Estimate of higher order derivatives like $D^{\alpha}(|\xi|^k)$ and $D^{\alpha}(e^{-t|\xi| / 2})$ In Torchinsky's harmonic analysis book, proof of Theorem 5.1, he had the following estimate (5.7), but he didn't write down the proof of this step. I am confused because I think we're differentiating with respect to $\xi$, so the chain rule must be used and some $\xi_j$ components has to appear. Is he differentiating with respect to $|\xi|$ or $\xi$? Why is that estimate true? $\xi\in\mathbb{R}^n$ and $\alpha$'s are multi-indices. Any ideas will be appreciated. $m$ is a function satisfies H"ormander multiplier condition of order $k$, i.e.,
$$\sup_{R>0}R^{2|\alpha|-n}\int_{R<|\xi|<2R}|D^{\alpha}m(\xi)|^2\, d\xi<C$$ for any multi-index $|\alpha|<k$. Not sure if it helps.
from the book:

In other words, it is bounded by expressions involving integrals of the form
$$
\int_{R^n}\left|D^\alpha\left(|\xi|^k e^{-t|\xi| / 2} m(\xi)\right)\right|^2 d \xi, \quad|\alpha|=k \tag{5.6}
$$
Let $\alpha=\alpha_1+\alpha_2+\alpha_3,\left|\alpha_1\right|+\left|\alpha_2\right|+\left|\alpha_3\right|=k$. Then the derivatives in (5.6) are linear combinations of monomials $D^{\alpha_1}(|\xi|^k) D^{\alpha_2}(e^{-t|\xi| / 2}) D^{\alpha_3} m(\xi)$ each of which can be dominated by
$$
c|\xi|^{k-\left|\alpha_1\right|} t^{|\alpha_2|} e^{-t|\xi| / 2}\left|D^{\alpha_3} m(\xi)\right| . \tag{5.7}
$$

 A: 
@CalvinKhor [...] So I'm confused why can he bound $D^\alpha(|\xi|^k)$ and $D^\alpha(e^{-t|\xi| / 2})$ by $|\xi|^{k-|\alpha|}$ and $t^{|\alpha|} e^{-t|\xi| / 2}$ respectively . - lzj 1 min ago

Lets say $f$ is $r$-homogeneous if $f(\lambda x) = \lambda^r f(x) $ for all $\lambda>0$. Then $D_if$ is $(r-1)$-homogeneous, where $D_i$ is the $i$th partial.
Proof:
$$ D_if(\lambda x) =\lim_{h\to 0} \frac{f(x + he_i/\lambda )\lambda ^r - f(x)\lambda ^r}{h} = \lambda^{r-1}\lim_{\eta\to 0} \frac{f(x + \eta e_i ) - f(x)}{\eta} = \lambda^{r-1} D_if(x). $$
Immediate corollary:
$$ D^\alpha(|\xi|^k) \lesssim_{\alpha,k,n} |\xi|^{k- |\alpha|}.$$
Proof: $|\xi|^k$ is $k$-homogeneous. Induction with the previous result shows that $g(\xi)=D^\alpha(|\xi|^k)$ is $(k-|\alpha|)$-homogeneous. Hence
$$ g(\xi) = g\left(|\xi| \frac{\xi}{|\xi|}\right)=|\xi|^{k-|\alpha|} g\left( \frac{\xi}{|\xi|}\right)$$
and therefore $|g(\xi)| \le |\xi|^{k-|\alpha|}\sup _{\sigma: |\sigma|=1} |g(\sigma)|$.
With this in hand, the second result (...once corrected) is not so hard.
Claim: for $|\alpha|\ge 1$, $D^\alpha e^{-|x|} = P_\alpha(x) e^{-|x|} $ where $P_\alpha(x)$ is a sum of $r$-homogeneous terms for $r=0,-1,\dots,1-|\alpha|$.
Proof: $D_i e^{-|x|} = -(D_i|x|) e^{-|x|}  $ so the result is true for $|\alpha|=1$. Then induction proves the claim.
Since $D^\alpha (f(tx)) = t^{|\alpha|} (D^\alpha f)(tx)$ by chain rule, we see that the true result should be
$$ |D^\alpha e^{-t|x|/2}| \lesssim_{\alpha,n} t^{|\alpha|}\max(1,|t x|^{1-|\alpha|})e^{-t|x|/2} $$
