Is $f(t)=t^\alpha$ for $\alpha\in(0,1)$ a sub-additive function?

Question

Possible Duplicate:
Does $|x|^p$ with $0<p<1$ satisfy the triangular inequality on $\mathbb{R}$?

Is the function

$$f(t)= t^{\alpha},\quad \alpha\in (0,1)$$

a subadditive function?

My teacher said categorically that this is true. But I'm not so sure.

EDIT: $~~~~~0\leq t\leq 1$

Try with $0\le t\le 1$. – Vittorio Patriarca Jun 05 '12 at 20:49 — Vittorio Patriarca, Jun 05 '12 at 20:49

Davide Giraudo · Accepted Answer · 2012-06-05T21:01:55.593

2

By homogeneity, in the inequality $(x+y)^{\alpha}\leq x^{\alpha}+y^{\alpha}$ just deal with the case $y=1$. Let $f(t):=t^{\alpha}+1-(t+1)^{\alpha}$. The derivative has the sign of $t^{\alpha-1}-(t+1)^{\alpha-1}$ which is non-negative since $\alpha<1$. Hence $f(t)\geq f(0)=0$ and we are done.

edited Jun 05 '12 at 21:01

answered Jun 05 '12 at 20:55

Davide Giraudo

181,608

2

The notation $a^\alpha$ is a bit confusing, you should consider using $x,y$ or $u,v$ instead methinks. – Asaf Karagila Jun 05 '12 at 21:00
That's true. Thanks! – Davide Giraudo Jun 05 '12 at 21:02

Is $f(t)=t^\alpha$ for $\alpha\in(0,1)$ a sub-additive function?

1 Answers1

Linked