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I am looking for an idea to prove if the inclusion map from $\ell^1(N)$ to $\ell^2(N)$ bounded and does it have a dense image.

And why is the set $A:=\{x: ||x||_1\le 1\} \subset \ell^2(N)$ closed and nowhere dense with the norm topology on $\ell^2(N)$?

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Bounded: Hint: $|x|^2 \leq |x|$ if $|x| \leq 1$. – user27126 Nov 15 '12 at 19:17
Dense image: Hint: What is the dual map for the inclusion $l^1(\mathbb{N}) \to l^2(\mathbb{N})$? – user27126 Nov 15 '12 at 19:17
up vote 1 down vote accepted

Let $x\in\ell^1$, then $|x(n)|<1$ for $n\geqslant n_0$, hence $|x(n)|^2<|x(n)|$ for these $n$ and this proves that $\ell^1\subset\ell^2$. We have that $$\lVert x\rVert_2^2=\sum_{j=0}^{+\infty}|x(n)|^2\leqslant\left(\sum_{j=0}^{+\infty}|x(j)|\right)^2=\sum_{j=0}^{+\infty}|x(n)|^2+2\sum_{l\leq k}|x(l)|\cdot |x(k)|,$$ which proves the boundedness of the inclusion.

As the sequences of finite support are dense in $\ell^2$, the image is dense.

$A$ is closed (in $\ell^2$). Indeed, if $x^n\to x$ in $\ell^2$ and $\lVert x^n\rVert_1\leqslant 1$ for all $n$, then for each $n$ and each integer $N$, $$\sum_{j=0}^N|x_j|\leqslant \sum_{j=0}^N|x_j^n|+\sum_{j=0}^N|x_j-x_j^n|\leqslant 1+\sum_{j=0}^N|x_j-x_j^n|\leqslant 1+\sqrt N\lVert x-x^n\rVert_2.$$ Take the $\lim_{n\to+\infty}$ on both sides to get $\sum_{j=0}^N|x_j|\leqslant 1$ for all $N$, giving the wanted result.

The fact that $A$ has non-empty interior is done here.

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we gave the same reference for the second part of the OP question! – Norbert Nov 15 '12 at 19:25

As for the proof of continuity of natural inclusion $i:x\mapsto x$ of $\ell_1(\mathbb{N})$ into $\ell_2(\mathbb{N})$ see this answer. In your case $p=1$, $q=2$. Note that $i$ maps finitely supported sequences of $\ell_1(\mathbb{N})$ onto finitely supported sequences of $\ell_2(\mathbb{N})$ so they are in the image . Recall this sequences are dense in $\ell_2(\mathbb{N})$. Hence $\mathrm{Im}(i)$ is dense in $\ell_2(\mathbb{N})$. As for the second question take a look at this answer.

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A comment: nowhere denseness of the $l^1$-sphere is actually weaker than $l^1$ being nowhere dense in $l^2$. Which allows an easier proof - if $l^1$ sphere is somewhere dense, it's not hard to see that $l^1 \to l^2$ is a surjection, which is absurd. – user27126 Nov 15 '12 at 19:22

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