4
$\begingroup$

Let $\Omega \subset \mathbb{R}^3$ be a lipschitz domain. We then have the trace operator $\tau : H^1(\Omega) \to L^2(\partial \Omega)$ and can define the space $H^{1/2}(\partial \Omega) := \tau(H^1(\Omega)),$ which is equipped with the norm $$\lVert g\rVert_{H^{1/2}(\partial \Omega)}:=\inf_{v\in H^1(\Omega), \tau(v)=g} \lVert v\rVert_{H^1(\Omega)}.$$

Finally, let $H^{-1/2}(\partial \Omega)$ denote the dual space of $H^{1/2}(\partial \Omega)$. We can view $L^2(\partial \Omega)$ as a subspace of $H^{-1/2}(\partial \Omega)$ via the linear map $$ \begin{split} L:L^2(\partial \Omega)&\to H^{-1/2}(\partial \Omega),\\ f &\mapsto L_f, \end{split} $$ where $$L_f(\tau(\phi)):= \int_{\partial \Omega} f\cdot \tau(\phi)\,dS,\quad \phi \in H^1(\Omega).$$

$L$ is injective by the fundamental lemma of the calculus of variations and also bounded: The inequality $$\lvert L_f(\tau(\phi))\rvert \leq \lVert f\rVert_{L^2(\partial \Omega)} \lVert \tau(\phi)\rVert_{L^2(\partial \Omega)}\leq C \lVert f\rVert_{L^2(\partial \Omega)} \lVert\phi\rVert_{H^1(\Omega)}$$ implies that $$\lvert L_f(\tau(\phi))\vert\leq C\lVert f\rVert_{L^2(\partial \Omega)}\lVert\tau(\phi)\rVert_{H^{1/2}(\partial \Omega)},$$ so $$\lVert L_f\rVert_{H^{-1/2}(\partial \Omega)}\leq C\lVert f\rVert_{L^2(\partial \Omega)}.$$

Does there exists a constant $C>0$ such that $$(+)\quad\lvert\lvert f\rvert\rvert_{L^2(\partial \Omega)}\leq C \lVert L_f\rVert_{H^{-1/2}(\partial \Omega)}\quad \forall f\in L^2(\partial \Omega)$$ or a similar inequality holds?

My question is motivated by the following: For $u\in L^2(\Omega)^3$, a function $v\in L^2(\Omega)^3$ is called the weak curl of $u$ if $$\int_{\Omega} u\cdot \text{curl }\phi\,dx = \int_{\Omega} v\cdot \phi\,dx\quad\forall \phi\in C_0^{\infty}(\Omega)^3.$$ We write $v=\text{curl } u$. We can introduce the hilbert space $H(\text{curl}, \Omega)$ of all functions $u\in L^2(\Omega)^3$ for which a weak curl exists with the norm $$\lvert\lvert u\rvert\rvert_{H(\text{curl},\Omega)}^2:= \lvert\lvert u\rvert\rvert_{L^2(\Omega)^3}^2 + \lvert\lvert \text{curl } u\rvert\rvert_{L^2(\Omega)^3}^2.$$ One can then show the following theorem (for a reference, see Theorem 3.29 in "Finite Element Methods for Maxwell's Equations" by Monk): $$ $$ Let $\Omega\subset \mathbb{R}^3$ be a bounded lipschitz domain. Then the trace map $\gamma_t$ which is defined classically by $$\gamma_t(v) := \nu \times (v\rvert_{\partial \Omega}),\quad v\in C^{\infty}(\overline{\Omega})^3$$ (where $\nu$ is the outward normal vector of $\Omega$) can be extended by continuity to a continuous linear map $\gamma_t:H(\text{curl},\Omega)\to H^{-1/2}(\partial \Omega)^3$. Furthermore, the following Green's theorem holds for any $v\in H(\text{curl},\Omega)$ and $\phi\in H^1(\Omega)^3$: $$\gamma_t(v) (\phi) = \int_{\Omega} \text{curl } v \cdot \phi - v\cdot \text{curl } \phi\,dx.$$ $$ $$ By the discussion above, we have $L^2(\partial \Omega)^3 \subset H^{-1/2}(\partial \Omega)^3$. In other words, for some $v\in H(\text{curl},\Omega)$, the functional $\gamma_t(v)$ can be expressed by a function $f\in L^2(\partial \Omega)^3$, i.e. $$\gamma_t(v)(\phi) = \int_{\partial \Omega} f\cdot \tau(\phi)\,dx\quad \forall \phi \in H^1(\Omega)^3.$$ Since $\gamma_t$ is bounded, we have $$\lvert\lvert \gamma_t(v)\rvert\rvert_{H^{-1/2}(\partial \Omega)^3}\leq C\lvert\lvert v\rvert\rvert_{H(\text{curl},\Omega)}.$$ My question is whether we have a similar inequality for $\lvert\lvert f\rvert\rvert_{L^2(\partial \Omega)^3}$: $$\lvert\lvert f\rvert\rvert_{L^2(\partial \Omega)^3} \leq C\lvert\lvert v\rvert\rvert_{H(\text{curl},\Omega)}.$$ This would follow immediatly from $(+)$.

$\endgroup$
4
  • 1
    $\begingroup$ Welcome to MO! Since I assume you plan to do something with this, could you maybe specify what that is; in particular, what exactly you mean by "bounded" here? (Or maybe kind of equivalently, how you want to understand $L^{-1}$.) $\endgroup$
    – Hannes
    Oct 10, 2023 at 14:31
  • $\begingroup$ Thank you for your comment. I have changed my question and added some context. $\endgroup$
    – Mandelbrot
    Oct 11, 2023 at 17:25
  • $\begingroup$ Starting from "In other words", why can $\gamma_t(v)$ be expressed by $f \in L^2(\partial\Omega)$ if it is generically in $H^{-1/2}(\partial\Omega)$? $\endgroup$
    – Hannes
    Oct 16, 2023 at 14:32
  • 1
    $\begingroup$ By "for some $v$" i mean that for certain $v$ (but not for all) the functional $\gamma_t(v)$ can be expressed by a function $f$, for example $f=0$ for $v=0$ (if that was your question) $\endgroup$
    – Mandelbrot
    Oct 16, 2023 at 20:35

1 Answer 1

2
+100
$\begingroup$

I think the last desired inequality cannot be right, but a direct counterexample escaped me, so here is an argument with a bit of a detour.

Let me maybe first frame this with an abstract result.

Lemma. If $A \colon X \supseteq D \to Y$ is a closed operator between Banach spaces with domain $D$, then $$\|Ax\|_Y \lesssim \|x\|_X \quad \text{for all}~x\in D$$ if and only if $D$ is a closed subspace of $X$. In particular, if $A$ is in fact densely defined, then the foregoing inequality is equivalent to $D = X$.

The "if" direction follows with the closed graph theorem. For the "only if" part, observe that if $D \supseteq (x_k) \to x$ in $X$, then by the assumed inequality, $(Ax_k)$ is a Cauchy sequence in $Y$ and thus convergent, hence the closedness of $A$ implies that $x \in D$.

That being said, consider $\gamma_t$ as an unbounded operator $D \subseteq H(\text{curl},\Omega) \to L^2(\partial\Omega)$ with the domain $$D = \Bigl\{ v \in H(\text{curl},\Omega) \colon \gamma_t(v) \in L^2(\partial\Omega)\Bigr\}.$$

Using the Green type formula, you can show that this is in fact a closed operator. Moreover, since, say, continuous differentiable functions on the closure of $\Omega$ are included in $D$, it is also densely defined. Hence, with the Lemma, it follows that the desired inequality $$\|\gamma_t(v)\|_{L^2(\partial\Omega)} \lesssim \|v\|_{H(\text{curl},\Omega)} \qquad \text{for all}~v \in D$$ is equivalent to $D = H(\text{curl},\Omega)$ which is, I suppose, certainly not correct.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.