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How to prove that $\ell^p \setminus \cup_{1\leq q <p}\ell^q$ is not empty?

Here $\ell^p=L^p(\mathbb{N})$.

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  • $\begingroup$ Try using Baire's category theorem, along with the inclusions $\ell^q \subset \ell^r$ for $q<r$. $\endgroup$ – Jose27 Feb 21 '12 at 18:46
  • $\begingroup$ Note that $l^q$ is not closed in $l^q$ because $\{e_{n}\}_{1}^{\infty}$ is Schauder basis for both. $\endgroup$ – checkmath Feb 21 '12 at 19:50
  • $\begingroup$ What if a ball of $l_q$ is closed in $l_p$? $\endgroup$ – GEdgar Feb 22 '12 at 18:04
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If $p=+\infty$, just take $x(n)=1$ for all $n\in \mathbb{N}$. If $p=1$, there is nothing to prove. If $p\in(1,+\infty)$ consider $$ x(n)=\frac{1}{(n+2)^{1/p}\log(n+2)} $$ Thus we have $$ \Vert x\Vert_p= \left(\sum\limits_{n=1}^\infty\frac{1}{(n+2)\log^p(n+2)}\right)^{1/p}\leq \left(\int\limits_{0}^\infty\frac{du}{(u+2)\log^p(u+2)}\right)^{1/p}= $$ $$ \left(\int\limits_{2}^\infty\frac{du}{u\log^p u}\right)^{1/p}= \left(\int\limits_{2}^\infty\frac{du}{u\log^p u}\right)^{1/p}= \left((1-p)^{-1}\log^{1-p} 2\right)^{1/p}< +\infty $$ So, $x\in l^p$. On the other hand for all $q\in[1,p)$ we have $$ \Vert x\Vert_q= \left(\sum\limits_{n=1}^\infty\frac{1}{(n+2)^{q/p}\log^q(n+2)}\right)^{1/q}\geq \left(\int\limits_{1}^\infty\frac{du}{(u+2)^{q/p}\log^q(u+2)}\right)^{1/q}= \left(\int\limits_{3}^\infty\frac{du}{u^{q/p}\log^q u}\right)^{1/q} $$ Now,I will prove that the last integral is diverges. Since $p>q$, then $$ \lim\limits_{u\to\infty}\frac{u^{\frac{p-q}{2p}}}{\log^q u}=+\infty $$ and as the consequence there exist $u_0>3$ such that $$\frac{u^{\frac{p-q}{2p}}}{\log^q u}>1 $$ Then $$ \int\limits_{u_0}^\infty\frac{du}{u^{q/p}\log^q u}= \int\limits_{u_0}^\infty \frac{u^{\frac{p-q}{2p}}}{\log^q u}\frac{du}{u^{1-\frac{p-q}{2p}}}>\int\limits_{u_0}^\infty \frac{du}{u^{1-\frac{p-q}{2p}}}=+\infty $$ $$ \Vert x\Vert_q\geq \left(\int\limits_{3}^\infty\frac{du}{u^{q/p}\log^q u}\right)^{1/q}> \left(\int\limits_{u_0}^\infty\frac{du}{u^{q/p}\log^q u}\right)^{1/q}=+\infty $$ So $x\notin l^q$ for all $q\in[1,p)$. Hence $x\in l^p\setminus\left(\bigcup\limits_{1\leq q<p} l^q\right)$

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    $\begingroup$ Could you explain better the first inequality when you are showing that $x$ is not in $l^q$? $\endgroup$ – checkmath Feb 21 '12 at 19:25
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    $\begingroup$ Nice answer Norbert! I was checking the maths it is a very tricky way! I was just curious about what do you do. You have the tricks with that questions about $l^p$ how you did in the question of "Some Guy". Can we correspond ? $\endgroup$ – checkmath Feb 23 '12 at 2:38
  • $\begingroup$ Sorry, I don't understand your question properly. If you meant by correspond is writing letters to each other, then you can use my e-mail (see my profile). If meant by correspond unifying approaches to yours question and Some Guys question, then I think GEdgars method would be nice. $\endgroup$ – Norbert Feb 23 '12 at 6:04
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Other users gave an explicit example. Baire theorem, as suggested by Jose27, can be also used. We argue by contradiction. Define $$F_n:=\left\{x\in\ell^p\mid \sum_{k=1}^\infty|x_k|^{p-\frac 1n}\leqslant n\right\}.$$ Since convergence in $\ell^p$ implies pointwise convergence, each $F_n$ is closed and $\ell^p=\bigcup_nF_n$ (under the assumption $\bigcup_{1\leqslant q\lt p}\ell^q=\ell^p$) because if $x\in \ell^{p-\frac 1{n_0}}$, then $\lVert x\rVert_{p-\frac 1n}\leqslant\lVert x\rVert_{p-\frac 1{n_0}}$ for $n\geqslant n_0$ so take $n$ such that $n^{p-\frac 1n}\geqslant\lVert x\rVert_{p-\frac 1{n_0}}$.

By Baire categories theorem, we would get that $\ell^p$ is contained in some $\ell^{p_0}$ for some $p_0\lt p$, which is not possible.

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chessmath, are you related to Some Guy?
If not, see question Is there a null sequence that is in not in $\ell_p$ for any $p<\infty$? and use the fact in my answer there.

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  • $\begingroup$ That one is the particular case when $p=\infty$. $\endgroup$ – checkmath Feb 22 '12 at 23:36
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Hint: Try to construct an element of the form $(n^\alpha)_{n\in\mathbb{N}}$ for a well chosen $\alpha\in\mathbb{R}$.

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  • $\begingroup$ It doesn't work properly. $\endgroup$ – checkmath Feb 21 '12 at 19:51

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