# A marginal space splitting $\{ \psi \}^{\perp}$

Let $$\psi \in L^2(\mathbb R^2,\mathbb C)$$. Is there a continuous projection from $$\{ \psi \}^{\perp}$$ onto $$\left\{ \varphi \in L^2(\mathbb R^2) \:\:\Big| \int \overline{\psi}(x,y) \varphi(x,y)\text{d}y = 0 \text{ a.e. in } x \in \mathbb R \right\}\:\: ?$$

Yes. The set you describe, call it $$E$$, is a closed linear subspace of $$L^2(\mathbb{R}^2)$$ and therefore we have an orthogonal projection from $$L^2(\mathbb{R}^2)$$ onto $$E$$. Explicitly, given $$\phi \in L^2(\mathbb{R}^2)$$, define $$g(x) = \int \overline{\psi}(x,y)\phi(x,y)\, dy$$ and $$h(x) = \int |\psi(x,y)|^2\, dy$$. Then the map $$\phi(x,y) \mapsto \phi(x,y) - \frac{g(x)}{h(x)}\psi(x,y)$$ is the orthogonal projection onto $$E$$. (The fact that $$\frac{g(x)}{h(x)}\psi(x,y)$$ belongs to $$L^2(\mathbb{R}^2)$$ is a little exercise in Fubini-Tonelli.)
• Thanks ! And would you know whether the operator $\varphi \mapsto \int \overline{\psi}(x,y)\varphi(x,y) \text{d} y$, from $\{ \psi \}^{\perp}$ to $L^2$, is surjective (or "almost surjective"), please ? ($\psi$ and its marginals are a.e non-vanishing) – Alfred Sep 4 '19 at 13:13