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Let $\Omega \subset \mathbb R^3$ denote an open, bounded and simply connected set with smooth boundary. The Helmholtz decomposition $$ L^2(\Omega) = \nabla H^1_0(\Omega) \oplus L^2(\operatorname{div}=0; \Omega)$$ entails the existence of two linear and bounded natural projections $\pi_1 \in \mathcal{L}(L^2(\Omega); \nabla H_0^1(\Omega))$, $\pi_2 \in \mathcal{L}(L^2(\Omega); L^2(\operatorname{div=0}; \Omega))$, where the divergence is to be understood in the weak sense. Such decomposition can be extended to $$ H_0(\operatorname{curl}; \Omega) = \nabla H_0^1(\Omega) \oplus (H_0(\operatorname{curl}; \Omega) \cap H^0(\operatorname{div}; \Omega)) \qquad (*), $$

where the zero subscript in the curl space indicates null tangential component while the zero superscript in the div space the divergence-free condition. Again we have at disposal two bounded projections $\widehat\pi_1, \widehat \pi_2$. I was wondering if we can establish some kind of non-trivial relationship between $\| \widehat \pi_1 \mathbf u \|$ and $\| \widehat \pi_2 \mathbf u \|$ for $\mathbf u \in H_0(\operatorname{curl}; \Omega)$, say a bound not involving $\| \mathbf u \|$ itself?

It is clear that something like $\| \widehat \pi_1 \mathbf u \| \le C \| \widehat \pi_2 \mathbf u \| $ (and/or viceversa) for an absolute constant $C>0$ is not possible, indeed if for instance $\operatorname{div} \mathbf u = 0$ then the gradient part vanishes and the bound would lead to a contradiction. To get around this we may assume that $\operatorname{div} \mathbf u, \operatorname{curl} \mathbf u $ are nonzero, but in this case I cannot either find a counterexample or figure out a result. Thanks in advance for any suggestion and insight.

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I don't know much about the curl spaces but I think I can answer in the negative for pure Hilbert space reasons.

If indeed $\hat{\pi}_1$ and $\hat{\pi}_2$ are projections, denote their ranges by $V_1$ and $V_2$, which we'll assume are also Hilbert subspaces of $V = V_1 + V_2$ ($= H_0(curl;\Omega)$ for your specific problem.)

For any $u \in V_1$, $\hat{\pi}_1(u) = u$ because it's a projection, and so it had better be that $\hat{\pi}_2(u) = 0$, and vice-versa if $u \in V_2$. So $\hat{\pi}_1(u)$ has no hope of being comparable to $\hat{\pi}_2(u)$ in any meaningful way.

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  • $\begingroup$ Thanks for you answer! In the last part of the question I am already mentioning the fact that a straightforward absolute estimate is not possible: I do so taking a divergence free vector, which is equivalent to say that it lies in the second subspace. In the very last lines I also suggest to assume $\operatorname{curl} \mathbf u \neq 0, \operatorname{div} \mathbf u \neq 0$ in order to precisely avoid the two limit cases... $\endgroup$
    – GaC
    Commented Jan 22, 2021 at 21:00
  • $\begingroup$ $div$ and $curl$ are linear maps, so their kernels are subspaces, but the complement of the kernel is an open and dense subset, so you'll get arbitrarily close to the behavior when $div u=0$ and/or $curl u = 0$. $\endgroup$
    – Sébastien Loisel
    Commented Jan 23, 2021 at 12:02

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