3

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

Thanks.

rschwieb
  • 153,510
RHS
  • 413

2 Answers2

6

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 $|x|^{r-1}\leq \max(1,|x|^r)\leq1+|x|^r$

SBF
  • 36,041
  • 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 09 '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 - Stop Putin - Jan 09 '13 at 15:34
  • BTW...... +1) :) – AD - Stop Putin - Jan 09 '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 09 '13 at 15:37
  • @RHS You will find this technique where estimates are needed e.g. in calculating various limits etc. Also, NO this cannot be proven using Jensen's inequality (joke) - I did not say that, but Jensen is a bit more subtle. – AD - Stop Putin - Jan 09 '13 at 15:57
  • 1
    I believe that $r\ge1$ is necessary. – robjohn Jan 09 '13 at 19:07
  • @robjohn: sure, I applied my inequality $r>0$ to $r-1$ in fact. Thanks for pointing this – SBF Jan 09 '13 at 19:14
  • I didn't accept your ans. because I have seen http://stats.stackexchange.com/questions/26454/proof-that-if-higher-moment-exists-then-lower-moment-also-exists before asking, which isn't what I look for. Sorry. And still the one I accepted isn't what I am looking for. But I think to be fair I will ask in a new question. – RHS Jan 10 '13 at 07:18
  • @RHS: well, it's only yours choice what to accept, I don't mind this at all. But I'm just interested how seen this another question is relevant :) – SBF Jan 10 '13 at 08:48
  • For more detail you can see Feller's Vol 1 p. 277. In discrete case, the rth moment $E(X^r) = \sum x_j^r f(x_j)$. If it exist the (r - 1)th moment exist by the inequality above. – RHS Jan 10 '13 at 10:47
  • @RHS: ok, but I just wondered how this fact affects accepting answers :) anyway, that was just a question – SBF Jan 10 '13 at 10:50
  • Sorry. Which fact do you refer to? The previous reply I ans. how to relate the inequality to the question in stats.stackexchange as you had asked. Not a reason for choosing an ans. – RHS Jan 10 '13 at 10:54
  • I think using max function is more like a trick. That is. So I want to look for another ans. But I was looking for one using Jensen's inequality ultimately. – RHS Jan 10 '13 at 11:10
0

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 $r\ge1$, $$ |x|^{r-1}\le|x|^r+1 $$

robjohn
  • 345,667