Proof. $(1)$: Let $R \ge 0$, then

where each $\bracs{\theta \in C([0, \infty); \real^d): \abs{\theta(t)}/(1+t) \le R}$ is a closed subset of $C([0, \infty); \real^{d})$, and their intersection is also closed. Thus $\norm{\cdot}_{\sps}$ is lower semicontinuous.

$(2)$: Let $K \subset [0, \infty)$ be compact, then $M = \sup_{t \in K}(1 + t) < \infty$. Thus $\norm{\theta|_K}_{u} \le M\norm{\theta}_{\sps}$.

$(3)$: Since $\theta(0) = 0$ and $\limv{t}\theta(t)/t = 0$ for all $\theta \in \sps$,

and $\norm{\Lambda \theta}_{u, (0, \infty)}= \norm{\theta}_{\sps}$. On the other hand, for any $f \in C_{0}((0, \infty); \real^{d})$, let $\Lambda^{-1}f$ be defined by $(\Lambda^{-1}f)(0) = 0$ and $(\Lambda^{-1}f)(t) = f(t)(1 + t)$ for $t > 0$. Since $\lim_{t \to 0}f(t) = \limv{t}f(t) = 0$,

so $\Lambda^{-1}f \in \sps$ with $\normn{\Lambda^{-1}f}_{\sps}= \norm{f}_{u, (0, \infty)}$.

As $C_{0}((0, \infty); \real^{d})$ is a separable Banach space, so is $\sps$.$\square$