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$\DeclareMathOperator{\curl}{curl}$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.

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

1 Answer 1

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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*}

Where:

$$ C=\max\left(\sup_{x\in D(x_{0},R)}|f(x)|,\sup_{x\in D(x_{0},R)}\left|\frac{\partial f}{\partial x_{1}}\right|,\sup_{x\in D(x_{0},R)}\left|\frac{\partial f}{\partial x_{2}}\right|\right) $$ 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*}

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  • $\begingroup$ Why we have $\|w_2\|+\|w_1\| \leq 2\|w\|$ ? $\endgroup$
    – SAKLY
    Commented Nov 23, 2022 at 19:48
  • $\begingroup$ because $\|w\|^2 =\|w_1\|^2 + \|w_2\|^2 $ so $\|w_i\| \leq \|w\|$ $\endgroup$
    – Porufes
    Commented Nov 24, 2022 at 2:29

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