Isometry between c and C(X)

Question

Let $c=\left\{(x_n)_{n\in\mathbb N} \in \ell^\infty: \lim_{n\to\infty}x_n\mbox{ exists}\right\}$ and $X = \{1/n : n \in \Bbb {N} \} \cup \{0\}$.

Define $T:c \to C(X)$ by $T((x_n)_{n\in\mathbb N}) = f$, where $f(1/n)= x_n$ and $f(0)$ = $\lim_{n \to \infty} x_n$.

How can I show that $T$ is an isometry?

Answer 1

Let $x=(x_n)_{n\in\mathbb N}$ and $y=(y_n)_{n\in\mathbb N}$ be elements of $c$.
We need to show that $\|T(x)-T(y)\|_{_{C(X)}}=\|x-y\|_{c}$.
We have that

$$ \|T(x)-T(y)\|_{_{C(X)}} = \sup_{u\in X} |f(u)-g(u)|, $$

where $f(1/n)=x_n$ and $f(0)= \lim_{n\to\infty}x_n$, and, $g(1/n)=y_n$ and $g(0)= \lim_{n\to\infty}g_n$.
Therefore,

\begin{align} \|T(x)-T(y)\|_{_{C(X)}}& = \max\{|f(0)-g(0)|, \sup_{n\in \mathbb N} |f(1/n)-g(1/n)|\}\\ & = \max\{|f(0)-g(0)|, \sup_{n\in \mathbb N} |x_n-y_n|\}\\& = \max\{|f(0)-g(0)|, \|x-y\|_{c}\}. \end{align}

Moreover, note that

\begin{align} |f(0)-g(0)|&=\left|\lim_{n\to\infty}x_n - \lim_{n\to\infty}y_n\right| = |\lim_{n\to\infty} x_n - y_n| =\lim_{n\to \infty} |x_n-y_n|, \end{align} and since $|x_n-y_n|\leq \|x-y\|_c$ for all $n\in\mathbb N$, it follows that $$ |f(0)-g(0)|\leq \|x-y\|_c. $$ So, according to the previous equation, we have that $$ \|T(x)-T(y)\|_{_{C(X)}} = \max\{|f(0)-g(0)|, \|x-y\|_{c}\} = \|x-y\|_{c}. $$

Answer 2

First show it is linear: for $x,y \in c$ and $a \in \mathbb{F}$, $T(ax+y)=f$, where $f(1/n)=ax_n+y_n$. Now, $T(x)=f_1$ where $f_1(1/n)=x_n$ and $T(y)=f_2$ where $f(1/n)=y_n$, so in total, $$T(ax+y)=f=af_1+f_2=aT(x)+T(y).$$

Next, show it preserves norm:

If $x\in c$, $\|x\|= \sup_n |x_n|=\sup_n |f(1/n)|=\|f\|=\|Tx\|$.