I have a question about how to prove a lemma such as this one,

For any $0<\alpha<1$ and $M_{0}>0$, there exists a $M_{1}>0$ such that $\left|z\right|^{\alpha}\leq M_{0}+M_{1}\left|z\right|$ holds for all $z\in\mathbb{R}$.

It is not too difficult to see that the inequality $\left|z\right|^{\alpha}\leq M_{0}+M_{1}\left|z\right|$ holds when you plot $\left|z\right|^{\alpha}$ and $M_{0}+M_{1}\left|z\right|$. I want to generalize this lemma to the case when $z\in\mathbb{R}^{n}$ for $n>1$. However, it is not possible to show by plotting that a similar inequality would hold when $n>2$. Therefore, I'm searching for another way to prove it. I thought about using a Taylor series expansion for the function $\left|z\right|^{\alpha}$ for positive real numbers and negative real numbers separately while considering terms higher order than the $2^{nd}$ degree in the remainder part. However, this approach becomes problematic when $z=0$. So, I wonder if there is another way to prove this inequality.

Thank you in advance.

Best regards,

  • $\begingroup$ Either you are confused, or I am. This estimate already applies to the norm (i.e. n > 1). Also you can't use Taylor series at a point that where a function fails to be differentiable. $\endgroup$ Jun 10, 2011 at 9:30
  • $\begingroup$ Thank you for your reply. I agree with you that this estimate already applies to the norm. I also agree with you about not being able to use a reasoning based on Taylor series. I'm not a mathematician, I'm a mechanical engineer working on control theory. I saw this inequality mentioned as a lemma on a paper which I read, where it was just mentioned that its proof was quite obvious. But I was just wondering how to show it. $\endgroup$
    – alperden15
    Jun 10, 2011 at 10:01
  • $\begingroup$ There are other sites where you can ask this level of question, such as math.stackexchange.com; please read the faq. As it happens, this site is for professional mathematicians, and generally questions at this level will be closed as not being of interest to mathematicians who are doing research. $\endgroup$
    – Todd Trimble
    Jun 10, 2011 at 10:44
  • $\begingroup$ Thank you for your reply. I'll check the website you mentioned, ask any question that I have at that website. $\endgroup$
    – alperden15
    Jun 10, 2011 at 11:54

1 Answer 1


The inequality is easy to see by noting that (a) you have a function bounded near the origin, and (b) when you divide by |z| it tends to 0 at infinity. So it can be proved by a case analysis into two cases. If you want a bound like A + B|z| you can divide into regions where A is enough (|z| small), and choose B as the maximum of the quotient outside that region.


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