show for$a\in\left(0,1\right)$ $x^{-a}$ is not self concordant s.t $0\leq x$ 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.

We have $f(x) = x^{-\alpha}, \alpha \in (0,1), x>0$. Therefore
$$\begin{align} 
& f'(x) = -\alpha x^{-\alpha-1}\\
& f''(x) = \alpha(\alpha + 1) x^{-\alpha-2} \\
& f'''(x) = -\alpha(\alpha + 1)(\alpha + 2)x^{-\alpha-3}
\end{align}$$
Base on the definition of self concordant according to wiki (basically the "convex optimization" book by Boyd and Vandenberghe, chapter 9), we must have
$$|f'''(x)|\le 2|f''(x)|^{\frac{3}{2}}$$
We form the quotient $Q(x)=\frac{|f'''(x)|}{ 2|f''(x)|^{\frac{3}{2}}}$ as follow
$$Q(x) = \frac{\frac{\alpha(\alpha + 1)(\alpha + 2)}{x^{\alpha+3}}}{2{\left(\frac{\alpha(\alpha + 1)}{x^{\alpha+2}}\right)^{\frac{3}{2}}}} = \frac{(\alpha + 2)x^{\frac{1}{2}\alpha}}{2\sqrt{\alpha(\alpha + 1)}} 
$$
By taking derivative of $Q(x)$, as $Q'(x) = \frac{(\frac{1}{2}\alpha)(\alpha + 2)x^{\frac{1}{2}\alpha-1}}{2\sqrt{\alpha(\alpha + 1)}} = 0$, we see that derivative of $Q$ at origin is unbounded below, in other words $x=0$ is asymptote of $Q(x)$ at origin and $Q(x) \ge 1$ for a sufficiently large $x$. $f(x)\; \color{red}{\text{is not self-concordant!}}$
However if you put $h(x) = -\ln(-f(x)) -\ln(x)$, then $h(x)$ is self-concordant.
