0
$\begingroup$

Denote $\|\cdot \|_p$ for the norm in $\ell_p^n$, where $1 \leq p \leq \infty$, and $n \geq 1$. Let $(x^\star_i)$ denote a nonincreasing arrangement of the sequence $(|x_i|) \in \mathbb{R}^n$.

We define $$h_{p, n, \delta}(x) = \sup_{\|y\|_2 \leq \delta, \|y\|_p \leq 1} \langle x, y \rangle, $$ which is the support functional of the convex body $\delta B_2^n \cap B_p^n$.

Fix $p \in [1, 2]$ and $\delta > 0$. Define \begin{gather*} C(p, n, \delta) = \inf_{x \neq 0} \frac{h_{p, n, \delta}(x)}{\delta \|x_{\leq m}^\star\|_2 + \|x_{\geq m}^\star\|_q}, \quad \mbox{where} \quad \\[1.5ex] \frac{1}{q} + \frac{1}{p} = 1, \quad \frac{1}{\alpha} = \frac{1}{p} - \frac{1}{2}, \quad \mbox{and} \quad m = [\delta^{-\alpha}]. \end{gather*} Here, we denote $v_{\leq t}$ (resp., $v_{\geq t}$) for the first $t$ (resp., last $n - t + 1$) coordinates of $v \in \mathbb{R}^n$.

Question: By Hölder's inequality, $C(p, n, \delta) \leq 1$. On the other hand, [1, Thms 4.1 and 4.2] imply $C(p, n, \delta) > 0$, independently of $n$. Are more explicit and quantitative estimate possible? Specifically:

(i) What is a (ideally, simple/quantitative) proof of $\inf_{n \geq 1} C(p, n, \delta) > 0$?

(ii) Can $C(p, n, \delta)$ or $\inf_{n \geq 1} C(p, n, \delta)$ be computed explicitly? If not, can we have more explicit, multiplicative approximations of them?

References:

[1] Holmstedt, Tord. Interpolation of quasi-normed spaces. Math. Scand. 26 (1970), 177--199. MR0415352

$\endgroup$
4
  • $\begingroup$ I don't really understand your question (i), surely by just picking literally any $x$ with all coordinates non-zero (e.g. $x = (1, 1, \ldots , 1)$), we get a positive lower bound for $C(p, n, \delta)$? $\endgroup$ Commented Jul 30 at 8:12
  • $\begingroup$ Well, I mean not that it changes much, both numerator and denominator are still proportional to $||x||$ (any norm since all norms are equivalent): $h_{p, \delta}(x)$ is literally a norm, and for the thing in the denominator it is easy to prove that it is equivalent to $\ell^n_1$ norm, say. Your second question is of course much harder. $\endgroup$ Commented Jul 30 at 8:52
  • $\begingroup$ Oh, well, you did not specify that you want a lower bound independent of $n$, this seems like a very big thing to omit from the post... This I don't know how to establish. $\endgroup$ Commented Jul 30 at 8:58
  • $\begingroup$ I did say that I was looking for a quantitative estimate, however. Does your approach yield such an estimate? $\endgroup$
    – Drew Brady
    Commented Jul 30 at 8:58

0

You must log in to answer this question.

Browse other questions tagged .