Let $k = (k_1, \cdots, k_d) \in \mathbb{N}^d$ and $s = (s_1, \cdots, s_d) \in \mathbb{R}_+^d$. define $$k^s = (k_1^{s_1},\cdots,k_d^{s_d})\in \mathbb{R}_+^d.$$ Let $|\cdot|_p$ denote the $p$-norm on Euclidean space $\mathbb{R}^d$. Under the constraint $|k|_2 \ge n>0$ ($p = 2$ is somehow conventional only, as all norms are equivalent), what is the upper bound (possibly with a dimension related constant) for $$\sup_{|k|_2 \ge n} |k^{-s}|_2,$$ where $k^{-s} = (k_1^{-s_1},\cdots,k_d^{-s_d})$ and $s_1, \cdots, s_d >0$.
A bit of background. The question rises in an attempt to prove some embedding theorem for (anisotropic) Sobolev space. When $s_1 = \cdots = s_d$, one can see that the upper bound actually gives the error bound on the approximation in the span of the first $n$ bases.
For each $s$ we have
$$|k^{-s}|_2=\left|\sum k_i^{-2s_i}\right|_2\le \left|\sum k_i^{0}\right|_2=\sqrt{d}.$$
On the other hand, put $k=(1,1,\dots,1, \lceil\sqrt{n}\rceil)$. Then $|k|_2\ge n$, but
$$|k^{-s}|_2>\left|\sum_{i=1}^{d-1} 1^{-2s_i}\right|_2=\sqrt{d-1}.$$