TITLE: show for$a\in\left(0,1\right)$ $x^{-a}$ is not self concordant s.t $0\leq x$
QUESTION [1 upvotes]: I want to show for $a\in\left(0,1\right)$ $x^{-a}$ is not self concordant s.t. $x > 0$. By definition, a function is called self-concordant function if $\left|\left(x^{-\alpha}\right)^{\prime\prime\prime}\right|\leq2\cdot\left(\left(x^{-\alpha}\right)^{\prime\prime}\right)^{\frac{3}{2}}$.
So what I need to prove is that $$\left|-ax^{-a-3}\left(-a-1\right)\left(-a-2\right)\right|\leq2\cdot\left(-ax^{-a-2}\left(-a-1\right)\right)^{\frac{3}{2}}$$
doesn't hold.
I got to $-x^{-a-3}\left(-a-1\right)\leq2\cdot\left(ax^{-a-2}\right)\sqrt{\left(-ax^{-a-2}\left(-a-1\right)\right)}$, but I don't know how to continue from here.

REPLY [1 votes]: Your approach is correct. In fact, the inequality you have obtained can be expressed as
$$
{
a(a+1)(a+2)x^{-a-3}\le 2a^\frac{3}{2}(a+1)^\frac{3}{2} x^{-\frac{3}{2}a-3}
\implies
\\
a+2\le 2a^\frac{1}{2}(a+1)^\frac{1}{2} x^{\frac{3}{2}a}
}
$$
which does not hold for all $x>0$ since for sufficiently small $x$, $LHS>RHS$ and for sufficiently large $x$, $LHS<RHS$. However, the function is self-concordant over
$
\left[\frac{(a+2)^\frac{2}{3a}}{(4a(a+1))^\frac{1}{3a}},\infty\right)
$.
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