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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 $\ell^q$?

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

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  • 3
    $\begingroup$ This is a great answer $\endgroup$ – leo Oct 1 '11 at 23:02
  • $\begingroup$ Where did you defined $ {x}_{j} \leq 1 $? $\endgroup$ – Royi Oct 27 '15 at 9:30
  • $\begingroup$ @Royi If a sum of non-negative terms is $1$ then every term is $0 \leq |x_j| \leq 1$. $\endgroup$ – rubik Oct 10 '16 at 20:38
  • $\begingroup$ Why does the summation go to infinity? $\endgroup$ – PaulrBear Feb 3 at 5:54

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