I'm trying to show that the sequence space $l^p$ is dense proper subspace of $l^q$ and is of first category, when $1\leq p To show that $l^p$ is of first category in $l^q$, I want to prove that $$\bar{B}(0,r)=\{\omega\in l^q:\lVert \omega\rVert_p\leq r\}$$ doesn't have interior points. Obviously these sets are closed. If that holds we can express $l^p$ in $l^q$ as $$l^p=\bigcup\limits_{n=1}^{\infty}\bar{B}(0, n).$$ So pick $x\in \bar{B}(0, r)$ and let $s>0$. We must find $y\in l^q$ s.t. $$\sum_{k=0}^{\infty}|y(k)-x(k)|^qr.$$ How can we find such $y$ and does the claim hold when $q=\infty$?
$l^p$ is dense proper subspace of first category in $l^q$ when $1\leq p
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functional-analysis
lp-spaces
baire-category
1 Answers
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The closure of $\ell^p$ for the $\ell^\infty$ norm is the space of sequences which go to zero as $n$ goes to infinity.
Suppose that the interior of $\overline{B}(0,n)$ is not empty: it contains a ball $B=\left\{x\in \ell^q,\lVert x-x_0\rVert\lt \delta\right\}$ for some $\delta>0$ and $x_0\in \ell^q$. This means that $$\tag{*} \forall x\in\ell^q,\quad \lVert x-x_0\rVert_q\lt \delta\Rightarrow\lVert x\rVert_p\leqslant n.$$ Let $y\in\ell^q$ be such that $\lVert y\rVert_q\lt \delta$. Then by (*) applied to $x=y+x_0$, we get that $\lVert y+x_0\rVert_p\lt n$. Since $x_0$ belong to $B$, it also belongs to $\overline{B}(0,n)$ hence $\lVert y\rVert_p\leqslant 2n$. But we can show that the norm $\ell^p$ and $\ell^q$ are not equivalent.