prime})<\delta\) and therefore \(\sum_{j\in J}q_{j}(f(x)-f(x^{\prime}))<\varepsilon\). Thus \(f(T)\) is a finitely enumerable \(\left( G,\varepsilon\right)\)-approximation to \(f(S)\).  

Proof. By Proposition 2, \(u\) is uniformly continuous on \(X\), so the restriction of \(u\) to \(S\) is uniformly continuous. It remains to apply Proposition 5.  

Proof. By Proposition 5 and the remark immediately preceding that proposition, \(f(S)\) is totally bounded in \(\mathbb{R}\) in the usual metric sense. Hence [1] can be applied.  

Proof. It follows from Proposition 1 and Corollary 2 that if \(x\in X\) and \(F\subset I\) is finitely enumerable, then \(\rho^{F}(x,K)\) exists.  

Proof. Let \(F\subset I\) be finitely enumerable and \(\varepsilon>0\). Construct a finitely enumerable \(\left( F,\varepsilon/3\right)\)-approximation \(\left\{ x_{1},\ldots,x_{n}\right\}\) to \(K\), and for each \(k\leq n\) let \(\rho _{k}=\rho^{F}(x_{k},S)\). Write \(\left\{ 1,\ldots,n\right\}\) as a union of two sets \(P\) and \(Q\) where \(\rho_{k}<2\varepsilon/3\) if \(k\in P\), and \(\rho_{k}>\varepsilon/3\) if \(k\in Q\). For each \(k\in P\) there exists \(s_{k}\in S\) such that \(\sum_{i\in F}p_{i}(x_{k}-s_{k})<2\varepsilon/3\). Given \(s\in S\), choose \(k\) such that \(\sum_{i\in F}p_{i}(s-x_{k})<\varepsilon/3\). Then \(\rho_{k}<\varepsilon/3\), so \(k\notin Q\) and therefore \(k\in P\); whence \[\begin{align} \sum_{i\in F}p_{i}(s-s_{k}) & \leq\sum_{i\in F}\left( p_{i}(s-x_{k})+p_{i}(x_{k}-s_{k})\right) \\ & \leq\sum_{i\in F}p_{i}(s-x_{k})+\sum_{i\in F}p_{i}(x_{k}-s_{k})<\frac{\varepsilon}{3}+\frac{2\varepsilon}{3}=\varepsilon. \end{align}\] Thus \(\left\{ s_{k}:k\in P\right\}\) is a finitely enumerable \(\left( F,\varepsilon\right)\)-approximation to \(S\). Since \(F\) and \(\varepsilon\) are arbitrary, it follows that \(S\) is totally bounded.  

If \(K\subset X\) is \(F\)-totally bounded, we define\[\mathsf{diam}^{F}(K)\equiv\sup\left\{ \sum_{i\in F}p_{i}(x-y):x,y\in K\right\} ,\] which exists by Proposition 1 and Corollary 2.

Observe here that if \(x,y\in X\), then for each \(z\in K\),\[\rho^{F}(x,K)\leq\sum_{i\in F}p_{i}(x-z)\leq\sum_{i\in F}p_{i}(x-y)+\sum_{i\in F}p_{i}(y-z),\] and therefore\[\begin{align} \rho^{F}(x,K) & \leq\sum_{i\in F}p_{i}(x-y)+\inf\left\{ \sum_{i\in F}p_{i}(y-z):z\in K\right\} \\ & =\sum_{i\in F}p_{i}(x-y)+\rho^{F}(y,K). \end{align}\]

The next three results and their proofs are locally-convex-space-analogues of ones for metric spaces (see [1]).

Proof. With \(S_{1}\equiv\{x_{0}\},\) we construct inductively a sequence \((S_{n})_{n\geqslant1}\) of finitely enumerable subsets of \(S\) such that

• \(\rho^{F}(x,S_{n})<2^{-n+1}r\) for each \(x\) in \(S\cap B^{F}(x_{0},r)\), and

• \(\rho^{F}(x,S_{n})<2^{-n+3}r\) for each \(x\) in \(S_{n+1}.\)

To that end, assume that \(S_{1},\ldots,S_{n}\) have been constructed with the appropriate properties, and let \(\{x_{1},\ldots,x_{N}\}\) be an \((F,2^{-n}r)\)-approximation to \(S.\) Write \(\{1,\ldots,N\}\) as a union of subsets \(A\) and \(B\) such that\[\begin{align} \rho^{F}(x_{k},S_{n}) & <2^{-n+3}r\quad\mathrm{if}\,\,k\in A,\\ \rho^{F}(x_{k},S_{n}) & >2^{-n+2}r\quad\mathrm{if}\,\,k\in B. \end{align}\] Let\[S_{n+1}\equiv\{x_{k}:k\in A\}.\] Clearly, \(S_{n+1}\) satisfies (b). Let \(x\) be any point of \(S\cap B^{F}(x_{0},r).\) By the induction hypothesis, there exists \(y\) in \(S_{n}\) with \(\sum_{i\in F}p_{i}(x-y)<2^{-n+1}r.\) Choosing \(k\) in \(\{1,\ldots,N\}\) such that \(\sum_{i\in F}p_{i}(x-x_{k})<2^{-n}r,\) we have \[\begin{align} \rho^{F}(x_{k},S_{n}) & \leqslant\sum_{i\in F}p_{i}(x_{k}-y)\\ & \leqslant\sum_{i\in F}p_{i}(x-x_{k})+\sum_{i\in F}p_{i}(x-y)\\ & <2^{-n}r+2^{-n+1}r\\ & <2^{-n+2}r. \end{align}\] Thus \(k\notin B\), so \(k\in A\), which is therefore (inhabited and) finitely enumerable. Moreover, \[\rho^{F}(x,S_{n+1})\leq\sum_{i\in F}p_{i}(x-x_{k})<2^{-(n+1)+1}r,\] so \(S_{n+1}\) satisfies the appropriate instance of (a). This completes the inductive construction. Letting \(K\) be the closure of \(\bigcup\limits_{n=1}^{\infty}S_{n}\) in \(X\), we see from (a) that \(S\cap B^{F}(x_{0},r)\subset K.\)

Next we prove:

• If \(m,n\)are positive integers with \(m\geq n\), and \(y\in S_{m}\), then there exists \(y_{n}\in S_{n}\)such that \(\sum_{i\in F}p_{i}(y-y_{n})<2^{-n+4}r\).

Given \(y\in S_{m}\) and letting \(y_{m}=y\), we see from (b) that for each \(k\) with \(n\leq k\leq m-1\) there exist points \(y_{k}\in S_{k}\) such that \(\sum_{i\in F}p_{i}(y_{k+1}-y_{k})<2^{-k+3}r\). Then\[\sum_{i\in F}p_{i}(y-y_{n})\leqslant\sum_{k=n}^{m-1}\sum_{i\in F}p_{i}(y_{k+1}-y_{k})<\sum_{k=n}^{\infty}2^{-k+3}r=2^{-n+4}r.\] In particular taking \(n=1\), we see that \(\rho^{F}(y,\left\{ x_{0}\right\} )=\sum_{i\in F}p_{i}(y-y_{1})<2^{3}r~\)for each \(y\in\bigcup\limits_{i=1}^{\infty}S_{i}\); whence \(\bigcup\limits_{n=1}^{\infty}S_{n}\subset S\cap B^{F}(x_{0},8r)\) and therefore \(K\subset\overline{B}^{F}(x_{0},8r)\). On the other hand, given \(x\in K\) and a positive integer \(n\), we can find \(m\) and \(y\in S_{m}\) such that \(\sum_{i\in F}p_{i}(x-y)<2^{-n+4}r\). If \(m<n\), then \(y\in\bigcup\limits_{k=1}^{n}S_{k\text{.}}\). If \(m\geq n\), then by the remark preceding this proposition,\[\rho^{F}(x,S_{n})\leqslant\sum_{i\in F}p_{i}(x-y)+\rho^{F}(y,S_{n})<2^{-n+4}r+2^{-n+4}r=2^{-n+5}r,\] the second inequality using (*). Thus there exists \(z\in S_{n}\subset \bigcup\limits_{k=1}^{n}S_{k\text{. }}\) such that \(\sum_{i\in F}p_{i}(x-z)<2^{-n+5}r\). Putting together the two alternatives for \(m\), we now see that the finitely enumerable set \(\bigcup\limits_{k=1}^{n}S_{k\text{. }}\)is a \(\left( F,2^{-n+5}r\right)\)-approximation to \(K\). Since \(n\) is arbitrary, we conclude that \(K\) is \(F\)-totally bounded.  

Proof. Given****\(\varepsilon>0\), construct an \(\left( F,\varepsilon /16\right)\)-approximation \(\left\{ x_{1},\ldots,x_{N}\right\}\) to \(S\). By Proposition 8, for each \(n\in\left\{ 1,\ldots,N\right\}\) there exists a closed, \(F\)-totally bounded set \(K_{n}\) such that \(S\cap B^{F}(x_{n},\varepsilon/16)\subset K_{n}\subset S\cap\overline{B}^{F}(x_{n},\varepsilon/2)\). Hence \[S\subset\bigcup_{n=1}^{N}\left( S\cap B^{F}\left( x_{n},\tfrac{\varepsilon }{16}\right) \right) \subset{\textstyle\bigcup_{n=1}^{N}} K_{n}.\] Also, for all \(x,y\in K_{n}\) we have\[\sum_{n=1}^{N}p_{n}(x-y)\leq\sum_{n=1}^{N}p_{n}(x-x_{n})+\sum_{n=1}^{N}p_{n}(y-x_{n})\leq\tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2}=\varepsilon \text{,}\] so \(\mathsf{diam}^{F}(K_{n})\leq\varepsilon\).  

Proof. By Corollary 3, for each positive integer \(k\) there exist a positive integer \(n_{k}\) and \(F\)-totally bounded sets \(S_{kj}\;\left( 1\leq j\leq n_{k}\right)\), with each \(\mathsf{diam}^{F}(S_{kj})<1/k\), whose union is \(S\). Let \(\left( r_{n}\right) _{n\geq1}\) be an enumeration of the real numbers\[c_{kj}\equiv\inf\left\{ f(x):x\in S_{kj}\right\} \;\;(1\leq k;\,1\leq j\leq n_{k}),\] each of which exists by Corollary 2. Consider any \(r\in\mathbb{R}\) such that \(r\neq r_{n}\) for each \(n\). For each positive integer \(k\), since \(r\;\)is distinct from each \(c_{kj}\), either \(c_{kj}>r\) for each \(j\leq n_{k}\), in which case \(S(f,r)={\textstyle\bigcup_{j=1}^{n_{k}}} S_{kj}(f,r)=\varnothing\), or else\[S_{k}\equiv\left\{ j:1\leq j\leq n_{k},\,c_{kj}<r\right\}\] is inhabited and therefore finitely enumerable. In the latter case, for each \(j\in S_{k}\) choose \(x_{kj}\in S_{kj}\). Given \(x\in S(f,r)\), choose \(j\leq n_{k}\) such that \(x\in S_{kj}\). Then \(c_{kj}\leq f(x)\leq r\), so \(c_{kj}<r\) (since \(r\neq c_{kj}\)) and therefore \(j\in S_{k}\). Hence\[\sum_{i\in F}p_{i}(x-x_{kj})\leq\mathsf{diam}^{F}(S_{kj})<\frac{1}{k}.\] Thus\[\left\{ x_{kj}:1\leq j\leq n_{k}\text{, }c_{kj}<r\right\}\] is a finitely enumerable \((F,1/k)\)-approximation to \(S\). Since \(k\) is arbitrary, \(S\) is \(F\)-totally bounded.  

Proof. By Theorem 9, for each positive integer \(N\) there exists a countable family \(\left( c_{N,k}\right) _{k\geq1}\) of real numbers such that if \(r\in\mathbb{R}\) and \(r\neq c_{N,k}\) for each \(k\), then \(S(f,r)\) is \(\left\{ 1,\ldots,N\right\}\)-totally bounded. Let \(\left( c_{k}\right) _{k\geq1}\) be an enumeration of all the numbers \(c_{N,k}\;(N,k\geq1)\), and let \(r\in\mathbb{R}\) satisfy \(r\neq c_{k}\) for each \(k\geq1\). If \(F\) is any finitely enumerable set of positive integers, then there exists \(N\) such that \(F\subset\left\{ 1,\ldots,N\right\}\). Given \(\varepsilon>0\), choose a finitely enumerable \((\left\{ 1,\ldots,N\right\} ,\varepsilon)\)-approximation \(Y\) to \(S\). For each \(x\in S\) there exists \(y\in Y\) such that \(\sum_{n\in F}p_{i}(x-y)\leq\sum_{n=1}^{N}p_{i}(x-y)<\varepsilon\). Hence \(Y\) is an \((F,\varepsilon)\)-approximation to \(S\). Since \(F\) and \(\varepsilon\;\)are arbitrary, it follows that \(S\) is totally bounded.  

The key application of these results is the following.⁸

Proof. Since, by Proposition 2, \(u\) is uniformly continuous on the totally bounded set \(K\), we see that\[C\equiv\sup\left\{ \left\vert u(x)\right\vert :x\in K\right\}\] exists by Corollary 2. Clearly, \(C\) is positive. Choose \(y_{1}\in K\) such that \(u(y_{1})>C/2\). Then\[y_{0}\equiv\frac{C}{2u(y_{1})}y_{1}\] belongs to the balanced set \(K\), and \(u(y_{0})=C/2\). Let \(\varepsilon>0\) and let \(F\) be a finitely enumerable subset of \(I\). By Propositions 1 and 5, each \(p_{i}(K)\) is a totally bounded subset of \(\mathbb{R}\), so by Proposition 4, there exists \(b>0\) such that \(\sum_{i\in F}p_{i}(x)\leq b\) for each \(x\in K\). By Theorem 9, there exists \(t\) such that\[0<t<\frac{C\varepsilon}{C+4b}\] and the set\[S_{t}\equiv\left\{ y\in K:\left\vert u(y)\right\vert \leq t\right\}\] is \(F\)-totally bounded. Pick an \(\left( F,t\right)\)-approximation \(\left\{ s_{1},\ldots,s_{n}\right\}\) to \(S_{t}\), and set\[y_{k}\equiv\frac{C}{C+2t}s_{k}-\frac{2}{C+2t}u(s_{k})y_{0}\;\;\;\left( 1\leq k\leq n\right) .\] Then \(y_{k}\in\ker u\). Since \(\left\vert u(s_{k})\right\vert \leq t\) and \(K\) is balanced,\[\frac{-u(s_{k})}{t}y_{0}\in K.\] Thus\[y_{k}=\frac{C}{C+2t}s_{k}+\left( 1-\frac{C}{C+2t}\right) \frac{-u(s_{k})}{t}y_{0}\in K\] and\[s_{k}-y_{k}=\left( 1-\frac{C}{C+2t}\right) \left( s_{k}+\frac{u(s_{k})}{t}y_{0}\right) =\frac{2t}{C+2t}\left( s_{k}+\frac{u(s_{k})}{t}y_{0}\right)\] Also,\[\begin{align} \sum_{i\in F}p_{i}(s_{k}-y_{k}) & =\frac{2t}{C+2t}\sum_{i\in F}p_{i}\left( s_{k}+\frac{u(s_{k})}{t}y_{0}\right) \\ & \leq\frac{2t}{C}\sum_{i\in F}\left( p_{i}(s_{k})+\frac{\left\vert u(s_{k})\right\vert }{t}p_{i}(y_{0})\right) \\ & \leq\frac{2t}{C}\left( \sum_{i\in F}p_{i}(s_{k})+\sum_{i\in F}p_{i}(y_{0})\right) \leq\frac{2t}{C}(b+b)=\frac{4tb}{C}. \end{align}\] If \(y\in K\cap\ker u\subset S_{t}\), then there exists \(k\) such that \(\sum_{i\in F}p_{i}(y-s_{k})<t\) and therefore\[\sum_{i\in F}p_{i}(y-y_{k})\leq\sum_{i\in F}p_{i}(y-s_{k})+\sum_{i\in F}p_{i}(s_{k}-y_{k})<t+\frac{4tb}{C}=t\frac{C+4b}{C}<\varepsilon.\] Thus \(\left\{ y_{1},\ldots,y_{n}\right\}\) is a finitely enumerable \(\left( F,\varepsilon\right)\)-approximation to \(K\cap\ker u\). Since \(F,\varepsilon\) are arbitrary, it follows that \(K\) is totally bounded.  

4 Applications↩︎

We conclude by dealing, but without proofs, with some theorems that depend on the work in our preceding sections. But first we have more definitions.

Let \(\left( X,\left( p_{i}\right) _{i\in I}\right)\) be a separable locally convex space. We say that a sequence \(\left( x_{n}\right) _{n\geq1}\) in \(X\)

• is a Cauchy sequence if for each \(\varepsilon>0\) and each finitely enumerable \(F\subset I\) there exists \(N\) such that \(\sum_{i\in F}p_{i}(x_{m}-x_{n})<\varepsilon\);

• converges to the (perforce unique) limit \(x_{\infty }\in X\) if for each \(\varepsilon>0\) and each finitely enumerable \(F\subset I\), there exists \(N\) such that \(\sum_{i\in F}p_{i}(x_{n}-x_{\infty })<\varepsilon\).

We say that the separable locally convex space \(X\) is complete if every Cauchy sequence in \(X\) converges to a limit in \(X\). If \(X\) and \(Y\) are normed linear spaces, then the (norm-) unit ball of \(\mathcal{B}(X,Y)\) is the set\[\mathcal{B}_{1}(X,Y)\equiv\left\{ T\in\mathcal{B}(X,Y):\left\Vert Tx\right\Vert \leq\left\Vert x\right\Vert \text{ for all }x\in X\right\} .\] The unit ball of the dual \(X^{\ast}\) of \(X\) is usually denoted by \(X_{1}^{\ast}\), and that of \(\mathcal{B}(H)\), where \(H\) is a Hilbert space, by \(\mathcal{B}_{1}(H)\).

The proof of Theorem 12 depends on Proposition 10 above, as well as a number of technical lemmas on pages 134–137 of [1].⁹

Another place where the results in Sections 1 and 2 are used is in the proof of [7]:

5 Concluding remarks↩︎

The theory could go from here in at least two directions: firstly, developing the more general theory of what Bishop calls uniform spaces (see [6] or [8]), of which theory the foregoing is a subset; secondly, investigating Bishop’s uniform spaces as apartness (uniform) spaces, as discussed in [1]. Be that as it may, what we have developed in Sections 1–3 above is enough of the constructive theory of locally convex spaces to enable results such as those in Section 4 (see also [9]).

Author’s address: Department of Mathematics & Statistics, University of Canterbury, Christchurch 8140, New Zealand Author’s email: dugbridges@gmail.com

References↩︎