Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to show the inclusion :

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

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?

share|cite|improve this question
A similar question is here – David Mitra Feb 29 '12 at 1:25
I think you want $\Vert\ \cdot\ \Vert_q\le \Vert\ \cdot\ \Vert_p$. – David Mitra Feb 29 '12 at 1:33
Right, thanks for the ref., let me rewrite. – 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 $|e_k|<1$ 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 series $x(k)=k^{-\frac{1}{p}}$ 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 series $$ x_n(k)= \begin{cases} 1,\qquad 1\leq k\leq n\\ 0,\qquad k>n \end{cases} $$ 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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.