Let $X$ be a vector space contained in $H^{1}(\mathbb R^d),$ then we can study 


$X^{\perp\_{L^2}}:=\left\{ \xi \in L^2; \langle \xi, x \rangle\_{L^2} =0 \ \forall x \in X \right\}$
and 


$X^{\perp\_{H^{-1}}}:=\left\{ \xi \in H^{-1}; \langle \widehat{\xi}, \widehat{x} \rangle\_{H^{-1},H^1} =0 \ \forall x \in X \right\}.$


Now assume that $X^{\perp\_{L^2}}$ is a finite-dimensional space. 


Does this imply that $X^{\perp\_{H^{-1}}} = X^{\perp\_{L^2}}$?


In general we only have $X^{\perp\_{H^{-1}}} \supset X^{\perp\_{L^2}}$, but I am wondering whether the assumption of $X^{\perp\_{L^2}}$ being finite-dimensional yields the converse implication as well?



Orthogonal complement vector space
Take $d=1$. In this case all functions in $H^1(\mathbb{R})$ are (absolutely) continuous, so evaluation at a point is well defined . Let $X = \{f \in H^1 : f(0) = 0\}$ which is a well-defined closed subspace of $H^1$ with codimension 1, but is dense in $L^2$. So $X^{\perp\_{L^2}} = 0$ but $X^{{\perp}\_{H^{-1}}}$ is one-dimensional, spanned by the Dirac mass at 0.


