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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?

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

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

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