# Bounds for $z\log(z)$ for $z\geq 0$

How to show that, for $\alpha<1$ it holds: $|\min\left(z\log z,0\right)|\leq Cz^{\alpha}$, $\forall z\geq 0$ for an appropriate $C<\infty$.

This pops up in the convergence results given in the JKO paper for gradient flows.

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Try to be precise: is $\alpha$ fixed? Do you want to show that there exists some $\alpha$ such that...? – Siminore Sep 17 '12 at 12:42
More like for any $\alpha<1$, we can find a a $C<\infty$ dependent on just $\alpha$ such that the inequality is satisfied for all $z\geq 0$. Hope that makes it clear. – Ush Sep 17 '12 at 12:54
I can't understand: $\lim_{z \to +\infty} z^{1-\alpha}\log z = +\infty$. – Siminore Sep 17 '12 at 12:57
@Siminore For any $z\ge 1$ the LHS is identically zero so $z\to\infty$ doesn't matter. – Erick Wong Sep 17 '12 at 13:04
@Siminore Your comment is not very clear to me. If you plot $|min(z \log z,0)|$ (note we are talking about minimum so the graph is like a inverted parabola from $x=0$ to $x=1$ and $0$ after $x=1$) the question becomes clear!! – Ush Sep 17 '12 at 13:06

So, as $z\ge 0$ and $\log z<0 \iff z\in (0,1)$, we want to prove that $$\forall \alpha<1 \ \exists C: \forall z\in (0,1): \ -z\log z \le Cz^\alpha$$ Set $x:=1/z$, now it is $>1$, then we want to prove $$\log x = -\log z \le C z^{\alpha-1} = Cx^{(1-\alpha)}$$ Set $\beta := 1-\alpha \ >0$, then using the fact that $\displaystyle{\lim_{x\to \infty}\frac{\log x}{x^\beta} = 0}$, we conclude that it is upper bounded.
For $z$ a positive real and $a$ a real $< 1$ , $z log(z) < C z^a$ because logs grow slower than roots. Analogue for complex z , write out the real and imaginary part and keep in mind that $exp$ has a period of $2 \pi i$ ( how does that make the branches of $log(z)$ look like ? ). That way you should be able to find the answer yourself.