Hints on showing Cauchy sequence converges

Question

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.

Answer 1

I won't insert all the details into a hint, but consider this oft-used technique:

$|x(t_2)-x(t_1)|<|x(t_2)-x_m(t_2)|+|x_m(t_2)-x_m(t_1)|+|x_m(t_1)-x(t_1)|$, 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 $C[0,T]$.

Answer 2

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 $X$.