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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$|x|^{r-1} \leq |x|^r + 1$ of a convex function?


share|cite|improve this question
What do you mean by a convex function here? – Ilya Jan 9 '13 at 15:24
I mean $|x|^{p}$ is a convex function. – RHS Jan 9 '13 at 15:25
Follow up question… – RHS Jan 10 '13 at 7:18
up vote 0 down vote accepted

First of all, if $r\lt1$, then multiplying the inequality by $|x|^{1-r}$ $$ 1\le|x|+|x|^{1-r} $$ which is false near $0$. So let's assume that $r\ge1$, then $$ |x|^{r-1}\le1\quad\text{if }|x|\le1 $$ and dividing both sides by $|x|^{r-1}$ yields $$ |x|^{r-1}\le|x|^r\quad\text{if }|x|\ge1 $$ Therefore, if $r\ge1$, $$ |x|^{r-1}\le|x|^r+1 $$

share|cite|improve this answer
It is hard to choose one, but I like this one with truncation better. – RHS Jan 10 '13 at 6:47

First of all, for $r>0$ we have that $|x|^r\leq 1$ iff $|x|\leq 1$. Moreover, if $|x|>1$ then $$ |x|^r = |x|\cdot|x|^{r-1}>|x|^{r-1}. $$ As a result, whenever $r-1>0$ (which is $r>1$) we have that $|x|^{r-1}\leq \max(1,|x|^r)\leq1+|x|^r$

share|cite|improve this answer
What is the above method you are using called? Is it very common? Where can I find more about it? Could you also show how to apply the Jensen's inequality here? – RHS Jan 9 '13 at 15:29
@RHS This is a common method that comes without name. Ilya splits the domain and estimate the expression $|x|^r$ in each domain. Also, this has nothing to do with Jensen's inequality - it is just a crude estimate. – AD. Jan 9 '13 at 15:34
BTW...... +1) :) – AD. Jan 9 '13 at 15:35
@AD. Where can I usually find this common method? And Did you mean the above inequality can't be proof with Jensen's inequality? – RHS Jan 9 '13 at 15:37
I believe that $r\ge1$ is necessary. – robjohn Jan 9 '13 at 19:07

Your Answer


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.