Let $H$ be a Hilbert space and $U$ a closed subspace of $H\times H$ .
Does then exist closed subspaces $V$ and $W$ of $H$ such that $H\times H =
U \oplus (V\times W)$ ? 


See also [Perturbations of an operator that disconnect the spectrum](https://mathoverflow.net/questions/24672/perturbations-of-an-operator-that-disconnect-the-spectrum)  .



Complement of a subspace which is a cartesian product
Let me expand my comment: the answer is yes, and in general $V$ and $W$ can be constructed as follows: let $p\_1,p\_2:U \to H$ denote the restrictions to $U$ of the two coordinate projections. Then $V$ is the image of the spectral projection $1\_{[0,1/2]}(p\_1p\_1^\*)$ and $V$ is the image of the spectral projection $1\_{[0,1/2[}(p\_2 p\_2^\*)$ (note that one interval is open at $1/2$, whereas the other is close). One can replace the two $1/2$'s by $\delta$ and $1-\delta$, but the choice of $\delta = 1/2$ gives the best constants.


This construction is perhaps clearer in the particular case when $U$ is a graph (i.e. $U \cap (\{0\} \times H) = \{0\}$, or equivalently $U = \{ (x,Tx),x \in D\}$ for a closed operator $T$ of domain $D \subset H$). One can then easily reduce to the case when $T$ is densely defined, and (using the polar decomposition), $T$ is self-adjoint. By the spectral theorem, we can assume that there is a measure space such that $H = L^2(X,\mu)$ and $T$ is the multiplication operator by some function $f:X \to \mathbb R^+$. Then $V$ (resp. $W$) is the space of functions in $L^2(X,\mu)$ that are zero outside of $I = f^{-1}([1,\infty[)$ (resp. $J = f^{-1}([0,1[) = X \setminus I$). For $(a,b) \in H \times H$ the corresponding decomposition in $U \oplus V \times W$ is then $(a 1\_J + b/f 1\_I , a f 1\_J + b 1\_I) + ((a-b/f)1\_I,(b- a f)1\_J)$.


