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Let $\phi \in L^{2}(\mathbb{R}^{d})$ be fixed. Denote by $P$ the orthogonal projection onto the subspace orthogonal to $\text{span}\{\phi\}$. In other words, for $f \in L^{2}(\mathbb{R}^{d})$ set: $$(Pf)(x) = f(x) - \phi(x)\int dy \overline{\phi(y)}f(y)$$

Suppose I want to consider functions on $L^{2}(\mathbb{R}^{d})\otimes L^{2}(\mathbb{R}^{d}) \cong L^{2}(\mathbb{R}^{d}\times \mathbb{R}^{d})$ and the projection $P \otimes P$. What is the correct formula for $P \otimes P$? I obtained the following one. For $f \in L^{2}(\mathbb{R}^{d}\times \mathbb{R}^{d})$, $$[(P\otimes P)f](x,y) = f(x,y)-\phi(x)\phi(y)\int dzdw \overline{\phi(z)}\overline{\phi(w)}f(z,w)$$ but I am not entirely sure if this is the correct formula. Is this correct?

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I should first point out the formula you wrote down for $P$ only works if $\phi$ is a unit vector, i.e., if $\int |\phi(x)|^2 \, dx = 1$. Assuming that is the case, the formula you wrote for $P \otimes P$ is still incorrect. The formula you wrote down is actually for $1 - (1 - P) \otimes (1 - P)$, which is a projection of co-rank $1$, whereas $P \otimes P$ has co-rank infinity. The correct formula can be obtained like this: let $f, g \in L^2(\mathbb{R}^d)$, then,

$$\begin{split} [(P \otimes P)(f \otimes g)](x, y) &= [Pf \otimes Pg](x, y)\\ &= Pf(x)Pg(y)\\ &= [f(x) - \phi(x) \int \overline{\phi(z)}f(z) \, dz][g(y) - \phi(y) \int \overline{\phi(w)}g(w) \, dw]\\ &= f(x)g(y) - \phi(x) \int \overline{\phi(z)}f(z)g(y) \, dz - \phi(y) \int \overline{\phi(w)}f(x)g(w) \, dw + \phi(x)\phi(y) \int \overline{\phi(z)}\overline{\phi(w)}f(z)g(w) \, dzdw \end{split}$$

Thus, by linearity and density of finite linear combinations of simple tensors in $L^2(\mathbb{R}^d \times \mathbb{R}^d) = L^2(\mathbb{R}^d) \otimes L^2(\mathbb{R}^d)$, we have, for $f \in L^2(\mathbb{R}^d \times \mathbb{R}^d)$,

$$[(P \otimes P)f](x, y) = f(x, y) - \phi(x) \int \overline{\phi(z)}f(z, y) \, dz - \phi(y) \int \overline{\phi(w)} f(x, w) \, dw + \phi(x)\phi(y) \int \overline{\phi(z)}\overline{\phi(w)}f(z, w) \, dzdw$$

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