# Proof of a product inequality in Sobolev spaces

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?

• Do you get the desired conclusion if you replace $m$ with $|\alpha|$ throughout? Sep 25, 2017 at 1:16
• @AnthonyCarapetis Indeed, starting from where I said "I was reading the proof..." replacing $m$ with $|\alpha|$ makes the inequalities hold, but the problem is I need to then sum over all $|\alpha|\le m$ to get the initially claimed result. Sep 25, 2017 at 4:11
• If you can bound each term in the sum by the desired RHS, then you can bound the total by some constant depending only on $n$ and $m$ (just the number of terms in the sum) times this same quantity. Sep 25, 2017 at 5:24
• @AnthonyCarapetis So that I agree with, it's just odd that we start with $|\alpha|\le m$, but then it seems like the author just treats everything as if it were $|\alpha|=m$, when there are certainly multi-indices that are not equal to $m$. I suppose in the first inequality, we could change the sum to be $|\beta|\le m$ since all the $\beta \le \alpha$ certainly satisfy that. Then we can just drop $\alpha$ all together. Sep 25, 2017 at 7:21
• I agree that (from what you've reproduced here) it seems the author has made a mistake, or at least made this difficult to follow. Sep 25, 2017 at 7:24

## 1 Answer

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.