Le $\lambda_1 \ge \lambda_2 \ge \ldots \ge \lambda_d$ be positive numbers. For any $x \in \mathbb R^d$ and $r \ge 0$, define $\gamma(x,r) := \sup_{z \in \mathcal E(r)}x^\top z$, where $$ \mathcal E(r) := \mathcal E \cap B_2^d(r) = \{z \in \mathbb R^d \mid \sum_j z_j^2/\lambda_j \le 1,\, \|z\|_2 \le r\}, $$ the intersection of a hyper-ellipsoid $\mathcal E := \{z \in \mathbb R^d \mid \sum_j z_j^2/\lambda_j \le 1\}$ and the Euclidean ball $B_2^d(r)$ of radius $r$. Note that for any $r \ne 0$, the mapping $x \mapsto \gamma(x,r)$ defines a norm on $\mathbb R^d$. Finally, from a statistical perspective, note that if $x \sim N(0,I_d)$, then $\mathbb E \gamma(x,r)$ is the Gaussian width of $\mathcal E(r)$.

Also define $\omega(x,r) := \sqrt{\sum_{j=1}^d x_j^2/\lambda_j(r)}$, where $\lambda_j(r) := \min(\lambda_j,r^2)$.

Question. Are there absolute positive constants $c_1$ and $c_2$ such that $$ \gamma(x,r) \le c_2\omega(x,c_1 r), $$ for all $x \in \mathbb R^d$ and $r \ge 0$.

Note that the reverse inequality holds. Indeed, if $\omega(x,r) \le 1$, then $\sum_{j=1}^d x_j^2/\min(\lambda_j,r^2) \le 1$ and so

• $\sum_{j=1}^d x_j^2/\lambda_j \le 1$, i.e $x \in \mathcal E$, and
• $\sum_{j=1}^d x_j^2/r^2 \le 1$, i.e $x \in B_2^d(r)$,

i.e, $x \in \mathcal E(r)$. We conclude that $\gamma \ge \omega$.