Let $K \subset \mathbb{R}^n$ denote a centrally symmetric convex body, with corresponding norm $\|\cdot\|_K$.
We define the parameters $$ M(K) = \mathbb{E} [\|x\|_K] \quad \mbox{and} \quad b(K) = \sup_{\|x\|_2 = 1} \|x\|_K. $$ The expectation is taken with respect to $x$ drawn uniformly on $\mathbb{S}^{n-1}$.
We say that a subspace $E$ is an $\epsilon$-approximately Euclidean subspace in $K$ if it holds that
$$
(1-\epsilon)\, B_E\subset M(K) \, [K \cap E] \subset (1+\epsilon) \, B_E, \qquad (*)
$$
where $B_E = B_2^n \cap E$ is the Euclidean ball in $E$.
The following result is a well-known version of Dvoretsky theorem.
Theorem.$~~~$ For every $\epsilon \in (0, 1)$, and every centrally symmetry convex body $K \subset \mathbb{R}^n$, with probability $1/2$, a uniformly random $k$-dimensional subspace $E$ satisfies (*), provided that $k \leq c \, \epsilon^2 n (M/b)^2$, where $c > 0$ is a universal constant.
This theorem motivates the following definition: for a convex body $K \subset \mathbb{R}^n$, we set $$ k(K, \varepsilon) = \max\Big\{1 \leq k \leq n \mid \nu_{n,k}\{E \in G_{n,k} : E~\text{satisfies}~(*)\} \geq 1/2\,\Big\}. $$ Above $G_{n,k}$ is the set of $k$-dimensional subspaces in $\mathbb{R}^n$, and $\nu_{n,k}$ the corresponding uniform measure. The randomized form of Dvoretzky above clearly implies that $$ k(K, \varepsilon) \geq c\, \varepsilon^2 n \Big(\frac{M(K)}{b(K)}\Big)^2 $$
Question: Is this sharp, for $K = B^n_1$? In this case, computing $M(K) \asymp b(K) =\sqrt{n}$, we see that $k(B^n_1, \varepsilon) \geq c_1 \, \varepsilon^2 n$, where $c_1 > 0$ is a universal constant But can this inequality be reversed up to universal constants: is it true that $k(B^n_1, \varepsilon) \leq c_2 \, \varepsilon^2 n$, for another universal constant $c_2 > 0$?
