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sheet2 B2 $l^p$ is dense subset of category 1 in $l^q$

 
  • Jan 19, 2024
Let $1≀pProof. since $|x_k|^q<|x_k|^p$ for $k$ large, $\sum_k|x_k|^p<∞\implies\sum_k|x_k|^q<∞$, so $l^p(ℝ)βŠ†l^q(ℝ)$. Consider $A_n≔\{x ∈ l^p:β€–xβ€–_p ≀ n\}$. Want to show $A_n$ is nowhere dense in $l^q$. Thus $l^p=\bigcup_{n=1}^∞A_n$ is of first category in $l^q$. Wlog $n=1$. To prove $A_1$ is closed, if a sequence $x_n∈A_1$ converges to $x∈l^q$, for any $iβˆˆβ„•$, ${|x^{(i)}-x_n^{(i)}|}≀\|x-x_n\|_q β†’ 0$, so $x_n^{(i)} β†’ x^{(i)}$ as $n β†’ ∞$. By Fatou's lemma [Proof: $βˆ€N,n^pβ‰₯β€–x_nβ€–_p^pβ‰₯\sum_{k≀N}|x_k^{(j)}|^pβ†’\sum_{k≀N}|x_k|^p$.] \[\|x\|_p≀\lim_{nβ†’βˆž}\|x_n\|_p≀1\]So $A_1$ is closed. To prove $\mathring{A_1}=βˆ…$ in $l^q$, for any $x ∈ A_1$, for any $r$, we show $B_r(x)βŠ„l^p$: Set $y_j=\frac1{j^{1/p}},y∈ l^q βˆ– l^p$, $β€–yβ€–_q=1$, then $\frac{r}{2}y+x$ is in $B_r(x)$ but not in $l^p$, since $yβˆ‰l^p$. As $r>0$ is arbitrary, $xβˆ‰\mathring{A_1}$ and thus $\mathring{A_1}=βˆ…$. In general every proper subspace has empty interior (2) $l^p$ is dense in $(l^q(ℝ),β€–β‹…β€–_{l^q})$. Proof: The space $c_{00}=\{(x_n):x_n=0 \text{ all but finitely many }n\}$ is dense in $l^q$ $βˆ€1≀q<∞$. Obviously $c_{00}βŠ‚ l^p$ so $l^p$ contains dense subset and is thus dense in $l^q$. %https://math.stackexchange.com/questions/2157682/