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I'm trying to show the inclusion :

$\ell^p\subseteq\ell^q$ for real-value sequences, and show that the norms satisfy: $\|\cdot\|_q<\|\cdot\|_p$.

I think I can show the first part without much trouble:

Take $a_n$ in $\ell^p$, then the partial sums are a Cauchy sequence, i.e., for any $\epsilon>0$ , there is a natural $N$ with $|S_{n,p}-S_{k,p}|<\epsilon$ for $n,k>N$, and $S_{n,p}$ the partial sums of $|a_n|^p$ and the individual terms go to $0$. So, we choose an index $J$ with $a_j<1$ for $j>J$. We then use that $f(x)=a^x$ decreases in $[0,1]$. This means that $|a_j|^p<|a_j|^q$.

So the tail of $S_{n,q}$, the partial sums of $|a_n|^q$ decrease fast-enough to converge, by comparison with the tail of $S_{n,p}$.

But I'm having trouble showing $\|\cdot\|_q<\|\cdot\|_p$ . Also, is there a specific canonical embedding between the two spaces?

  • 3
    A similar question is [here](http://math.stackexchange.com/questions/4094/how-do-you-show-that-l-p-subset-l-q-for-p-leq-q)2012-02-29
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    I think you want $\Vert\ \cdot\ \Vert_q\le \Vert\ \cdot\ \Vert_p$.2012-02-29
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    Right, thanks for the ref., let me rewrite.2012-02-29

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Let $x\in \ell^p$ and $0

It is worth to note that the inequality $\Vert\cdot\Vert_p\leq C\Vert\cdot\Vert_q$ is impossible for any constant $C\geq 0$. Indeed consider sequences $$ x_n(k)= \begin{cases} 1,\qquad 1\leq k\leq n\\ 0,\qquad k>n \end{cases} $$ Then $$ C\geq\lim\limits_{n\to\infty}\frac{\Vert x_n\Vert_p}{\Vert x_n\Vert_q}=\lim\limits_{n\to\infty}n^{\frac{1}{p}-\frac{1}{q}}=+\infty. $$ Therefore such a constant $C>0$ doesn't exist.

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    Just a remark regarding the example showing $\ell^q\ne\ell^p$, I think $x(k)=k^{-1/p}$ is more transparent, no?2014-02-20