In mathematics, the Hilbert projection theorem is a famous result of convex analysis that says that for every vector x {\displaystyle x} in a Hilbert space H {\displaystyle H} and every nonempty closed convex C ⊆ ⊆ --> H , {\displaystyle C\subseteq H,} there exists a unique vector m ∈ ∈ --> C {\displaystyle m\in C} for which ‖ ‖ --> c − − --> x ‖ ‖ --> {\displaystyle \|c-x\|} is minimized over the vectors c ∈ ∈ --> C {\displaystyle c\in C} ; that is, such that ‖ ‖ --> m − − --> x ‖ ‖ --> ≤ ≤ --> ‖ ‖ --> c − − --> x ‖ ‖ --> {\displaystyle \|m-x\|\leq \|c-x\|} for every c ∈ ∈ --> C . {\displaystyle c\in C.}
Some intuition for the theorem can be obtained by considering the first order condition of the optimization problem.
Consider a finite dimensional real Hilbert space H {\displaystyle H} with a subspace C {\displaystyle C} and a point x . {\displaystyle x.} If m ∈ ∈ --> C {\displaystyle m\in C} is a minimizer or minimum point of the function N : C → → --> R {\displaystyle N:C\to \mathbb {R} } defined by N ( c ) := ‖ ‖ --> c − − --> x ‖ ‖ --> {\displaystyle N(c):=\|c-x\|} (which is the same as the minimum point of c ↦ ↦ --> ‖ ‖ --> c − − --> x ‖ ‖ --> 2 {\displaystyle c\mapsto \|c-x\|^{2}} ), then derivative must be zero at m . {\displaystyle m.}
In matrix derivative notation[1]
Hilbert projection theorem — For every vector x {\displaystyle x} in a Hilbert space H {\displaystyle H} and every nonempty closed convex C ⊆ ⊆ --> H , {\displaystyle C\subseteq H,} there exists a unique vector m ∈ ∈ --> C {\displaystyle m\in C} for which ‖ ‖ --> x − − --> m ‖ ‖ --> {\displaystyle \lVert x-m\rVert } is equal to δ δ --> := inf c ∈ ∈ --> C ‖ ‖ --> x − − --> c ‖ ‖ --> . {\displaystyle \delta :=\inf _{c\in C}\|x-c\|.}
If the closed subset C {\displaystyle C} is also a vector subspace of H {\displaystyle H} then this minimizer m {\displaystyle m} is the unique element in C {\displaystyle C} such that x − − --> m {\displaystyle x-m} is orthogonal to C . {\displaystyle C.}
Let δ δ --> := inf c ∈ ∈ --> C ‖ ‖ --> x − − --> c ‖ ‖ --> {\displaystyle \delta :=\inf _{c\in C}\|x-c\|} be the distance between x {\displaystyle x} and C , {\displaystyle C,} ( c n ) n = 1 ∞ ∞ --> {\displaystyle \left(c_{n}\right)_{n=1}^{\infty }} a sequence in C {\displaystyle C} such that the distance squared between x {\displaystyle x} and c n {\displaystyle c_{n}} is less than or equal to δ δ --> 2 + 1 / n . {\displaystyle \delta ^{2}+1/n.} Let n {\displaystyle n} and m {\displaystyle m} be two integers, then the following equalities are true:
By giving an upper bound to the first two terms of the equality and by noticing that the middle of c n {\displaystyle c_{n}} and c m {\displaystyle c_{m}} belong to C {\displaystyle C} and has therefore a distance greater than or equal to δ δ --> {\displaystyle \delta } from x , {\displaystyle x,} it follows that:
The last inequality proves that ( c n ) n = 1 ∞ ∞ --> {\displaystyle \left(c_{n}\right)_{n=1}^{\infty }} is a Cauchy sequence. Since C {\displaystyle C} is complete, the sequence is therefore convergent to a point m ∈ ∈ --> C , {\displaystyle m\in C,} whose distance from x {\displaystyle x} is minimal. ◼ ◼ --> {\displaystyle \blacksquare }
Let m 1 {\displaystyle m_{1}} and m 2 {\displaystyle m_{2}} be two minimum points. Then:
Since m 1 + m 2 2 {\displaystyle {\frac {m_{1}+m_{2}}{2}}} belongs to C , {\displaystyle C,} we have ‖ m 1 + m 2 2 − − --> x ‖ 2 ≥ ≥ --> δ δ --> 2 {\displaystyle \left\|{\frac {m_{1}+m_{2}}{2}}-x\right\|^{2}\geq \delta ^{2}} and therefore
Hence m 1 = m 2 , {\displaystyle m_{1}=m_{2},} which proves uniqueness. ◼ ◼ --> {\displaystyle \blacksquare }
Assume that C {\displaystyle C} is a closed vector subspace of H . {\displaystyle H.} It must be shown the minimizer m {\displaystyle m} is the unique element in C {\displaystyle C} such that ⟨ ⟨ --> m − − --> x , c ⟩ ⟩ --> = 0 {\displaystyle \langle m-x,c\rangle =0} for every c ∈ ∈ --> C . {\displaystyle c\in C.}
Proof that the condition is sufficient: Let z ∈ ∈ --> C {\displaystyle z\in C} be such that ⟨ ⟨ --> z − − --> x , c ⟩ ⟩ --> = 0 {\displaystyle \langle z-x,c\rangle =0} for all c ∈ ∈ --> C . {\displaystyle c\in C.} If c ∈ ∈ --> C {\displaystyle c\in C} then c − − --> z ∈ ∈ --> C {\displaystyle c-z\in C} and so
Proof that the condition is necessary: Let m ∈ ∈ --> C {\displaystyle m\in C} be the minimum point. Let c ∈ ∈ --> C {\displaystyle c\in C} and t ∈ ∈ --> R . {\displaystyle t\in \mathbb {R} .} Because m + t c ∈ ∈ --> C , {\displaystyle m+tc\in C,} the minimality of m {\displaystyle m} guarantees that ‖ ‖ --> m − − --> x ‖ ‖ --> ≤ ≤ --> ‖ ‖ --> ( m + t c ) − − --> x ‖ ‖ --> . {\displaystyle \|m-x\|\leq \|(m+tc)-x\|.} Thus
It suffices to prove the theorem in the case of x = 0 {\displaystyle x=0} because the general case follows from the statement below by replacing C {\displaystyle C} with C − − --> x . {\displaystyle C-x.}
Hilbert projection theorem (case x = 0 {\displaystyle x=0} )[2] — For every nonempty closed convex subset C ⊆ ⊆ --> H {\displaystyle C\subseteq H} of a Hilbert space H , {\displaystyle H,} there exists a unique vector m ∈ ∈ --> C {\displaystyle m\in C} such that inf c ∈ ∈ --> C ‖ ‖ --> c ‖ ‖ --> = ‖ ‖ --> m ‖ ‖ --> . {\displaystyle \inf _{c\in C}\|c\|=\|m\|.}
Furthermore, letting d := inf c ∈ ∈ --> C ‖ ‖ --> c ‖ ‖ --> , {\displaystyle d:=\inf _{c\in C}\|c\|,} if ( c n ) n = 1 ∞ ∞ --> {\displaystyle \left(c_{n}\right)_{n=1}^{\infty }} is any sequence in C {\displaystyle C} such that lim n → → --> ∞ ∞ --> ‖ c n ‖ = d {\displaystyle \lim _{n\to \infty }\left\|c_{n}\right\|=d} in R {\displaystyle \mathbb {R} } [note 1] then lim n → → --> ∞ ∞ --> c n = m {\displaystyle \lim _{n\to \infty }c_{n}=m} in H . {\displaystyle H.}
Let C {\displaystyle C} be as described in this theorem and let
Lemma 1 — If c ∙ ∙ --> := ( c n ) n = 1 ∞ ∞ --> {\displaystyle c_{\bullet }:=\left(c_{n}\right)_{n=1}^{\infty }} is any sequence in C {\displaystyle C} such that lim n → → --> ∞ ∞ --> ‖ c n ‖ = d {\displaystyle \lim _{n\to \infty }\left\|c_{n}\right\|=d} in R {\displaystyle \mathbb {R} } then there exists some c ∈ ∈ --> C {\displaystyle c\in C} such that lim n → → --> ∞ ∞ --> c n = c {\displaystyle \lim _{n\to \infty }c_{n}=c} in H . {\displaystyle H.} Furthermore, ‖ ‖ --> c ‖ ‖ --> = d . {\displaystyle \|c\|=d.}
Because C {\displaystyle C} is convex, if m , n ∈ ∈ --> N {\displaystyle m,n\in \mathbb {N} } then 1 2 ( c m + c n ) ∈ ∈ --> C {\displaystyle {\frac {1}{2}}\left(c_{m}+c_{n}\right)\in C} so that by definition of the infimum, d ≤ ≤ --> ‖ 1 2 ( c m + c n ) ‖ , {\displaystyle d\leq \left\|{\frac {1}{2}}\left(c_{m}+c_{n}\right)\right\|,} which implies that 4 d 2 ≤ ≤ --> ‖ c m + c n ‖ 2 . {\displaystyle 4d^{2}\leq \left\|c_{m}+c_{n}\right\|^{2}.} By the parallelogram law,
Since H {\displaystyle H} is complete, there exists some c ∈ ∈ --> H {\displaystyle c\in H} such that lim n → → --> ∞ ∞ --> c n = c {\displaystyle \lim _{n\to \infty }c_{n}=c} in H . {\displaystyle H.} Because every c n {\displaystyle c_{n}} belongs to C , {\displaystyle C,} which is a closed subset of H , {\displaystyle H,} their limit c {\displaystyle c} must also belongs to this closed subset, which proves that c ∈ ∈ --> C . {\displaystyle c\in C.} Since the norm ‖ ‖ --> ⋅ ⋅ --> ‖ ‖ --> : H → → --> R {\displaystyle \|\,\cdot \,\|:H\to \mathbb {R} } is a continuous function, lim n → → --> ∞ ∞ --> c n = c {\displaystyle \lim _{n\to \infty }c_{n}=c} in H {\displaystyle H} implies that lim n → → --> ∞ ∞ --> ‖ c n ‖ = ‖ ‖ --> c ‖ ‖ --> {\displaystyle \lim _{n\to \infty }\left\|c_{n}\right\|=\|c\|} in R . {\displaystyle \mathbb {R} .} But lim n → → --> ∞ ∞ --> ‖ c n ‖ = d {\displaystyle \lim _{n\to \infty }\left\|c_{n}\right\|=d} also holds (by assumption) so that ‖ ‖ --> c ‖ ‖ --> = d {\displaystyle \|c\|=d} (because limits in R {\displaystyle \mathbb {R} } are unique). ◼ ◼ --> {\displaystyle \blacksquare }
Lemma 2 — A sequence ( c n ) n = 1 ∞ ∞ --> {\displaystyle \left(c_{n}\right)_{n=1}^{\infty }} satisfying the hypotheses of Lemma 1 exists.
The existence of the sequence follows from the definition of the infimum, as is now shown. The set S := { ‖ ‖ --> c ‖ ‖ --> : c ∈ ∈ --> C } {\displaystyle S:=\{\|c\|:c\in C\}} is a non-empty subset of non-negative real numbers and d := inf c ∈ ∈ --> C ‖ ‖ --> c ‖ ‖ --> = inf S . {\displaystyle d:=\inf _{c\in C}\|c\|=\inf S.} Let n ≥ ≥ --> 1 {\displaystyle n\geq 1} be an integer. Because inf S < d + 1 n , {\displaystyle \inf S<d+{\frac {1}{n}},} there exists some s n ∈ ∈ --> S {\displaystyle s_{n}\in S} such that s n < d + 1 n . {\displaystyle s_{n}<d+{\frac {1}{n}}.} Since s n ∈ ∈ --> S , {\displaystyle s_{n}\in S,} d = inf S ≤ ≤ --> s n {\displaystyle d=\inf S\leq s_{n}} holds (by definition of the infimum). Thus d ≤ ≤ --> s n < d + 1 n {\displaystyle d\leq s_{n}<d+{\frac {1}{n}}} and now the squeeze theorem implies that lim n → → --> ∞ ∞ --> s n = d {\displaystyle \lim _{n\to \infty }s_{n}=d} in R . {\displaystyle \mathbb {R} .} (This first part of the proof works for any non-empty subset of S ⊆ ⊆ --> R {\displaystyle S\subseteq \mathbb {R} } for which d := inf s ∈ ∈ --> S s {\displaystyle d:=\inf _{s\in S}s} is finite).
For every n ∈ ∈ --> N , {\displaystyle n\in \mathbb {N} ,} the fact that s n ∈ ∈ --> S = { ‖ ‖ --> c ‖ ‖ --> : c ∈ ∈ --> C } {\displaystyle s_{n}\in S=\{\|c\|:c\in C\}} means that there exists some c n ∈ ∈ --> C {\displaystyle c_{n}\in C} such that s n = ‖ c n ‖ . {\displaystyle s_{n}=\left\|c_{n}\right\|.} The convergence lim n → → --> ∞ ∞ --> s n = d {\displaystyle \lim _{n\to \infty }s_{n}=d} in R {\displaystyle \mathbb {R} } thus becomes lim n → → --> ∞ ∞ --> ‖ c n ‖ = d {\displaystyle \lim _{n\to \infty }\left\|c_{n}\right\|=d} in R . {\displaystyle \mathbb {R} .} ◼ ◼ --> {\displaystyle \blacksquare }
Lemma 2 and Lemma 1 together prove that there exists some c ∈ ∈ --> C {\displaystyle c\in C} such that ‖ ‖ --> c ‖ ‖ --> = d . {\displaystyle \|c\|=d.} Lemma 1 can be used to prove uniqueness as follows. Suppose b ∈ ∈ --> C {\displaystyle b\in C} is such that ‖ ‖ --> b ‖ ‖ --> = d {\displaystyle \|b\|=d} and denote the sequence
Proposition — If C {\displaystyle C} is a closed vector subspace of a Hilbert space H {\displaystyle H} then[note 3]
Proof that C ∩ ∩ --> C ⊥ ⊥ --> = { 0 } {\displaystyle C\cap C^{\bot }=\{0\}} :
If c ∈ ∈ --> C ∩ ∩ --> C ⊥ ⊥ --> {\displaystyle c\in C\cap C^{\bot }} then 0 = ⟨ ⟨ --> c , c ⟩ ⟩ --> = ‖ ‖ --> c ‖ ‖ --> 2 , {\displaystyle 0=\langle \,c,\,c\,\rangle =\|c\|^{2},} which implies c = 0. {\displaystyle c=0.} ◼ ◼ --> {\displaystyle \blacksquare }
Proof that C ⊥ ⊥ --> {\displaystyle C^{\bot }} is a closed vector subspace of H {\displaystyle H} :
Let P := ∏ ∏ --> c ∈ ∈ --> C F {\displaystyle P:=\prod _{c\in C}\mathbb {F} } where F {\displaystyle \mathbb {F} } is the underlying scalar field of H {\displaystyle H} and define
Proof that C + C ⊥ ⊥ --> = H {\displaystyle C+C^{\bot }=H} :
Let x ∈ ∈ --> H . {\displaystyle x\in H.} The Hilbert projection theorem guarantees the existence of a unique m ∈ ∈ --> C {\displaystyle m\in C} such that ‖ ‖ --> x − − --> m ‖ ‖ --> ≤ ≤ --> ‖ ‖ --> x − − --> c ‖ ‖ --> for all c ∈ ∈ --> C {\displaystyle \|x-m\|\leq \|x-c\|{\text{ for all }}c\in C} (or equivalently, for all x − − --> c ∈ ∈ --> x − − --> C {\displaystyle x-c\in x-C} ). Let p := x − − --> m {\displaystyle p:=x-m} so that x = m + p ∈ ∈ --> C + p {\displaystyle x=m+p\in C+p} and it remains to show that p ∈ ∈ --> C ⊥ ⊥ --> . {\displaystyle p\in C^{\bot }.} The inequality above can be rewritten as:
Expression as a global minimum
The statement and conclusion of the Hilbert projection theorem can be expressed in terms of global minimums of the followings functions. Their notation will also be used to simplify certain statements.
Given a non-empty subset C ⊆ ⊆ --> H {\displaystyle C\subseteq H} and some x ∈ ∈ --> H , {\displaystyle x\in H,} define a function
Effects of translations and scalings
When this global minimum point m {\displaystyle m} exists and is unique then denote it by min ( C , x ) ; {\displaystyle \min(C,x);} explicitly, the defining properties of min ( C , x ) {\displaystyle \min(C,x)} (if it exists) are:
If C ⊆ ⊆ --> H {\displaystyle C\subseteq H} is a non-empty subset, s {\displaystyle s} is any scalar, and x , x 0 ∈ ∈ --> H {\displaystyle x,x_{0}\in H} are any vectors then
Examples
The following counter-example demonstrates a continuous linear isomorphism A : H → → --> H {\displaystyle A:H\to H} for which min ( A ( C ) , A ( x ) ) ≠ ≠ --> A ( min ( C , x ) ) . {\displaystyle \,\min(A(C),A(x))\neq A(\min(C,x)).} Endow H := R 2 {\displaystyle H:=\mathbb {R} ^{2}} with the dot product, let x 0 := ( 0 , 1 ) , {\displaystyle x_{0}:=(0,1),} and for every real s ∈ ∈ --> R , {\displaystyle s\in \mathbb {R} ,} let L s := { ( x , s x ) : x ∈ ∈ --> R } {\displaystyle L_{s}:=\{(x,sx):x\in \mathbb {R} \}} be the line of slope s {\displaystyle s} through the origin, where it is readily verified that min ( L s , x 0 ) = s 1 + s 2 ( 1 , s ) . {\displaystyle \min \left(L_{s},x_{0}\right)={\frac {s}{1+s^{2}}}(1,s).} Pick a real number r ≠ ≠ --> 0 {\displaystyle r\neq 0} and define A : R 2 → → --> R 2 {\displaystyle A:\mathbb {R} ^{2}\to \mathbb {R} ^{2}} by A ( x , y ) := ( r x , y ) {\displaystyle A(x,y):=(rx,y)} (so this map scales the x − − --> {\displaystyle x-} coordinate by r {\displaystyle r} while leaving the y − − --> {\displaystyle y-} coordinate unchanged). Then A : R 2 → → --> R 2 {\displaystyle A:\mathbb {R} ^{2}\to \mathbb {R} ^{2}} is an invertible continuous linear operator that satisfies A ( L s ) = L s / r {\displaystyle A\left(L_{s}\right)=L_{s/r}} and A ( x 0 ) = x 0 , {\displaystyle A\left(x_{0}\right)=x_{0},} so that min ( A ( L s ) , A ( x 0 ) ) = s r 2 + s 2 ( 1 , s ) {\displaystyle \,\min \left(A\left(L_{s}\right),A\left(x_{0}\right)\right)={\frac {s}{r^{2}+s^{2}}}(1,s)} and A ( min ( L s , x 0 ) ) = s 1 + s 2 ( r , s ) . {\displaystyle A\left(\min \left(L_{s},x_{0}\right)\right)={\frac {s}{1+s^{2}}}\left(r,s\right).} Consequently, if C := L s {\displaystyle C:=L_{s}} with s ≠ ≠ --> 0 {\displaystyle s\neq 0} and if ( r , s ) ≠ ≠ --> ( ± ± --> 1 , 1 ) {\displaystyle (r,s)\neq (\pm 1,1)} then min ( A ( C ) , A ( x 0 ) ) ≠ ≠ --> A ( min ( C , x 0 ) ) . {\displaystyle \,\min(A(C),A\left(x_{0}\right))\neq A\left(\min \left(C,x_{0}\right)\right).}
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