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 $\gamma(x,r) := \sup_{z \in E(r)}x^\top z$, where $$ E(r) := E \cap B_2^d(r) = \{z \in \mathbb R^d \mid \sum_j z_j^2/\lambda_j \le 1,\, \|z\|_2 \le r\}, $$ the intersection of a hyper-ellipsoid $E := \{z \in \mathbb R^d \mid \sum_j z_j^2/\lambda_j \le 1\}$ 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 $E(r)$.
Also define $D(r) := \{x \in \mathbb R^d \mid \sum_{j=1}^d x_j^2/\lambda_j(r) \le 1\}$, where $\lambda_j(r) := \min(\lambda_j,r^2)$. Let $x \mapsto \omega(x,r)$ be the support function of $D(r)$.
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 $x \in D(r)$, 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 $x \in E$, and
- $\sum_{j=1}^d x_j^2/r^2 \le 1$, i.e $x \in B_2^d(r)$,
i.e, $x \in E(r)$. Thus, $D(r) \subseteq E(r)$, and we conclude that $\gamma \ge \omega$.
Edit: Solution
We show that the claimed bound holds with $c_1=1$ and $c_2=\sqrt 2$. For this, it suffices to show that $E(r) \subset \sqrt{2} D(r)$. Indeed, let $x \in E(r)$. Then, $$ \begin{split} \sum_j x_j^2/\lambda_j(r) &= \sum_j x_j^2/\min(r^2,\lambda_j) = \sum_j x_j^2 \max(1/r^2,1/\lambda_j)\\ & \le \sum_j x_j^2 (1/r^2 + 1/\lambda_j) = \sum_j x_j^2/r^2 + \sum_j x_j^2/\lambda_j \le 1 + 1 = 2. \end{split} $$ Thus, $x \in \sqrt{2} D(r)$.
