Section 5.4 Locally Convex spaces
We met so far several spaces whose natural topology cannot be defined via a single norm but rather via a family of seminorms.Definition 5.4.1. Seminorm.
A seminorm on a vector space \(X\) is a function \(p:X\to[0,\infty)\) such that- \(p(x)\geq0\) for all \(x\in X\text{;}\)
- \(p(\lambda x)=|\lambda|p(x)\) for all \(\lambda\in\mathbb R\) and \(x\in X\text{;}\)
- \(p(x+y)\leq p(x)+p(y)\) for all \(x,y\in X\)
Definition 5.4.2. Locally Convex space.
By locally convex vector space we mean a vector space \(X\) whose topology is generated by a family of seminorms \(p_\alpha\text{,}\) \(\alpha\in I\text{,}\) such that, for every \(v\in X\text{,}\) there is some \(\alpha_0\in I\) such that \(p_{\alpha_0}(x)\neq0\text{.}\)Definition 5.4.3. Cauchy sequences in locally convex spaces.
A sequence \(x_n\) is cauchy if, for every \(p_\alpha\) and \(\eps>0\text{,}\) there is a \(N>0\) such that \(p_\alpha(x_n-x)<\eps\) for every \(n\geq N\text{.}\)- a sequence \(x_n\) converges to \(x\) in \(X\) if and only if \(p_\alpha(x_n-x)\to0\) for every \(\alpha\in I\text{;}\)
- every converging sequence is Cauchy.
Definition 5.4.4. Frechet space.
By Frechet space we mean a complete locally convex vector space \(X\) whose topology is generated by a countable family of seminorms \(p_n\text{.}\)
\begin{equation*}
d(x,y) = \sum_{n=1}^\infty \frac{1}{2^n}\frac{p_n(x-y)}{1+p_n(x-y)}.
\end{equation*}
A space whose topology is induced by a distance function that makes it a complete metric space is called a completely metrizable space. Hence we can summarize the observation above by the following: Theorem 5.4.5.
Every Frechet space is a completely metrizable space.Theorem 5.4.6. Topology induced by a family of seminorms.
The topology induced by a family of seminorms \(p_\alpha, \alpha\in I\) is generated by the basic neighborhoods of zero of the form
\begin{equation*}
\{p_{\alpha_1}(x)<\eps\}\cap\dots\cap\{p_{\alpha_k}(x)<\eps\}
\end{equation*}
where \(\alpha_1,\dots\alpha_k\) are finitely many indices in \(I\) and \(\eps>0\text{.}\)Proof.
\begin{equation*}
\{p_{\alpha_1}(x)\leq\eps\}\cap\dots\cap\{p_{\alpha_k}(x)\leq\eps\}\subset\cal U
\end{equation*}
Hence, \(\cal U_0\subset\cal U\text{.}\)
\begin{equation*}
p_x(f) = |f(x)|,\;x\in\bS^1.
\end{equation*}
This family is uncountable and convergence in this topology means pointwise convergence:
\begin{equation*}
f_n\to f\text{ if and only if }f_n(x)\to f(x)\text{ for every }x\in\bS^1.
\end{equation*}
Now, consider the set
\begin{equation*}
A = \{f\in C^0(\bS^1)\,:\, f(q)=0\text{ for infinitely many }q\in\bQ\}.
\end{equation*}
1. \(A\) is sequencially closed. Indeed, assume that \(f_n\to f\) and that \(f\) has only finitely many zeros in \(\bQ\text{.}\) Denote by \(Z\) be the set of these zeros. Since \(f_n\to f\text{,}\) for each \(q\in\bQ\setminus Z\text{,}\) there is a \(N(q)\) such that \(f_n(q)\neq0\) for all \(n\geq N(q)\text{.}\) This, though it is not possible since, by definition, given any \(f_n\text{,}\) there is some \(q\in\bQ\setminus Z\) such that \(f_n(q)=0\text{.}\) Hence, \(f\) must have infinitely many zeros in \(\bQ\text{,}\) namely \(f\in A\text{.}\)
2. \(A\) is not closed. Take \(f(x)=1\text{.}\) By the argument above, \(f\) cannot be the sequence of elements of \(A\text{.}\) Yet, \(f\) lies in the closure of \(A\text{.}\) Indeed, basic neighborhoods of \(f\) are given by sets
\begin{equation*}
U_{x_1,\dots,x_k}(\eps) = \{g\in C^0(\bS^1) \,:\, |g(x_i)-1|<\eps\text{ for }i=1,\dots,k\}.
\end{equation*}
Let \(W=\{x_1,\dots,x_k\}\) and \(x\in\bS^1\setminus W\text{,}\) choose a sequence \(q_n\to x\text{,}\) \(q_n\in\bQ\) and set
\begin{equation*}
Z = \{x\}\cup\{q_1,q_2,\dots\}.
\end{equation*}
Since \(W\) and \(Z\) are closed and disjoint, by Urysohn's Lemma there exist a continuous function \(g:\bS^1\to[0,1]\) such that \(g|_Z=0\) and \(g|_W=1\text{.}\) Then \(g\in A\) and \(g\in U_{x_1,\dots,x_k}(\eps)\) for all \(\eps>0\text{.}\) Hence, \(f\) lies in the closure of \(A\text{.}\) This example shows clearly that sequences are not able to fully encode the topology induced by an uncountable family of seminorms. A tool that can be used in these cases is defined below. Definition 5.4.7.
Directed set A directed set is a non-empty set \(A\) together with a preorder, namely a binary relation \(\succ\) such that, for each \(a,b\in A\text{,}\) there is a \(c\in A\) such that \(c\succ a\) and \(c\succ b\text{.}\)Definition 5.4.8.
Net A net on a set \(X\) is a map \(n:A\to X\) whose domain \(A\) is a direct set.Definition 5.4.9.
Convergence of a net We say that a net \(x_\alpha\text{,}\) \(\alpha\in I\text{,}\) in \(X\) converges to a point \(x\in X\) if, for every neighborhood \(U\) of \(x\text{,}\) there is an \(a\in I\) such that \(x_b\in U\) for every \(b\succ a\text{.}\)
\begin{equation*}
Z_F = \{r\}\cup\{q_1,q_2,\dots\}.
\end{equation*}
Since \(F\) and \(Z_F\) are closed and disjoint, there is a \(g_F\in C^0(\bS^1)\) such that
\begin{equation*}
g_F|_F=1\text{ and }g_F|_{Z_F}=0.
\end{equation*}
Hence, \(g_F, F\in I\) is a net in \(A\text{.}\) We claim that \(g_F\to f\text{.}\) Indeed, given any \(x_0\in\bS^1\text{,}\) set \(F_0=\{x_0\}\in I\text{.}\) Then, for every \(F\succ F_0\) (i.e. for every \(F\subset\bS^1\) such that \(x_0\in F\)), we have that \(g_F(x_0)=1\) (since \(g_F|_F=1\)). Hence, let \(U\) be a basic neighborhood of \(f\text{,}\) namely
\begin{equation*}
U = \{g\in C^0(\bS^1) \,:\, |g(x_i)-1|<\eps\text{ for }i=1,\dots,k\}.
\end{equation*}
for some points \(x_1,\dots,x_k\) and some \(\eps>0\text{,}\) and set \(G=\{x_1,\dots,x_k\}\text{.}\) Then, for every \(F\succ G\text{,}\) we have that \(g_F(x_i)=1\) for every \(i=1,\dots,k\text{,}\) so that \(g_F\in U\text{.}\) This means precisely that \(g_F\to f\text{.}\) This long example on pointwise convergence topology suggests that nets plays precisely the role of sequences even when the topology is generated by uncountably many seminorms. In fact, we have the following results, of which we omit the proofs. Theorem 5.4.10. Nets encode the topology.
Let \(X,Y\) be topological spaces. Then:- \(C\subset X\) is closed if and only if every converging net inside \(C\) converges to an element of \(C\text{.}\)
- A map \(f:X\to Y\) is continuous if and only if, given any convergent net \(x_\alpha\in X, \alpha\in I\text{,}\) \(f(\lim x_\alpha) = \lim f(x_\alpha)\text{.}\)
- \(X\) is compact if and only if every net \(x_\alpha\in X, \alpha\in I\) has a converging subnet.
