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I need help with the following exercise of an introductory course in functional analysis.

Let $\mathbb{K}=\mathbb{R}\text{ or }\mathbb{C}$. Consider $l^1_N=\left\{x\in\mathbb{K}^N\colon||x||_1=\sum_{n=1}^N|x_n|< \infty\right\}$. We need to show that the closed unit ball ($B_1=\left\{x\in l_N^1\colon ||x||_1\leq 1\right\}$) is compact.

(I know that if a space is finite dimensional then it unit closed ball is compact. But here we are asked for a direct proof in this space without using this general result).

As a suggestion, it's enough to prove tha $B_1$ is sequentially compact (i.e. every sequence in $B_1$ has a convergent subsequence). I don't know how to construct such a subsequence exactly.

Thanks in advance.

Jesús Carretero Rodríguez
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    The condition that $\sum_{n=1}^N |x_n|$ be finite is vacuous (this is a finite sum). The $|\cdot|_1$-function is a norm on the finite dimensional vector space $l^1_N$ and the closed unit ball of a finite dimensional vector space under any norm is compact. So the key observation here may simply be that $|\cdot|_1$ is a norm. See e.g. https://math.stackexchange.com/questions/287360/is-it-true-that-the-unit-ball-is-compact-in-a-normed-linear-space-iff-the-space – leslie townes Mar 05 '23 at 12:56
  • Isn't $l^1_N$ exactly $\mathbb{R}^N?$. Then you could use Heine-Borel – Luis Dias Mar 05 '23 at 13:06
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    To construct directly a convergent subsequence, perhaps, the classic idea will work: divide $B_1$ into several closed subsets with a radius half of the initial one (https://mathoverflow.net/questions/98007/covering-a-unit-ball-with-balls-half-the-radius). At least one will have an infinite part of the initial sequence. Continuing the process will give a point of convergence. – zkutch Mar 05 '23 at 13:30

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Since we are working with $\mathbb R$ or $\mathbb C$ and those spaces are complete, it is sufficient to show that $B_1$ is closed and bounded.

For every $x\in B_1$, $\lVert x\rVert\le 1$ so $B_1$ is bounded.

To show that $B_1$ is closed we will show that its complement is open.

Let $p=(p_1,\dots ,p_N)$ be a point in $l_{N}^{1}$ that is not in $B_1$. Then $\lVert p\rVert\gt 1$ so $\lVert p\rVert=1+d$ for some $d\gt 0$.

Let $B_{\text{d/2}}(p)$ be the open ball of radius $\frac{d}{2}$ around $p$ and let $q\in B_{\text{d/2}}(p)$.

Then $$\lVert p\rVert=\lVert p-q+q\rVert\le\lVert p-q\rVert+\lVert q\rVert\implies\lVert q\rVert\ge\lVert p\rVert-\lVert p-q\rVert=1+d-\lVert p-q\rVert$$

Therefore, $$\lVert q\rVert\ge 1 + d - \lVert p-q\rVert$$

We chose $q$ so that $\lVert p-q\rVert\lt\frac{d}{2}$ so $$-\lVert p-q\rVert\gt -\frac{d}{2}\implies 1+d-\lVert p-q\rVert\gt 1+d-\frac{d}{2}=1+\frac{d}{2}\gt 1$$

Therefore, $\lVert q\rVert\gt 1$ so $q$ is in the complement of $B_1$ and since that is true for every $q$ in the open ball of radius $\frac{d}{2}$, the open ball around $p$ is contained in the complement of $B_1$. Since that is true for all $p\not\in B_1$, the complement of $B_1$ is open and $B_1$ is closed.

John Douma
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