Take the 2-minute tour ×
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.

Suppose that a sequence $x=(x_n)$ belongs both to $\ell^p$ and $\ell^q$ ($p,q>1$, $p\neq q$). Is there any inequality between $\|x\|_p$ and $\|x\|_q$. Can one $\ell^p$ be continuously embedded into another $\ell^q$?

share|improve this question
    
The map $p\mapsto ||x||_p$ is decreasing. –  leo Oct 1 '11 at 22:24
    
You do not need the restriction $p,q>1$. See math.stackexchange.com/questions/4094/… –  AD. Oct 26 '11 at 8:53

1 Answer 1

up vote 13 down vote accepted

If $1 \leq p \leq q \lt \infty$ then $\|x\|_{q} \leq \|x\|_{p}$ and clearly $\|x\|_p \geq \|x\|_\infty$. In particular, $\ell^p$ embeds continuously into $\ell^q$ whenever $p \leq q$.

To see this, note that both sides of the inequality $\|x\|_{q} \leq \|x\|_{p}$ are homogeneous in $x$ (multiplying $x$ with a positive real number multiplies both sides with the same positive factor), so we may take without loss of generality an $x$ with $\|x\|_{p} = 1$. Then $\|x\|_{q}^{q} = \sum_{j = 1}^{\infty} |x_{j}|^{q} \leq \sum_{j = 1}^{\infty} |x_{j}|^{p} = 1$, and this is because for $t \leq 1$ and $p \leq q$ we have $t^{q} \leq t^{p}$.

This means that $p \mapsto \|x\|_p$ is decreasing. In terms of spaces, we have the inclusions $\ell^1 \subset \ell^p \subset \ell^q \subset \ell^{\infty}$ whenever $1 \lt p \lt q \lt \infty$ and it is not hard to show that the inclusions are all strict.

Note that this is opposite to the case of finite measure spaces $(\Omega,\mu)$, where the inclusions go the other way around: $L^1(\Omega,\mu) \supset L^p(\Omega,\mu) \supset L^{q}(\Omega,\mu) \supset L^{\infty}(\Omega,\mu)$.

See also the Wikipedia page on $L^p$-spaces.

share|improve this answer
2  
This is a great answer –  leo Oct 1 '11 at 23:02
    
Well, thanks! :) –  t.b. Oct 1 '11 at 23:07

Your Answer

 
discard

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.