# Upper bound on the size of vectors contained in an ellipsoid?

Crossposted at Math SE

Consider the diagonal matrix

$$D=\left[\begin{array}{cccc} 1^{- 2 p} & 0 & \cdots & 0 \\ 0 & 2^{-2 p} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & n^{-2 p} \end{array}\right]$$

and the ellipsoid

$$\mathcal{E}_D = \left\{\theta \in \mathbb{R}^{n}: \theta^{\top} D^{-1} \theta \leq 1\right\}$$

I need to prove the following bound. I am pretty confused on how to start proving it.

For all $$\theta \in \mathcal{E}_D$$, $$\forall j \in [n]$$, we have $$\left|\theta_{[j]}\right| \lesssim \frac{1}{j^{p+\frac{1}{2}}}$$

where $$\theta_{[j]}$$ is $$j-th$$ the largest entry in absolute value, i.e., $$\left|\theta_{[1]}\right| \geq \cdots \geq \left|\theta_{[n]}\right|$$.

Denote $$|\theta_{[j]}|=s$$. Replace $$j$$ largest (in absolute value) coordinates of $$\theta$$ to $$s$$, other coordinates to 0. The vector remains in ellipsoid $$\mathcal{E}_D$$. If the coordinates equal to $$s$$ have indices $$m_1, we get $$s^2(m_1^{2p}+\ldots+m_j^{2p})\leqslant 1.$$ If $$p\geqslant 0$$, this yields $$s^2(1^{2p}+\ldots+j^{2p})\leqslant 1,$$ and since $$1^{2p}+\ldots+j^{2p}\geqslant\frac1{2p+1} j^{2p+1}=\int_0^j x^{2p}dx$$ we get $$s\leqslant \sqrt{2p+1}j^{-p-1/2}.$$
If $$p<0$$, you can not get the bound which does not depend on $$n$$, look at the vector $$(0,0,\ldots,n^{-p})$$.