Q:

$C^{-1} (1+|x|^{2})^{\frac{s}{2}} \leq (1+|x|)^{\frac{s}{2}} \leq C (1+|x|^{2})^{\frac{s}{2}}$?

Let $s\in \mathbb R,$ and 
define $f: \mathbb R^{n}\to [0, \infty)$ such that $f(x)= (1+|x|^{2})^{\frac{s}{2}}, (x\in \mathbb R^{n})$ and $g:\mathbb R^{n}\to [0, \infty)$ such that $g(x)= (1+|x|)^{\frac{s}{2}}, (x\in \mathbb R^{n}).$
My Question: Can we expect  to get  a constant $C>0$ such that, $C^{-1} f(x) \leq g(x) \leq C f(x);$  for every $x\in \mathbb R^{n}$, that is, $f$ and $g$ is equivalent ? If yes, How ? (At lest for some specific values of $s$ and $d$, if required )
Thanks,

A:

As far as I can see, you are trying to find and upper bound for
$$\frac{f(x)}{g(x)}=\left(\frac{1+|x|^2}{1+|x|}\right)^{\frac s2}$$ 
and a lower positive bound for its inverse:
$$\frac{g(x)}{f(x)}=\left(\frac{1+|x|}{1+|x|^2}\right)^{\frac s2}$$
If $s>0$, the first tends to infinity and the second to $0$ when $|x|\rightarrow \infty$, so there is no such $C$.
If $s=0$, $f=g=1$, so $C=1$. But I think that this is too trivial.
If $s<0$ the limits are reversed: $0$ for $f/g$ and infinity for $g/f$, so there is neither any $C$ in this case.
By the way, what is $d$?

