Simplified estimate for derivatives of $f\circ g$?

Solution 1:

It turns out, the proof is just one more application of the interpolation inequalities.

For B1, we start from A1 and use the interpolation inequality $[\Psi ]_{C^p} \lesssim \|D\Psi\|_{C^1}^{(m-p)/(m-1)} \|D\Psi\|^{(p-1)/(m-1)}_{C^{m-1}}$: \begin{align*} [ \Psi\circ u ]_m &\lesssim [u]_m \sum_{p=1}^m \|D\Psi\|_{C^1}^{(m-p)/(m-1)} \|D\Psi\|^{(p-1)/(m-1)}_{C^{m-1}} \|u\|_{C^0}^{p-1} \\&= [u]_m \sum_{p=1}^m \|D\Psi\|_{C^1}^{(m-p)/(m-1)} \left( \|u\|_{C^0}^{m-1}\|D\Psi\|_{C^{m-1}}\right)^{(p-1)/(m-1)} \\&\lesssim_m [\Psi]_{C^1}[u]_m + \|\nabla\Psi\|_{C^{m-1}}\|u\|_{C^0}^{m-1}[u]_{C^m}, \end{align*} where we used Young's inequality for products $a^{\theta}b^{1-\theta} \lesssim a + b$ which is valid for all $\theta\in[0,1]$ (including the endpoints $\theta=0,1$), in particular for $\theta = \frac{p-1}{m-1}$ with $1-\theta=\frac{m-p}{m-1}$.

$$ \sum_{p=2}^m [\Psi]_{C^p} \|u\|_{C^0}^{p-1} \le \max_{k\in[1,m-1]} \|u\|_{C^0}^{k-1} \|\nabla\Psi\|_{C^{m-1}} \lesssim (\|u\|_{C^0} + \|u\|_{C^0}^{m-1})\|\nabla\Psi\|_{C^{m-1}}.$$

For B2, we instead start from A2, and proceed similarly: \begin{align*} [\Psi\circ u ]_{C^m} &\lesssim \sum_{p=1}^m \|D\Psi\|_{C^1}^{(m-p)/(m-1)} \|D\Psi\|^{(p-1)/(m-1)}_{C^{m-1}} [u]_{C^1}^{(p-1)\frac{m}{m-1}} [u]_{C^m}^{\frac{m-p}{m-1}} \\ &= \sum_{p=1}^m \left([\Psi]_{C^1} [u]_{C^m}\right)^{(m-p)/(m-1)} \left( \|D\Psi\|_{C^{m-1}} [u]_{C^1}^m\right)^{(p-1)/(m-1)} \\&\lesssim_m [\Psi]_{C^1} [u]_{C^m} + \|D\Psi\|_{C^{m-1}} [u]_{C^1}^m. \end{align*}