Inequality involving a curl in two dimension

$$\DeclareMathOperator{\curl}{curl}$$Let $$\Omega\subset\mathbb{R}^2$$ be bounded lipschitz domain, $$x_0\in\Omega$$ a fixed point and denote by $$D(x_0,R)$$ the disk centered at $$x_0$$ with radius $$R>0$$.

Define the Hilbert space $$H(\curl,\Omega)=\left\{\mathbf u\in (L^2(\Omega))^2;\ \curl \mathbf u :=\frac{\partial u_2}{\partial x_1}-\frac{\partial u_1}{\partial x_2}\in L^2(\Omega)\right\}.$$ Let $$A$$ denote a bounded linear operator from $$H(\curl,\Omega)$$ to itself and $$f\in \mathscr{C}^{\infty}(\Omega)$$ supported on $$D(x_0,R)$$.

I want to find a constant $$C$$ that depend on $$f$$ such that we have $$\|fA\|_{\mathscr{L}(H(\curl,\Omega))}\leq C \|A\|_{\mathscr{L}(H(\curl,\Omega))}.$$ What I do is the following: for $$\mathbf u\in H(\curl,\Omega)$$ such that $$\|\mathbf u\|_{H(\curl,\Omega)}\leq 1$$ we have $$\|(fA)\mathbf u\|^2_{H(\curl,\Omega)}=\|(fA)\mathbf u\|^2_{(L^2(\Omega))^2}+\|\curl((fA)\mathbf u)\|^2_{L^2(\Omega)}$$ We then obtain for the first norm: $$\|(fA)\mathbf u\|_{(L^2(\Omega))^2}\leq \sup_{x\in D(x_0,R)} |f(x)| \|Au\|_{(L^2(\Omega))^2}.$$ But, I did not find a way to treat the second norm.

• Is it possible that multiplication by $f$ is a bounded operator on $H(curl, \Omega)$? Commented Nov 17, 2022 at 22:16
• @Mason $H(curl, \Omega)$ is stable by the multiplication by $f$. Commented Nov 21, 2022 at 10:40

Notice that $$curl(fw)=\frac{\partial f}{\partial x_{1}}w_{2}-\frac{\partial f}{\partial x_{2}}w_{1}+fcurl(w)$$ thus: \begin{align*} \left\Vert curl(fw)\right\Vert _{L^{2}(\Omega)} & \leq\left\Vert \frac{\partial f}{\partial x_{1}}w_{2}-\frac{\partial f}{\partial x_{2}}w_{1}\right\Vert _{L^{2}(\Omega)}+\left\Vert fcurl(w)\right\Vert _{L^{2}(\Omega)}\\ & \leq\left\Vert \frac{\partial f}{\partial x_{1}}w_{2}\right\Vert _{L^{2}(\Omega)}+\left\Vert \frac{\partial f}{\partial x_{2}}w_{1}\right\Vert _{L^{2}(\Omega)}+\left\Vert fcurl(w)\right\Vert _{L^{2}(\Omega)}\\ & \leq C\left[\|w_{2}\|_{L^{2}(\Omega)}+\|w_{1}\|_{L^{2}(\Omega)}+\|curl(w)\|_{L^{2}(\Omega)}\right]\\ & \leq C\left[2\|w\|_{\left(L^{2}(\Omega)\right)^{2}}+\|curl(w)\|\right]\\ & \leq C\left[2\|w\|_{H(curl(\Omega))}+\|w\|_{H(curl(\Omega))}\right]\\ & \leq3C\|w\|_{H(curl(\Omega))} \end{align*}
$$C=\max\left(\sup_{x\in D(x_{0},R)}|f(x)|,\sup_{x\in D(x_{0},R)}\left|\frac{\partial f}{\partial x_{1}}\right|,\sup_{x\in D(x_{0},R)}\left|\frac{\partial f}{\partial x_{2}}\right|\right)$$ Therefore by setting $$w=Au$$: \begin{align*} \left\Vert curl(fAu)\right\Vert _{L^{2}(\Omega)} & \leq3C\|Au\|_{H(curl(\Omega))}\\ & \leq3C\|A\|\|u\| \end{align*}
• Why we have $\|w_2\|+\|w_1\| \leq 2\|w\|$ ? Commented Nov 23, 2022 at 19:48
• because $\|w\|^2 =\|w_1\|^2 + \|w_2\|^2$ so $\|w_i\| \leq \|w\|$ Commented Nov 24, 2022 at 2:29