Prove C[0, 1] is complete with metric a different metric would any one tell me whether C[0,1] is complete under the following metric
$$
\sup_{t\in [0, T]}e^{-Lt}|x(t)-y(t)|
$$
and how to prove the claim
I know some reasoning on how to prove C[0, 1] is complete with the usual sup norm. Just wondering whether this new metric would make any difference.
 A: Let $C[0,1]$ be equipped with the metric $d_1$, where $d_1 = \sup_{t \in [0,T]} e^{-Lt} \left|x(t) - y(t) \right|.$
Suppose $(x_n(t))$ is a Cauchy sequence in $(C[0,1],d_1)$, then $\forall \epsilon > 0 \ \exists N \in \mathbb{N}$ such that $$n,m > N \implies \sup_{t \in [0,T]} e^{-Lt} \left| x_n(t)-x_m(t) \right| < \epsilon.$$ Observe that $$\forall n,m > N, \sup_{t \in [0,T]} e^{-Lt} \left| x_n(t) - x_m(t) \right| \leq \sup_{t \in [0,T]} \left| x_n(t) - x_m(t) \right|.$$ We can therefore conclude that $(C[0,1], d_1)$ is a complete metric space given that $(C[0,1], d_u)$ is a complete metric space. However, to be more explicit, consider that $$\sup_{t \in [0,T]} e^{-Lt} \left| x_n(t) - x_m(t) \right| = \sup_{t \in [0,T]} e^{-Lt} \left| x_n(t) - x(t) + x(t) - x_m(t) \right|$$ $$   \leq \sup_{t \in [0,T]} e^{-Lt} \left| x_n(t) - x(t) \right| + \sup_{t \in [0,T]} e^{-Lt} \left| x(t) - x_m(t) \right|$$ $$< \sup_{t\in [0,T]} \left| x_n(t) - x(t) \right| + \sup_{t \in [0,T]} \left| x(t) - x_m(t) \right| < 2 \epsilon.$$ Therefore, $(x_n(t))$ converges to $x(t)$ in $(C[0,1],d_1)$. 
