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In particular, I'm interested in the subset $\{x : \lVert x\rVert_1 \ge 1\}$ inside $\ell^2$. Is this a closed subset?

thank you!

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You can typeset your formulas in LaTeX. – Giuseppe Negro Feb 3 '13 at 21:54
For the question, the inclusion cannot be closed since $\ell^1$ is $\lVert\cdot\rVert_2$-dense in $\ell^2$. To wit, note that $c_{00}\subset \ell^1$, and $c_{00}$ is dense in $\ell^2$. – Giuseppe Negro Feb 3 '13 at 21:55
On the other hand, The set $B_1=\{x :\Vert x\Vert_1\le 1\}$ is closed in $\ell_2$. – David Mitra Feb 3 '13 at 22:22
up vote 3 down vote accepted

Consider the vectors $$ x_n=(\textstyle\underbrace{{1\over n},{1\over n},\ldots{1\over n}}_{n-\text {terms}},0,0,\ldots). $$

We have $\Vert x_n\Vert_1=1$ for each $n$. Also, $\Vert x_n\Vert_2=1/\sqrt n$ for each $n$. It follows that $x_n\rightarrow 0$ in $\ell_2$. Since $\Vert{ 0}\Vert_2=0$, it follows that the set $\{x :\Vert x\Vert_1\ge 1\}$ is not closed in $\ell_2$. In particular, the inclusion map from $\ell_1$ to $\ell_2$ is not a closed map.

(Incidentally, the set $B_1=\{x :\Vert x\Vert_1\le 1\}$ is closed in $\ell_2$:

Suppose $(x_n)$ is a sequence in $B_1$ that converges in the $\ell_2$-norm to $x\in\ell_2$. Then $(x_n)$ converges coordinatewise. Given a positive integer $n$, we have for each $k$ that $\sum\limits_{i=1}^n|x_k(i)|\le 1$. Letting $k$ tend to infinity, we deduce $\sum\limits_{i=1}^n|x (i)|\le 1 $. This is true for all $n$. Thus $x\in \ell_1$ and $\Vert x\Vert_1\le1$.)

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Thank you David Mitra, Jacob Schlather and brom for your answers – guest Feb 4 '13 at 7:50

Here's another way of looking at it that helped me when I first thought about these things (though of course this answer is equivalent to Jacob Schlather's):

Recall that for $s = (1, 1/2, 1/3, ..., 1/k, ...)$ you have $s \in \ell^2$ but $s \notin \ell^1$. Now if your set $S = \{x: \|x\|_1 \geq 1\}$ were closed in $\ell^2$ then its complement $S^c = \{x: \|x\|_1 < 1\}$ would have to be open in $\ell^2$. Since $0 \in S^c$ this means there would have to exist $r > 0$ such that $S^c$ contains the $\ell^2$-ball around $0$, that is,

$$B_r = \{x: \|x\|_2 < r\} \subset S^c$$

But for every $r > 0$ the set $B_r$ contains $u = rs/2\|s\|_2$ since $$\|u\|_2 = \|\frac{rs}{2\|s\|_2}\|_2 = \frac{r}{2\|s\|_2}\|s\|_2 = r/2 < r. $$

But also $$\|u\|_1 = \|\frac{rs}{2\|s\|_2}\|_1 = \frac{r}{2\|s\|_2}\|s\|_1 = \infty > 1$$

so $u \notin S^c$ and hence it is never true that

$$B_r \subset S^c$$

Hence $S^c$ is not open and therefore $S$ is not closed.

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Let $x=\sum_{n=1}^\infty e_n/n$ and $x_n=\sum_{k=1}^n e_k/k$ then $x_k \rightarrow x$ in $\ell^2$, but not in $\ell^1$.

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