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Given $n$, is there a $C(n)$-isomorphic embedding of $l^n_{\infty}$ into $l^m_1$ for sufficiently large $m$ and $C(n)<<\log(n)$?

For $n=2$ this can done with $m=2$. There are some results about $(1+\epsilon)$-isometric embedding of $l^n_p$ into $l^m_1$ for $p\leq 2$ but I couldn't find anything for $p>2$.

We have an easy isometric embedding in the reverse direction though, $l^n_1 \to l^{2^n}_{\infty}$.

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    $\begingroup$ I guess you want to say $l_1^n\to l_\infty^{2^n}$. $\endgroup$
    – Dirk Werner
    Commented Sep 10, 2022 at 18:48
  • $\begingroup$ @Dirk Werner Yes, thanks. I have edited it. $\endgroup$
    – Arun
    Commented Sep 11, 2022 at 6:14

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No. Best embedding constant of $\ell^k_{\infty}$ into $L_1$ is of the order $\sqrt{k}$. This follows from the facts that $L_1$ has cotype 2 while the cotype 2 constant of $\ell^k_{\infty}$ is $\sqrt{k}$. See Tomczak-Jaegermann's book for these notions and more.

Once we have these, the argument is a simple calculation. Let $T:\ell^k_{\infty}\to L_1$ be a linear embedding, and $(e_i)_1^k$ denote the unit vectors in $\ell^k_{\infty}$, and $T(e_i)=x_i$. Then $\frac{1}{\|T^{-1}\|}\le\|x_i\|\le \|T\|$. We have $$\frac{1}{2^k}\sum_{\pm}\|\sum_{i=1}^k \pm e_i\|=1$$ But $$\frac{1}{2^k}\sum_{\pm}\|\sum_{i=1}^k \pm x_i\|\ge \frac{\sqrt k}{C\cdot \|T^{-1}\|}$$ where $C$ is the cotype 2 constant of $L_1$.

It follows that $\|T\|\|T^{-1}\|\ge \frac{\sqrt k}{C}$.

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    $\begingroup$ Meanwhile, I learned from Gideon Schechtman that the best possible value for the constant C is $\sqrt{2}$, and further, the lower bound $\sqrt(k$)/C$ can be achieved. $\endgroup$
    – Arun
    Commented Sep 14, 2022 at 9:02

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