# Hints on showing Cauchy sequence converges

Let $T>0$ and $L\geq0$. Let $C[0,T]$ be the space of all continuous real valued functions on $[0,T]$ with the metric $\rho$ defined by

$$\rho(x,y)=\sup_{0\leq t\leq T}e^{-Lt}\left|x(t)-y(t)\right|$$

How can we verify that $\left(C[0,T],\rho\right)$ is a complete metric space?

My working:

Let $\{x_n(t)\}$ be an arbitrary Cauchy sequence in $C[0,T]$. We need to show that $\{x_n(t)\}$ converges to say $x(t)\in C[0,T]$.

The definition of Cauchy sequence states that $\{x_n(t)\}$ is Cauchy if $\forall\epsilon>0,\exists N$ such that $m,n\geq N\implies\rho(x_m(t),x_n(t))<\epsilon$.

But $\rho(x_m(t),x_n(t))=\sup_{0\leq t\leq T}e^{-Lt}\left|x_m(t)-x_n(t)\right|$.

Since every Cauchy sequence is bounded, then $\forall t\in[0,T]$, $\rho\left(x_m(t),x_n(t)\right)<K$ for a constant $K$.

Then I am stuck, I am not sure how to show $\{x_n(t)\}$ converges.

Could anybody please give some hints?

Thanks.

$|x(t_2)-x(t_1)|<|x(t_2)-x_m(t_2)|+|x_m(t_2)-x_m(t_1)|+|x_m(t_1)-x(t_1)|$, the RHS of which is arbitrarily small.
The existence of the function $x(t)$, the 'limit' of the Cauchy sequence, or more formally that function for which $\sup_{0\leq t\leq T}|x(t)-x_m(t)|$ is arbitrarily small, is obvious - what need be shown is that it is continuous.
EDIT: On second thoughts, perhaps it will clear the fog a little to demonstrate the existence of $x(t)$.
$\epsilon>\rho(x_m(t),x_n(t))=\sup_{0\leq t\leq T}e^{-Lt}\left|x_m(t)-x_n(t)\right|$$>e^{-LT}\sup_{0\leq t\leq T}|x_m(t)-x_n(t)| Since e^{-LT} is a constant, you can immediately see that \sup_{0\leq t\leq T}|x_m(t)-x_n(t)| itself tends to zero. Now fix t=t_0; then {x_n(t_0)} is a Cauchy sequence on the reals (why?) and tends to a limit, which we define to be x(t_0) for all t_0 \in [0,T]. Afterwards, show as above that x(t), which at this point is merely known to be a real function, is also a continuous function and thus belongs to C[0,T]. • Thanks. Yes, I am not very sure when you said the existence of the 'limit' is obvious. Could you please clarify that a bit more? – user338393 Jun 25 '16 at 14:27 • there you go @user338393 – Yon Teh Jun 25 '16 at 14:36 Denote by d(x,y):=\sup_{0\leq t\leq T}|x(t)-y(t)| the "usual" distance in X:=C([0,T]). Then$$\rho(x,y)\leq d(x,y),\quad d(x,y)\leq e^{LT}\rho(x,y)\qquad\forall x,\>y\in X\ .$$This shows the Cauchy sequences in (X,\rho) and (X,d) are the same. Since (X,d) is known to be complete (see below) we can at once infer that (X,\rho) is complete as well. The completeness of (X,d) is shown as follows: If (x_n)_{n\geq1} is a Cauchy sequence in (X,d) then for each fixed t\in[0,T] the sequence \bigl(x(t)\bigr)_{n\geq1} is a Cauchy sequence of real numbers, hence convergent to a \xi(t)\in{\mathbb R}. Now let an \epsilon>0 be given. Since (x_n)_{n\geq1} is a Cauchy sequence in (X,d) there is an n_0 such that$$|x_n(t)-x_m(t)|\leq\epsilon$$for all t\in[0,T] and all n, m\geq n_0. Letting m\to\infty here shows that$$|x_n(t)-\xi(t)|\leq\epsilon\qquad\forall t\in[0,T], \quad \forall n\geq n_0 \ .$$This says that$x_n\to\xi$uniformly when$n\to\infty$, and this proves that$\xi$is continuous, hence an element of$X\$.