7.3 The Uniform Topology
Definition 7.3.1 (Uniform Topology).label Let $T$ be a set and $(X, \fU)$ be a uniform space. For each $U \in \fU$, let
then the uniform topology on $X^{T}$ is the topology induced by the uniformity generated by $\bracs{E(U)| U \in \fU}$.
Remark 7.3.1.label Properties such as completeness require the presence of a uniform structure. However, referring to it as the uniform uniformity is a little ridiculous. As such, in this section, completeness with respect to the uniform topology corresponds to completeness with respect to the uniform uniformity.
Proposition 7.3.2.label Let $X$ be a topological space and $Y$ be a uniform space, then:
- (1)
$C(X; Y) \subset Y^{X}$ is closed with respect to the uniform topology.
- (2)
If $X$ is a uniform space, then $UC(X; Y) \subset Y^{X}$ is closed with respect to the uniform topology.
In particular, if $Y$ is complete, then the above spaces are complete.
Proof. Let $x \in X$ and $V$ be an entourage of $Y$, then there exists a symmetric entourage $W$ of $Y$ such that $W \circ W \circ W \subset V$.
(1): Let $f \in \ol{C(X; Y)}$, then there exists $g \in C(X; Y)$ with $(f, g) \in E(W)$. Let $U \in \cn(x)$ such that $g(U) \subset W(g(x))$, then for any $y \in U$,
so $(f(x), f(y)) \in W \circ W \circ W \subset V$.
Therefore $f$ is continuous at $x$.
(2): Let $f \in \ol{UC(X; Y)}$, then there exists $g \in UC(X; Y)$ with $(f, g) \in E(W)$. Let $U$ be an entourage of $X$ such that $(g \times g)(U) \subset W$, then for any $(x, y) \in U$,
so $(f(x), f(y)) \in W \circ W \circ W \subset V$.
If $Y$ is complete, then $Y^{T}$ with the uniform topology is complete by Proposition 7.2.5. Thus $C(T; X)$ and $UC(T; X)$ are both complete subspaces by Proposition 6.7.3.$\square$
Corollary 7.3.3.label Let $X$ be a topological space, $\sigma \subset 2^{X}$ be an ideal such that $X$ is $\sigma$-generated, and $Y$ be a uniform space, then $C(X; Y) \subset Y^{X}$ is closed with respect to the $\sigma$-uniformity.
Proof. Let $f \in \overline{C(X; Y)}\subset Y^{X}$ with respect to the $\sigma$-uniformity. By Proposition 7.3.2, $f \in C(S; Y)$ for all $S \in \sigma$, so $f \in C(X; Y)$ by (3) of Definition 5.1.10.$\square$
Proposition 7.3.4.label Let $X$ be a compact topological space and $E$ be a TVS over $K \in \RC$, then:
- (1)
$C(X; E)$ equipped with the uniform topology is a topological vector space over $K$.
- (2)
If $E$ is locally convex and $\rho: E \to [0, \infty)$ is a continuous seminorm, then
\[[\cdot]_{u, \rho}: C(X; E) \to [0, \infty) \quad f \mapsto \sup_{x \in X}\rho(f(x))\]is a continuous seminorm on $C(X; E)$ with respect to the uniform topology.
The uniform topology on $C(X; E)$ is locally convex, and induced by seminorms of the form $[\cdot]_{u, \rho}$, where $\rho$ ranges over all continuous seminorms on $E$.
- (3)
If $E$ is normed, then
\[[\cdot]_{u}: C(X; E) \quad f \mapsto \sup_{x \in X}\norm{f(x)}_{E}\]is a norm on $C(X; E)$ that induces the uniform topology.
- (4)
If $E$ is complete, then so is $C(X; E)$.
As such, the uniform topology is the canonical topology on $C(X; E)$.
Proof. (1): Since $X$ is compact, for any $f \in C(X; E)$, $f(X) \in \mathfrak{B}(E)$, so (1) holds by Proposition 10.12.1.
(2): By Proposition 11.10.1.
(4): By Proposition 7.3.2.$\square$
Theorem 7.3.5 (Dini).label Let $X$ be a compact topological space, $\angles{f_\alpha}_{\alpha \in A}\subset C(X; \real)$, and $f \in C(X; \real)$ such that $f_{\alpha} \upto f$ pointwise, then $f_{\alpha} \upto f$ uniformly.
Proof. Let $\eps > 0$, then for each $\alpha \in A$, $\bracs{f - f_\alpha < \eps}\subset X$ is open, and $X = \bigcup_{\alpha \in A}\bracs{f - f_\alpha < \eps}$. By compactness, there exists $B \subset A$ finite such that $X = \bigcup_{\beta \in B}\bracs{f - f_\beta < \eps}$. Let $\alpha_{0} \in A$ with $\alpha_{0} \ge \beta$ for all $\beta \in B$, then since $f_{\alpha} \upto f$ pointwise,
for any $\alpha \ge \alpha_{0}$.$\square$
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