Why is $\|\operatorname{Co}(D_{\alpha}X)\|_{\gamma} \|\det [D_{\alpha}X(\alpha)]^{-1}\Vert_{\gamma}\le c\| D_{\alpha}X\|_{\gamma}^{2n-1}$? I am studying particle-trajectory method for solution to the Euler equation, chapter four of Majda and Bertozzi's book. Let $$X:\mathbb{R}^n\longrightarrow\mathbb{R}^n$$
be a smooth, invertible transformation with
$$|\det D_{\alpha}X(\alpha)|\ge c_1>0~,$$
being $0<\gamma\le 1$. Then, I do not know why the second inequality is true in the following expression:
$$
\Vert(D_{\alpha}X)^{-1}\Vert_{\gamma}\le c\Vert\operatorname{Co}(D_{\alpha}X)\Vert_{\gamma}
\Vert\det [D_{\alpha}X(\alpha)]^{-1}\Vert_{\gamma}\le c\Vert D_{\alpha}X\Vert_{\gamma}^{2n-1}.
$$
Here, $\operatorname{Co}(D_{\alpha}X)$ is the cofactor matrix of $D_\alpha X$. This appears in Lemma 4.2 (page 159).  Thanks a lot!
 A: As far as I remember, this is never actually used in the book. We use $c$ to mean a positive constant that can change from line to line.
By the definition of cofactor matrices, each component $C_{ij}:=(\operatorname{Co} D_\alpha X)_{ij}$ of the cofactor matrix is (up to sign) a minor, i.e. a determinant of an $(n-1)\times(n-1)$ submatrix. A determinant of a $d\times d$ matrix is a certain sum of products of $d$ components of the matrix (times a sign). In the present case, our matrix is $D_\alpha X$, so these components are some derivatives of $X$. Hence, we have
$$ \| C_{ij} \|_\gamma = \left\| \sum_{\sigma,\mu} c_{\sigma,\mu} (\partial_{\sigma(1)} X^{\mu(1)}) (\partial_{\sigma(2)} X^{\mu(2)}) \dots (\partial_{\sigma(n-1)} X^{\mu(n-1)})  \right\|_\gamma \le c \|D_\alpha X\|_{\gamma}^{n-1} $$
For the inverse determinant term $\|(\det D_\alpha X)^{-1}\|_\gamma$, we first use that composing Lipschitz maps with Holder maps gives Holder maps:
$$ \| f(g)\|_{\gamma} \le \|f\|_{C^0} + [f]_{\mathrm{Lip}}[g]_{\gamma}$$
In our case $f(x)=1/|x|$ on $[c_1,\infty)$, $\|f\|_{C^0} = 1/c_1$, and $[f]_{\operatorname{Lip}}=1/c_1^2$.
So
$$ \|(\det D_\alpha X)^{-1}\|_\gamma \le \frac1{c_1}+\frac1{c_1^2}[\det D_\alpha X]_\gamma$$
and $\|\det D_\alpha X\|_\gamma \le c\|D_\alpha  X\|_{\gamma}^n$ by the same reasoning as the first term. Now we bound the constant term by using the assumption again:
$ 1 \le \frac1{c_1} \|\det \nabla X\|_{C^0} $ implies that
$$\|(\det D_\alpha X)^{-1}\|_\gamma \le \frac1{c_1^2}\|\det D_\alpha X\|_\gamma \le c  \|D_\alpha X\|_{\gamma}^{n}$$
Putting it all together gives
$$ \lVert\operatorname{Co} D_\alpha X\|_{\gamma} \|(\det D_\alpha X)^{-1}\|_{\gamma} \le c \|D_\alpha X\|_{\gamma}^{n-1} \|D_\alpha X\|_{\gamma}^{n} = c  \|D_\alpha X\|_{\gamma}^{2n-1}$$
as desired.
