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 $m_1<m_2<\ldots<m_j$, 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 $(0,0,\ldots,n^{-p})$.


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