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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.

Arthur
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  • Welcome to Math.SE! Please consider updating your question with some information about what you have tried or where you are getting stuck. You will find people are much more willing to help if you do! – Nick Peterson Oct 09 '13 at 12:44
  • generic argument would be every Cauchy sequence converges to some function in this space. So first construct a candidate and then prove the Cauchy sequence does converge to this candidate. I suppose I can take limit of x$_n$ as the limit. – 123 Oct 09 '13 at 12:49
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    "I know some reasoning on how to prove $C[0, 1]$ is complete with the usual sup norm." Good. Now try to prove that the given norm is equivalent to the ordinary sup norm. – Daniel Fischer Oct 09 '13 at 12:51

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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 $(C[0,1],d_1)$.