An inequality for a continuous non-smooth function

Question

Hello,

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,

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.