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For any sequence $\omega\in[1, \infty)^{\mathbb{N}}$, define the weighted $\ell^p_\omega$-norm of the sequence $v$ by $$\Vert v\Vert_{\ell^p_\omega} := \left(\sum_{k=1}^\infty \omega_k^p |v|_k^p\right)^{1/p} .$$

My question is: If $v\in\ell^q$ and $p>q$, under which conditions on $\omega$ will $v\in\ell^p_\omega$?

Obviously, since $p>q$, we can use Hölder‘s inequality to bound $$ \Vert v\Vert_{\ell^p_\omega}^p = \Vert \omega^p v^{p-q}v^q\Vert_{\ell^1} \le \Vert \omega^p v^{p-q}\Vert_{\ell^\infty}\Vert v^q\Vert_{\ell^1} = \Vert \omega^{p/(p-q)} v\Vert_{\ell^\infty}^{p-q} \Vert v\Vert_{\ell^q}^q . $$ This gives the condition $\Vert\omega^{p/(p-q)}v\Vert_{\ell^\infty} <\infty$. (Here, all operations on the sequences have to be understood element-wise.)

  • But is this condition necessary?
  • Moreover, can we find an unbounded increasing sequence $\omega$ (preferably algebraically increasing like $\omega_k=k^s$ for $s=\frac1q-\frac1p$) such that $\ell^q\subseteq\ell^p_\omega$?
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  • $\begingroup$ I don’t understand how. Since $p>q$, it always holds that $\Vert v\Vert_{\ell^p_1} = \Vert v\Vert_{\ell^p} \le \Vert v\Vert_{\ell^q}$. $\endgroup$
    – Philipp Trunschke
    Commented Oct 12, 2022 at 8:01

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The answer to both questions is no.

  • Taking $q=1$, $p=2$, $v_k = k^{-(1+\varepsilon)}$ and $\omega_k = k^{(1+3/2\varepsilon)/2}$ for some $\varepsilon>0$ provides a counter-example to the first assertion.
  • This answer provides a counter-example for the second assertion when $\omega$ increases algebraically. It can be generalized to any explicitly given sequence $\omega$.
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