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

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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.)

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  • $\begingroup$ 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) $\endgroup$ – Alfred Sep 4 '19 at 13:13

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