> [!definition] > > Let $U \subset \real^d$ be [[Open Set|open]] and bounded, then $UC(U)$ is the space of all [[Space of Bounded Continuous Functions|bounded and continuous]] functions that are also [[Uniform Continuity|uniformly continuous]]. > > If $U \subset \real^d$ is open, then $UC(U) \subset BC(U)$ is the space of functions that are uniformly continuous on bounded subsets of $U$. > [!definition] > > Let $U$ be an open set, then $UC^k(U)$ is the space of all functions in $f \in UC(U)$ such that $\partial^\alpha f \in UC(U)$ for each $\alpha$ with $\abs{\alpha} \le k$. Define the following seminorms > $ > \norm{f}_{\alpha} = \norm{\partial^\alpha f}_u \quad \abs{\alpha} \le k > $ > then $UC^k(U)$ is a [[Banach Space|Banach space]]. > > *Proof*. Uniform limits of uniformly continuous functions are uniformly continuous. > [!theorem] > > Let $f \in BC(U)$, then $f \in UC(U)$ if and only if $f$ admits a continuous extension to $\ol{U}$. > > *Proof*. First suppose that $U$ is bounded. For any $p \in \partial U$, let $\seq{x_n} \subset U$ such that $x_n \to p$, and define $\ol{f}(p) = \limv{n}f(p)$. To see that this is well-defined, let $\seq{y_n} \subset U$ be another such sequence. For any $\eps > 0$, there exists $\delta > 0$ such that $f(x, y) < \eps$ for all $x, y$ with $\abs{x - y} < \delta$. Let $N \in \nat$ such that $\abs{x_n - y_n} < \delta$ for all $n \ge N$, then $\abs{f(x_n)- f(y_n)} < \eps$ for all $n \ge N$. Therefore $\abs{f(x_n) - f(y_n)} \to 0$ and $f(p)$ is well-defined. > > On the other hand, if $f$ admits a continuous extension to $\ol{U}$, then $\ol{f}$ is continuous on a [[Compactness|compact]] set, so $f$ is uniformly continuous. > > Now suppose that $U$ is arbitrary. For any $p \in \partial U$, let $V \in \cn^o(p)$ be a neighbourhood, then by the above $f$ admits an extension to $V \cap \partial U$. For the same reasons, this extension is unique and can be taken to all of $\partial U$. On the other hand, if $f$ admits a continuous extension to $\ol{U}$, then taking bounded sets yields that the function is uniformly continuous.