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
    $\begingroup$ A similar question is here $\endgroup$ – David Mitra Feb 29 '12 at 1:25
  • $\begingroup$ I think you want $\Vert\ \cdot\ \Vert_q\le \Vert\ \cdot\ \Vert_p$. $\endgroup$ – David Mitra Feb 29 '12 at 1:33
  • $\begingroup$ Right, thanks for the ref., let me rewrite. $\endgroup$ – AQP Feb 29 '12 at 2:34

Let $x\in \ell^p$ and $0<p<q<+\infty$. If $x=0$, then everything is obvious. Otherwise consider $e=\frac{x}{\Vert x\Vert_p}$. For all $k\in\mathbb{N}$ we have $|e_k|\leq 1$ and $\Vert e\Vert_p=1$. Now since $p<q$ we get $$ \Vert e\Vert_q= \left(\sum\limits_{k=1}^\infty |e_k|^q\right)^{1/q}\leq \left(\sum\limits_{k=1}^\infty |e_k|^p\right)^{1/q}= \Vert e\Vert_p^{p/q}=1 $$ Then we can write $$ \Vert x\Vert_q=\Vert \Vert x\Vert_p e\Vert_q=\Vert x\Vert_p\Vert e\Vert_q\leq\Vert x\Vert_p $$ In fact this inequality means that $\ell^p\subseteq \ell^q$. Also we can exclude the equality sign in this inclusion, because the sequence $x(k)=k^{-\frac{1}{p}}$ belongs to $\ell^q$ but not to $\ell^p$. If we assume that $p\geq 1$, we can speak of normed spaces $\ell^p$ and $\ell^q$. Then the last inequality means that the natural inclusion $i:\ell^p\to \ell^q:x\mapsto x$ is a continuous linear operator.

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.

  • 2
    $\begingroup$ Just a remark regarding the example showing $\ell^q\ne\ell^p$, I think $x(k)=k^{-1/p}$ is more transparent, no? $\endgroup$ – AD. Feb 20 '14 at 21:27

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