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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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    Sorry, I had read $\max$ instead of $\min$. You actually want an estimate for *small* $z$, while I had thought you wanted it for *large* $z$.2012-09-17

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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.

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    Perfect. This is rigorously put!!2012-09-17
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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.