Prove $\mathbb R^n$ is complete with respect to non-standard norms

Question

Here is a proof (page 6) for the completeness of $\mathbb R^n$ with respect to the standard Euclidean norm.

I do understand the proof but I want to try to do the proof by myself for the completeness of $\mathbb R^n$ with respect to the following norm:

$||(\lambda_1, \lambda_2, \dots , \lambda_n)||_∞ := \max(|\lambda_1|, |\lambda_2|, ...|\lambda_n|)$

May I please ask someone to briefly tell me how to prove that with respect to this particular norm? (by using the definition of completeness, namely, every Cauchy sequence converges) Thanks so much.

I do understand that all norms are equivalent But I just want to know how to prove that with respect to this particular norm directly using the method of showing every Cauchy sequence converges.

Answer 1

Suppose that $(x^k)_{k \ge 0}$ is a Cauchy sequence in $(\mathbb{R}^n, \|-\|_\infty)$. Then for all $\epsilon >0$ there is some $N$ such that for all $k, l \ge N$, $\|x^k - x^l\|_\infty < \epsilon$. Then by definition of $\|-\|_\infty$, for all coordinates $1 \le i \le n$, $|x^k_i - x^l_i| < \epsilon$ for $k,l \ge N$. It follows that $\{x^k_i\}_{k \ge 0}$ is a Cauchy sequence in $\mathbb{R}$, and hence converges to some $x_i$. It follows easily that $\lim_{k \to \infty} x^k = (x_1, \dots, x_n)$.

Answer 2

Note that $\lVert \mathbf{x}\rVert_2 = \left(\displaystyle\sum_{i=1}^n x_i^2\right)^{1/2}$ satisfies the inequalities

$$\lVert \mathbf{x}\rVert_\infty\le \lVert\mathbf{x}\rVert_2\le \sqrt{n}\cdot \lVert\mathbf{x}\rVert_\infty$$

The first inequality is just replacing the sum with only the one term $\sqrt{x_i^2}$ where $x_i=\displaystyle \max_{1\le j\le n} x_j$, and the second follows from replacing all of the summands with the max.

But then if the distance in the normal way goes to $0$, then by the first inequality so too does the "new" norm you have listed, because that distance is smaller than the original one. So in fact you can prove that $\Bbb R^n$ is complete in the new norm if and only if it is in the way you are used to.

Answer 3

Here's a guideline for how to show directly that $\mathbb{R}^n$ is complete with any given norm:

Step 1: Show that if a sequence of vectors $v_k = (v_k^1, \ldots, v_k^n)$ is a Cauchy sequence, then the components of these vectors form $n$ Cauchy sequences $(v_k^1), \ldots, (v_k^n)$.

Step 2: Use the fact that $\mathbb{R}$ is complete to show that these $n$ Cauchy sequences $(v_k^1),\ldots,(v_k^n)$ converge to limits $v^1, \ldots, v^n$.

Step 3: Show that the sequence of vectors $(v_k)$ converges to the vector $v := (v^1, \ldots, v^n)$.