Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
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

2 Answers 2

up vote 0 down vote accepted

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.

share|improve this answer
    
I can't understand why every answer is downvoted. –  Siminore Sep 17 '12 at 16:38
    
Perfect. This is rigorously put!! –  Ush Sep 17 '12 at 16:45

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.