Definition 8.11.3 (Space of Bounded Continuous Functions). Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
$BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$.
$BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
Proof. (1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by Definition 8.11.2.
(2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^{X}$ by Definition 8.11.2, $BC(X; E)$ is a closed subspace.
If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by Definition 8.11.2. Therefore $BC(X; E)$ is also complete.$\square$