Show that $C(I)^n$ is a Banach space under the following norm

Question

Apologies if this question has been asked beforehand, the use of different norms in different (but similar) questions makes it really hard to filter out the relevant ones. Anyway:

Let $C(I)^n$ denote the set of functions $y:\mathbb{R} \rightarrow \mathbb{R}^n$ that are continuous in every coordinate in the open interval $I$ (which is what I take it to be, guessing from the lecture notes). Define the norm

$$||y|| := \max\limits_{i = \{1, \ldots, n\}} \max\limits_{x \in I} |y_i(x)|$$

on this space. To show this is a Banach space, we can rewrite the Cauchy condition as follows:

$$\forall \varepsilon > 0, k,l \geq N(\varepsilon): ||y^{k}-y^{l}|| < \varepsilon \iff \forall i \in \{1,\ldots,n\}, x \in I: |y^{(k)}_i(x)-y^{(l)}_i(x)| < \varepsilon$$

and because $\mathbb{R}$ is a complete space with respect to the norm $|\cdot|$, we know that the sequence $\{y^{(n)}_i(x)\}_{n \in \mathbb{N}}$ converges and we can define the limit functions $y_i$ for each $i \in \{1,\ldots,n\}$ and $y = (y_1,\ldots,y_n)^T$. My goal was to show that $\{y^{(n)}\}_{n \in \mathbb{N}}$ must necessarily converge to $y$ in $C(I)^n$ under this definition, but at the moment it is only pulling the knots in my brain even tighter.

Any help is appreciated!

Answer 1

Let me start from scratch. Let $I=[0,1]$. We have $C(I)^n:=C(I,\mathbb{R}^n)$, the continuous functions from $I$ to $\mathbb{R}^n$. Each continuous function $I\to\mathbb{R}^n$ can be identified with a $n$-tuple of functions $I\to\mathbb{R}$, namely the coordinate functions; i.e. if $f\in C(I,\mathbb{R}^n)$, then there exist unique continous functions $f_1,\dots,f_n\in C(I,\mathbb{R})$ such that $f(t)=(f_1(t),\dots,f_n(t))$ for all $t\in I$. Conversely, given any $n$-tuple of continous functions $f_1,\dots,f_n\in C(I,\mathbb{R}))$, the function defined by $f(t):=(f_1(t),\dots,f_n(t))$ is a function of $C(I,\mathbb{R}^n)$. We thus have an identification of $C(I,\mathbb{R}^n)$ with $C(I)\times\dots\times C(I)$ ($n$-times) as sets and this identification is easily seen to be linear, so this is an isomorphism of vector spaces.

The norm defined on on $C(I,\mathbb{R}^n)$ is given by $\|f\|=\max_{i=1,\dots,n}\|f_i\|_\infty$, where $f=(f_1,\dots,f_n)$ and for $g\in C(I)$, $\|g\|_\infty:=\max_{x\in I}|g(x)|$. So let's break the problem in 2 easier problems:

1. Assuming that $(X_1,\|\cdot\|_1),\dots, (X_n,\|\cdot\|_n)$ are Banach spaces, consider their cartesian product $Y:=X_1\times \dots\times X_n$. This is trivially a vector space. Define the norm $\|(x_1,\dots,x_n)\|:=\max_{i=1,\dots,n}\|x_i\|_i$ for all $(x_1,\dots,x_n)\in Y$. Show that $(Y,\|\cdot\|)$ is also a Banach space (i.e. complete). I leave this as an exercise to you, I am pretty sure you can do it (do it for $n=2$ for simplicity, then induction proves it for a general $n$ as well)

2. Prove that $(C(I),\|\cdot\|_\infty)$ is a Banach space, i.e. complete. But this is standard and well-covered fact, see for example the accepted answer on this post Space of bounded continuous functions is complete (and use the fact that continous functions over a compact set are automatically bounded to apply it to your case).

Combining (1) and (2), what you are trying to prove follows immediately.