Let $C[0,1]$ be the space of continuous $\mathbb{C}$ - valued functions on $[0,1]$. Let $||\{a_n\}||_{\infty} = \textrm{sup}_{n}\ |a_n|$ be the supremum norm where $a_n \in \mathbb{C}$. I'm assuming that for $a_n \in \mathbb{C}$ the author means $\textrm{sup}_n \ |a_n| = \textrm{sup}_n \ ||a_n||_{\mathbb{C}}$.
- Show that $C[0,1]$ is a Banach Space under the supremum norm.
I understand that I have to show that every Cauchy sequence of $C[0,1]$ converges in $C[0,1]$ where the convergence is given by the sup norm i.e $||f_n(x) - L||_{\infty} = \sup_n |f_n(x) - L| < \epsilon$ for some $L \in C[0,1]$. I started by trying to get a candidate for $L$. Below is a result I can use.
- If $\{f_n\}$ is Cauchy in $C[0,1]$ then for each fixed $x \in [0,1]$ we have $\{f_n(x)\}$ is Cauchy in $\mathbb{C}$ - which is complete.
The latter means $\{f_n(x)\}$ converges to some $L_x \in \mathbb{C}$ and so we can write $L_x = \lim_n f_n(x)$, i.e for $n \geq N$ we have $||f_n(x) - L(x)||_{\mathbb{C}} < \epsilon$. Letting $L(x) = L_x$, it follows that $L(x)$ is continuous and defined on $[0,1]$. Moreover, for any $\epsilon >0$ and $n \geq N$ we have,
$$||f_n(x) - L(x)||_{\infty} = \sup_{n \geq N} ||f_n(x) - L(x)||_{\mathbb{C}} < \epsilon$$
Does this seem right ?
