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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., $\left|\theta_{[1]}\right| \geq \cdots \geq \left|\theta_{[n]}\right|$.

_{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., $\left|\theta_{[1]}\right| \geq \cdots \geq \left|\theta_{[n]}\right|$.

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

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., $\left|\theta_{[1]}\right| \geq \cdots \geq \left|\theta_{[n]}\right|$.

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. 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 the largest entry in absolute value, i.e., $\left|\theta_{[1]}\right| \geq \cdots \geq \left|\theta_{[n]}\right|$. I am pretty confused on how to start showing this.

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