Closest point theorem in a Hilbert space

Every nonempty closed convex subset $C$ of a Hilbert space contains a unique point nearest to each $x$. A minimizing sequence is Cauchy by the parallelogram identity, and uniqueness follows by applying the same identity to two minimizers and their midpoint.

For a closed subspace $F$ of a Hilbert space,The closest point $P_{F}x$ gives the first component, and differentiating $\Vert x-P_{F}x-tu{\Vert }^2$ at zero shows that the remainder is orthogonal to every $u\in F$.

The orthogonal complement of a subset $F$ of an inner-product space is

For a linear subspace $F$ of a Hilbert space,Thus $F=F^{\perp \perp }$ exactly when $F$ is closed.

The orthogonal projection $P_F$ onto a closed subspace $F$ sends $x$ to the component in the decomposition $x=P_{F}x+(I-P_F)x$ with $(I-P_F)x\in F^{\perp }$. It is a contraction: $\Vert P_{F}x\Vert \le \Vert x\Vert$.

For a nonzero closed subspace of a Hilbert space $V$ and a closed subspace of a Hilbert space $W$, this angle measures the worst loss under projection from $V$ to $W$. Its positive cosine is a uniform lower bound for $P_W{|}_V$. The order of the subspaces matters in general. This is distinct from the smallest angle between two subspaces, whose cosine is a supremum. For equal finite dimensions the cosine is the smallest singular value of the cross Gram matrix of orthonormal bases. When $V$ is zero, the infimum is over an empty unit sphere and does not define an angle in $[0,\pi /2]$; use a lower-bound formulation instead.

For closed subspaces of a Hilbert space, set $T=P_{V^{\perp }}{|}_W$. The two positive cosines bound $T$ and its adjoint operator $P_W{|}_{V^{\perp }}$ below. The first bound gives a closed range and injectivity; the second makes the range dense by image-kernel orthogonality for an adjoint. Thus $T$ is a bounded bijection, and $w=T^{-1}{P}_{V^{\perp }}f$ yields the unique direct sum decomposition $f=w+v$. In equal finite dimensions, a lower bound on $T$ alone suffices by the rank-nullity theorem.

For two finite-dimensional subspaces with projections $P,Q$, define $\cos \vartheta ={\sup }_{\Vert u\Vert =\Vert v\Vert =1}|⟨u,v⟩|=\Vert PQ\Vert$, with vectors drawn from the respective subspaces. The angle is zero when they intersect nontrivially. To measure separation after removing a common intersection, restrict both subspaces to its orthogonal complement. The resulting angle controls sums of positive operators in the Kitaev geometrical lemma.

Suppose positive operators $A,B$ have positive eigenvalues at least $\gamma$ and their nullspaces meet only at zero. If $\vartheta$ is their smallest angle between two subspaces, then $A+B\ge 2\gamma {\sin }^2(\vartheta /2)I$. Indeed $A+B\ge \gamma (2I-P-Q)$, while $(P+Q{)}^2\le (1+\cos \vartheta )(P+Q)$ gives $P+Q\le (1+\cos \vartheta )I$. With a common nullspace, apply the same proof on its orthogonal complement to bound the positive spectral gap.

For orthogonal projections $P,Q$ with $\Vert PQ\Vert \le q<1$, the identity $(PQ{)}^r=PQ(QPQ{)}^{r-1}$ and $\Vert QPQ\Vert =\Vert PQ{\Vert }^2$ imply $\Vert (PQ{)}^r\Vert \le q^{2r-1}$ for $r\ge 1$. When the projections have a common fixed subspace, first remove its orthogonal projection. This estimate is stronger than generic submultiplicativity and is useful for alternating local projection layers.