3
$\begingroup$

For nonnegative $x \in \mathbb{R}_+^n$, $p \geq 1$ I want to find $\epsilon_{n,p}$ such that

$ \frac{n^{1/p}}{||1/x||_p} \geq \min (x) \geq \frac{n^{1/p}}{||1/x||_p} - \epsilon_{p,n} ||x||_\infty $

Here is how far I got. Without loss of generality, take $||x||_\infty=1$, furthermore, let $x$ be the vector $(m,1,...,1)$, where $m$ is the minimum element of $x$. Then our goal is to find the maximum value of

$\epsilon_{n,p}(m) = \left(\frac{n}{p - 1 + m^{-p}}\right)^{1/p} - m$

What is an upper bound on $\epsilon_{n,p}(m)$ on $m \in [0,1]$?

See this plot for n=3, p=3

1 Answers 1

3

[Edit to answer the modified question:]

For $p=1$, $\epsilon_{n,p}(m)=(n-1)m$ and there is no upper bound on $\epsilon$, so I will assume $p>1$.

The derivative of this new version of $\epsilon$ with respect to $m$ is

\epsilon'_{n,p}(m) = n^{1/p}m^{-(p+1)}\left(p - 1 + m^{-p}\right)^{-(p+1)/p} - 1\;.

Setting this to zero yields

$n^{1/p}m^{-(p+1)}\left(p - 1 + m^{-p}\right)^{-(p+1)/p}=1\;,$

$n^{-1/(p+1)}m^{p}\left(p - 1 + m^{-p}\right)=1\;,\tag{1}$

$(p - 1)m^p + 1=n^{1/(p+1)}\;,$

$m=\left(\frac{n^{1/(p+1)}-1}{p-1}\right)^{1/p}\;.\tag{2}$

Then using (1) to substitute $p - 1 + m^{-p}$ and (2) to substitute $m$ yields

$ \begin{eqnarray} \epsilon_{n,p}(m) &=& \left(n^{1/p}n^{-1/(p(p+1))}-1\right)m \\ &=& \left(n^{1/(p+1)}-1\right)\left(\frac{n^{1/(p+1)}-1}{p-1}\right)^{1/p} \\ &=& \left(\frac{\left(n^{1/(p+1)}-1\right)^{p+1}}{p-1}\right)^{1/p}\;. \end{eqnarray} $

For $n=3$, $p=3$, this is about $0.171$, in agreement with your plot. To show that this is always a global maximum, note that $\epsilon_{n,p}(m)$ goes to $0$ for $m\to0$ and to $-\infty$ for $m\to\infty$, so if the derivative vanishes at a single point, it must be a global maximum.

  • 0
    Noted, and thanks for the answer.2011-04-29