Non compacity of a subset of $\ell^1$

Question

Consider $l^1 = \left\{ \{a_n\}_{n=1}^\infty : a_n \in \mathbb{R}, \ \ \sum_{n=1}^\infty |a_n| < \infty \right\}$ and let $K = \left\{f \in l^1 : |f(k)| \leq \frac{1}{k} \ \ \forall k\right\}$. Show that $K$ is not compact in $l^1$.

I can't quite finish this up. I'm using the following sequence of sequences: $\{ \{1, 0,0, \dots \}, \{1,\frac{1}{2}, 0, \dots\}, \{1,\frac{1}{2}, \frac{1}{3},0, \dots \}, \dots \}$ i.e, $f_n(x) = \frac{1}{k}$ if $n \le k$ and $f_n(x) = \frac{1}{k}$ if $n > k$.

I'm trying to show that this goes to infinity in $l^1$, and then showing that no subsequence of it converges to a sequence in $l^1$, but I can't seem to find the right approach. Any nudge in the right direction would be great.

Answer 1

You were starting well.

As $\ell^1$ is a complete normed space, it is enough to prove that $K$ is not totally bounded to prove that it is not compact.

In fact $K$ is not even bounded as $\Vert f_n \Vert_1 \to \infty$ as $n \to \infty$. This is true as $\Vert f_n \Vert_1$ is the partial sum of the harmonic series that diverges.

Therefore $K$ is not totally bounded and not compact.

Answer 2

Suppose $f_n$ has a convergent subsequence $f_{n_k}$, then $\sum_{j=1}^{n_k}\frac{1}{j}e_j$ is a Cauchy sequence with respect to $||\cdot||_1$, that is to say given $\epsilon>0,$ there is $N$ such that whenever $n_h,n_k>N$, one has$$\sum_{j=n_h}^{n_k}\frac{1}{j}<\epsilon,$$ which is not true because if one fixes the smallest $n_h>N$, then taking $n_k\to\infty$ would lead to a contradiction.