Definition 8.11.2 (Space of Bounded Functions). Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is bounded if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^{T}$ is the space of all bounded functions from $T$ to $E$, and:
$B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$.
$B(T; E)$ is a closed subset of $E^{T}$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
Proof. (1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by Proposition 8.3.2, so $B(T; E)$ is closed under addition and scalar multiplication.
Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by Proposition 8.11.1.
(2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_{E}(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
so $f \in B(T; E)$.
If $E$ is complete, then $E^{T}$ with the uniform topology is complete by Proposition 6.1.5. Thus $B(T; E)$ is also complete by Proposition 5.5.3.$\square$