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.



Is there any quantitative relationship between the two terms of a Helmholtz decomposition?
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.


