$\Bbb R^d$ with norm is a Banach

Another problem I found in a textbook when brushing up on my real analysis. Was given a hint too.

Prove that $$\Bbb R^d$$ with the norm

$$||x||_1=\sum_{i=1}^d |x_i|, \ x ∈ \Bbb R^d$$

is a Banach space.

Hint: Consider a Cauchy sequence $$(x_n) \text{ in }(\Bbb R^d, || · ||_1)$$ and show that it is Cauchy coordinate-wise. What could the limit of $$(x_n)$$ then be?

• Thanks for the hint, but I'd rather do it on my own. – Saucy O'Path Jan 20 at 17:29
• Every finite dimensional space is a Banach space, Proof: Use that all norms on finite dimensional spaces are equivalent. – user408856 Jan 20 at 17:34
• Excellent hint. I suggest following it and attempting the problem. – Nap D. Lover Jan 20 at 18:03

To show that $$\Bbb R^d$$ with the norm

$$\Vert x \Vert_1 = \displaystyle \sum_1^d \vert x_i \vert, \tag 1$$

where

$$x = (x_1, x_2, \ldots, x_d) \tag 2$$

is Banach we need to prove it is Cauchy-complete with respect to this norm; that is, if

$$y_i \in \Bbb R^d \tag 3$$

is a $$\Vert \cdot \Vert_1$$-Cauchy sequence, then there exists

$$y \in \Bbb R^d \tag 4$$

with

$$y_i \to y \tag 5$$

in the $$\Vert \cdot \Vert_1$$ norm.

Now if $$y_i$$ is $$\Vert \cdot \Vert_1$$-Cauchy, for every real $$\epsilon > 0$$ there exists $$N \in \Bbb N$$ such that, for $$m, n > N$$,

$$\Vert y_m - y_n \Vert_1 < \epsilon; \tag 6$$

if we re-write this in terms of the defiinition (1) we obtain

$$\displaystyle \sum_{k = 1}^d \vert y_{mk} - y_{nk} \vert < \epsilon, \tag 7$$

and we observe that, for every $$l$$, $$1 \le l \le d$$, this yields

$$\vert y_{ml} - y_{nl} \vert \le \displaystyle \sum_{k = 1}^d \vert y_{mk} - y_{nk} \vert < \epsilon; \tag 8$$

thus the sequence $$y_{ml}$$ for fixed $$l$$ is Cauchy in $$\Bbb R$$ with respect to the usual norm $$\vert \cdot \vert$$; and since $$\Bbb R$$ is Cauchy-complete with respect to $$\vert \cdot \vert$$ we infer that for each $$l$$ there is a $$y^\ast_l$$ with

$$y_{ml} \to y_l^\ast \tag 9$$

in the $$\vert \cdot \vert$$ norm on $$\Bbb R$$; thus, taking $$N$$ larger if necessary, we have

$$\vert y_{ml} - y_l^\ast \vert < \dfrac{\epsilon}{d}, \; 1 \le l \le d, \; m > N; \tag{10}$$

setting

$$y^\ast = (y_1^\ast, y_2^\ast, \ldots, y_d^\ast) \in \Bbb R^d, \tag{11}$$

we further have

$$\Vert y_m - y^\ast \Vert_1 = \displaystyle \sum_{l = 1}^d \vert y_{ml} - y_l \vert < d \dfrac{\epsilon}{d} = \epsilon, \tag{12}$$

that is,

$$y_m \to y^\ast \tag{13}$$

in the $$\Vert \cdot \Vert_1$$ norm on $$\Bbb R^d$$; thus $$\Bbb R^d$$ is $$\Vert \cdot \Vert_1$$-Cauchy complete; hence $$\Vert \cdot \Vert_1$$-Banach.

Use this: $$\|x - y\|_1 \ge |x_k - y_k|, 1\le k \le d$$ and the fact that the real numbers are complete.