Let $||\cdot||_m$ denote the $H^m$ norm (in particular, $H^0 = L^2$). For $u,v \in L^{\infty}\cap H^m(\mathbb{R}^n)$, $m \in \mathbb{Z}_{\ge 0}$, we have
$$
||uv||_m \lesssim |u|_{L^\infty}||D^mv||_0 + |v|_{L^\infty}||D^mu||_0
$$
(here, $\lesssim$ means "$\le$ a constant times...").
I was reading through the proof and there were a few lines I could not verify:
Fix a multiindex $\alpha$ with $|\alpha|\le m$. Then we have
$$
||D^\alpha(uv)||_0 \lesssim \sum_{\beta \le \alpha} ||D^{\beta}uD^{\alpha-\beta}v||_0 \lesssim \sum_{\beta \le \alpha} ||D^\beta u||_{L^{2m/|\beta|}}||D^{\alpha-\beta}v||_{L^{2m/|\alpha-\beta|}}.
$$
They claim the last inequality follows from Holder's Inequality, but those two aren't conjugates, so perhaps there's a step I'm missing?
The next one is as follows. First, they use a form of Gagliardo-Nirenberg I'm not familiar with:
$$
||D^iu||_{L^{2r/i}} \lesssim |u|_{L^\infty}^{1-i/r}||D^ru||_0^{i/r} = (*).
$$
I'll take that as fact, but the following is a bit confusing:
Continuing from the above, we have
\begin{align}
(*) &\lesssim \sum_{\beta \le \alpha} |u|_{L^\infty}^{1-|\beta|/m}||D^mu||_0^{|\beta|/m}|v|_{L^\infty}^{1-|\alpha-\beta|/m}||D^mv||_0^{|\alpha-\beta|/m} \\ &\lesssim \sum_{\beta\le \alpha} (|u|_{L^\infty}||D^mv||_0)^{|\alpha-\beta|/m}(|v|_{L^\infty}||D^mu||_0)^{|\beta|/m}.
\end{align}
I'm not sure how they combine the terms in the last step.
All of the above makes sense if $|\alpha| = m$, but otherwise I can't see it. Could anyone give some hints?
Replacing the sum over $\beta \le \alpha$ with sum over $\beta \le m$ in the second string of inequalities at the beginning of the post will give us the result, since then $|\alpha-\beta| = m-|\beta|$, and we can apply Holder as they mentioned as well as combine the terms in the second part of my question.