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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 .

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If you take $H = \mathbb{R}$, $H \times H = \mathbb{R}^2$, and $U$ a line through the origin that isn't an axis, doesn't this give a counterexample? – Christopher A. Wong Nov 22 '11 at 21:05
@Christopher: in your example, take $V=H$ and $W=\{0\}$. – Yemon Choi Nov 22 '11 at 21:28
Does $\oplus$ mean orthogonal sum of Hilbert spaces, or just direct sum of topological vector spaces? – Sergei Akbarov Nov 22 '11 at 22:48
@Sergei : just direct sum of topological vector spaces – jjcale Nov 23 '11 at 0:38
Something like this should work (I am too lazy to write a proof): Denote by $p_1,p_2:U\to H$ the restrictions to $U$ of the two coordinate projections. Let $V$ be the image of the spectral projection $1_{[0,1/2]}(p_1 p_1^*)$, and W the image of the spectral projection $1_{[0,1/2[}(p_2 p_2^*)$. Then $U\oplus (V\times W)=H \times H$. It works with $1/2$ replaced by any $0<\delta<1$. – Mikael de la Salle Nov 23 '11 at 16:09
up vote 8 down vote accepted

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 $(a 1_J + b/f 1_I , a f 1_J + b 1_I) + ((a-b/f)1_I,(b- a f)1_J)$.

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Thanks for adding some details! – Matthew Daws Nov 23 '11 at 21:55
Beautiful solution, Mikael. – Pietro Majer Nov 24 '11 at 7:11
This is a nice and, to me, quite surprising, result. – Bill Johnson Dec 3 '11 at 15:38

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