Show that $\|fu\|_{W^1}\leq C\|f\|_{C^1}\|u\|_{W^1}$

Question

Let $W^{1}(\Omega):=\{u\in L^{2}(\Omega): \partial^{\alpha}u\in L^{2}(\Omega), \forall \alpha\in\mathbb{N}\cup\{0\}\mbox{ with } |\alpha|\leq1\}$ with norm, $$\|u\|^{2}_{W^{1}(\Omega)} = \sum_{|\alpha|\leq1}{\|\partial^{\alpha}u\|^{2}_{L^{2}(\Omega)}}$$

If $f\in C^{1}(\Omega)$ such that $\|f\|_{C^1}=\displaystyle\max_{\alpha\in\{0,1\}}\sup_{x\in\Omega}{|\partial^{\alpha}f|}<\infty$ and for any $u\in W^{1}(\Omega)$ (i.e, $u\in L^{2}(\Omega)$ and $\partial^{\alpha}u\in L^{2}(\Omega)$ for $\alpha\in\{0,1\}$), then $fu\in W^{1}(\Omega)$ and $$\|fu\|_{W^1}\leq C\|f\|_{C^1}\|u\|_{W^1}$$

My approach: If I want to prove that $fu\in W^{1}(\Omega)$, then I need to prove that $fu\in L^{2}(\Omega)$ such that $\partial^{\alpha}fu\in L^{2}(\Omega)$. So, we know that the sum of two functions in $L^2$ is in $L^2$, then $$\int_{\Omega}{|\partial^{\alpha}f|^{2}d\mu}\leq\sup_{\alpha\in\{0,1\}}{|\partial^{\alpha}f|^{2}}\int_{\Omega}d\mu=\|f\|_{C^1(\Omega)}\mu(\Omega)<\infty$$

But I don't know if the measure of the set $\Omega\subset\mathbb{R}^n$ is finite. If the above is true, with $\partial^{\alpha}u\in L^{2}$ it follows that $\partial^{\alpha}(fu)=f\partial u+u\partial f\in L^{2}(\Omega)$ but I also don't know if the function $f$ is in $L^2(\Omega)$. So,

$$\|\partial^{\alpha}(fu)\|^{2}_{W^1}\leq\|f\partial u\|^{2}_{W^1}+\|u\partial f\|^{2}_{W^1}$$

Answer 1

By definition, $$\|fu\|_{W^1}=\|fu\|_{L^2}+\sum\limits_{|\alpha|=1}\|\partial^\alpha (fu)\|_{L^2}.$$ Now, note that $$\|fu\|_{L^2}^2=\int |fu|^2\, dx\leq \|f\|_\infty^2\|u\|_{L^2}^2,$$ so that $$\|fu\|_{L^2}\leq \|f\|_\infty\|u\|_{L^2}\leq\|f\|_{C^1}\|u\|_{L^2}.$$ Now, just do the same for the $\sum\limits_{|\alpha|=1}\|\partial^\alpha (fu)\|_{L^2}$ part. Each term in the sum contributes two pieces (due to the product rule), and these pieces are split into two cases (again, by the product rule):

($1$) a derivative lands on $f$: Then, you sup out a derivative of $f$, so the term is bounded by $\|f\|_{C^1}\|u\|_{L^2}$

($2$) a derivative lands of $u$: Then, you sup out $f$, so the term is bounded by $\|f\|_{C^1}\|u\|_{W^1}$.

Finally, you just combine everything together.

To illustrate the described process, one bounds a term in the sum as \begin{align*} \|\partial_j (fu)\|_{L^2}&=\|u\partial_jf+f\partial_j u\|_{L^2}\leq \|u\partial_jf\|_{L^2}+\|f\partial_j u\|_{L^2}\\ &\leq \|\partial_j f\|_\infty \|u\|_{L^2}+\|f\|_{\infty}\|\partial_j u\|_{L^2}\\ &\leq \|f\|_{C^1}\|u\|_{L^2}+\|f\|_{C^1}\|u\|_{W^1}\\ &\leq 2\|f\|_{C^1}\|u\|_{W^1} \end{align*}

Answer 2

For simplicity, assume $n=1$. Then we see that \begin{align} \|fu\|_{W^1}^2 =& \|fu\|_{L^2(\Omega)}^2+\|\nabla(fu)\|^2_{L^2(\Omega)}\\ \lesssim &\ \|fu\|_{L^2(\Omega)}^2+\|\nabla f\cdot u\|^2_{L^2(\Omega)}+\|f\cdot \nabla u\|^2_{L^2(\Omega)}\\ \le&\ \|f\|_{L^\infty(\Omega)}^2\|u\|_{L^2(\Omega)}^2+ \|\nabla f\|_{L^\infty(\Omega)}^2\|u\|^2_{L^2(\Omega)}+\|f\|_{L^\infty(\Omega)}^2\|\nabla u\|^2_{L^2(\Omega)}\\ \le&\ \|f\|_{C^1}^2\|u\|_{W^1}^2. \end{align}