Hilbert projection theorem

Hilbert projection theorem This article relies largely or entirely on a single source. Relevant discussion may be found on the talk page. Please help improve this article by introducing citations to additional sources. Find sources: "Hilbert projection theorem" – news · newspapers · books · scholar · JSTOR (February 2020) In mathematics, the Hilbert projection theorem is a famous result of convex analysis that says that for every vector {displaystyle x} in a Hilbert space {displaystyle H} and every nonempty closed convex {displaystyle Csubseteq H,} there exists a unique vector {displaystyle min C} for which {displaystyle |c-x|} is minimized over the vectors {displaystyle cin C} ; that is, such that {displaystyle |m-x|leq |c-x|} for every {displaystyle cin C.} Contents 1 Finite dimensional case 2 Statement 2.1 Detailed elementary proof 2.2 Proof by reduction to a special case 3 Consequences 4 Properties 5 See also 6 Notes 7 References 8 Bibliography Finite dimensional case 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 {displaystyle H} with a subspace {displaystyle C} and a point {displaystyle x.} If {displaystyle min C} is a minimizer or minimum point of the function {displaystyle N:Cto mathbb {R} } defined by {displaystyle N(c):=|c-x|} (which is the same as the minimum point of {displaystyle cmapsto |c-x|^{2}} ), then derivative must be zero at {displaystyle m.} In matrix derivative notation[1] {displaystyle {begin{aligned}partial lVert x-crVert ^{2}&=partial langle c-x,c-xrangle \&=2langle c-x,partial crangle end{aligned}}} Since {displaystyle partial c} is a vector in {displaystyle C} that represents an arbitrary tangent direction, it follows that {displaystyle m-x} must be orthogonal to every vector in {displaystyle C.} Statement Hilbert projection theorem — For every vector {displaystyle x} in a Hilbert space {displaystyle H} and every nonempty closed convex {displaystyle Csubseteq H,} there exists a unique vector {displaystyle min C} for which {displaystyle lVert x-mrVert } is equal to {displaystyle delta :=inf _{cin C}|x-c|.} If the closed subset {displaystyle C} is also a vector subspace of {displaystyle H} then this minimizer {displaystyle m} is the unique element in {displaystyle C} such that {displaystyle m-x} is orthogonal to {displaystyle C.} Detailed elementary proof Proof that a minimum point {displaystyle y} exists Let {displaystyle delta :=inf _{cin C}|x-c|} be the distance between {displaystyle x} and {displaystyle C,} {displaystyle left(c_{n}right)_{n=1}^{infty }} a sequence in {displaystyle C} such that the distance squared between {displaystyle x} and {displaystyle c_{n}} is less than or equal to {displaystyle delta ^{2}+1/n.} Let {displaystyle n} and {displaystyle m} be two integers, then the following equalities are true: {displaystyle left|c_{n}-c_{m}right|^{2}=left|c_{n}-xright|^{2}+left|c_{m}-xright|^{2}-2leftlangle c_{n}-x,,,c_{m}-xrightrangle } and {displaystyle 4left|{frac {c_{n}+c_{m}}{2}}-xright|^{2}=left|c_{n}-xright|^{2}+left|c_{m}-xright|^{2}+2leftlangle c_{n}-x,,,c_{m}-xrightrangle } Therefore {displaystyle left|c_{n}-c_{m}right|^{2}=2left|c_{n}-xright|^{2}+2left|c_{m}-xright|^{2}-4left|{frac {c_{n}+c_{m}}{2}}-xright|^{2}} (This equation is the same as the formula {displaystyle a^{2}=2b^{2}+2c^{2}-4M_{a}^{2}} for the length {displaystyle M_{a}} of a median in a triangle with sides of length {displaystyle a,b,} and {displaystyle c,} where specifically, the triangle's vertices are {displaystyle x,c_{m},c_{n}} ).

By giving an upper bound to the first two terms of the equality and by noticing that the middle of {displaystyle c_{n}} and {displaystyle c_{m}} belong to {displaystyle C} and has therefore a distance greater than or equal to {displaystyle delta } from {displaystyle x,} it follows that: {displaystyle |c_{n}-c_{m}|^{2};leq ;2left(delta ^{2}+{frac {1}{n}}right)+2left(delta ^{2}+{frac {1}{m}}right)-4delta ^{2}=2left({frac {1}{n}}+{frac {1}{m}}right)} The last inequality proves that {displaystyle left(c_{n}right)_{n=1}^{infty }} is a Cauchy sequence. Since {displaystyle C} is complete, the sequence is therefore convergent to a point {displaystyle min C,} whose distance from {displaystyle x} is minimal. {displaystyle blacksquare } Proof that {displaystyle m} is unique Let {displaystyle m_{1}} and {displaystyle m_{2}} be two minimum points. Then: {displaystyle |m_{2}-m_{1}|^{2}=2|m_{1}-x|^{2}+2|m_{2}-x|^{2}-4left|{frac {m_{1}+m_{2}}{2}}-xright|^{2}} Since {displaystyle {frac {m_{1}+m_{2}}{2}}} belongs to {displaystyle C,} we have {displaystyle left|{frac {m_{1}+m_{2}}{2}}-xright|^{2}geq delta ^{2}} and therefore {displaystyle |m_{2}-m_{1}|^{2}leq 2delta ^{2}+2delta ^{2}-4delta ^{2}=0.} Hence {displaystyle m_{1}=m_{2},} which proves uniqueness. {displaystyle blacksquare } Proof of characterization of minimum point when {displaystyle C} is a closed vector subspace Assume that {displaystyle C} is a closed vector subspace of {displaystyle H.} It must be shown the minimizer {displaystyle m} is the unique element in {displaystyle C} such that {displaystyle langle m-x,crangle =0} for every {displaystyle cin C.} Proof that the condition is sufficient: Let {displaystyle zin C} be such that {displaystyle langle z-x,crangle =0} for all {displaystyle cin C.} If {displaystyle cin C} then {displaystyle c-zin C} and so {displaystyle |c-x|^{2}=|(z-x)+(c-z)|^{2}=|z-x|^{2}+|c-z|^{2}+2langle z-x,c-zrangle =|z-x|^{2}+|c-z|^{2}} which implies that {displaystyle |z-x|^{2}leq |c-x|^{2}.} Because {displaystyle cin C} was arbitrary, this proves that {displaystyle |z-x|=inf _{cin C}|c-x|} and so {displaystyle z} is a minimum point.

Proof that the condition is necessary: Let {displaystyle min C} be the minimum point. Let {displaystyle cin C} and {displaystyle tin mathbb {R} .} Because {displaystyle m+tcin C,} the minimality of {displaystyle m} guarantees that {displaystyle |m-x|leq |(m+tc)-x|.} Thus {displaystyle |(m+tc)-x|^{2}-|m-x|^{2}=2tlangle m-x,crangle +t^{2}|c|^{2}} is always non-negative and {displaystyle langle m-x,crangle } must be a real number. If {displaystyle langle m-x,crangle neq 0} then the map {displaystyle f(t):=2tlangle m-x,crangle +t^{2}|c|^{2}} has a minimum at {displaystyle t_{0}:=-{frac {langle m-x,crangle }{|c|^{2}}}} and moreover, {displaystyle fleft(t_{0}right)<0,} which is a contradiction. Thus {displaystyle langle m-x,crangle =0.} {displaystyle blacksquare } Proof by reduction to a special case It suffices to prove the theorem in the case of {displaystyle x=0} because the general case follows from the statement below by replacing {displaystyle C} with {displaystyle C-x.} Hilbert projection theorem (case {displaystyle x=0} )[2] — For every nonempty closed convex subset {displaystyle Csubseteq H} of a Hilbert space {displaystyle H,} there exists a unique vector {displaystyle min C} such that {displaystyle inf _{cin C}|c|=|m|.} Furthermore, letting {displaystyle d:=inf _{cin C}|c|,} if {displaystyle left(c_{n}right)_{n=1}^{infty }} is any sequence in {displaystyle C} such that {displaystyle lim _{nto infty }left|c_{n}right|=d} in {displaystyle mathbb {R} } [note 1] then {displaystyle lim _{nto infty }c_{n}=m} in {displaystyle H.} Proof Let {displaystyle C} be as described in this theorem and let {displaystyle d:=inf _{cin C}|c|.} This theorem will follow from the following lemmas. Lemma 1 — If {displaystyle c_{bullet }:=left(c_{n}right)_{n=1}^{infty }} is any sequence in {displaystyle C} such that {displaystyle lim _{nto infty }left|c_{n}right|=d} in {displaystyle mathbb {R} } then there exists some {displaystyle cin C} such that {displaystyle lim _{nto infty }c_{n}=c} in {displaystyle H.} Furthermore, {displaystyle |c|=d.} show Proof of Lemma 1 Lemma 2 — A sequence {displaystyle left(c_{n}right)_{n=1}^{infty }} satisfying the hypotheses of Lemma 1 exists. show Proof of Lemma 2 Lemma 2 and Lemma 1 together prove that there exists some {displaystyle cin C} such that {displaystyle |c|=d.} Lemma 1 can be used to prove uniqueness as follows. Suppose {displaystyle bin C} is such that {displaystyle |b|=d} and denote the sequence {displaystyle b,c,b,c,b,c,ldots } by {displaystyle left(c_{n}right)_{n=1}^{infty }} so that the subsequence {displaystyle left(c_{2n}right)_{n=1}^{infty }} of even indices is the constant sequence {displaystyle c,c,c,ldots } while the subsequence {displaystyle left(c_{2n-1}right)_{n=1}^{infty }} of odd indices is the constant sequence {displaystyle b,b,b,ldots .} Because {displaystyle left|c_{n}right|=d} for every {displaystyle nin mathbb {N} ,} {displaystyle lim _{nto infty }left|c_{n}right|=lim _{nto infty }d=d} in {displaystyle mathbb {R} ,} which shows that the sequence {displaystyle left(c_{n}right)_{n=1}^{infty }} satisfies the hypotheses of Lemma 1. Lemma 1 guarantees the existence of some {displaystyle xin C} such that {displaystyle lim _{nto infty }c_{n}=x} in {displaystyle H.} Because {displaystyle left(c_{n}right)_{n=1}^{infty }} converges to {displaystyle x,} so do all of its subsequences. In particular, the subsequence {displaystyle c,c,c,ldots } converges to {displaystyle x,} which implies that {displaystyle x=c} (because limits in {displaystyle H} are unique and this constant subsequence also converges to {displaystyle c} ). Similarly, {displaystyle x=b} because the subsequence {displaystyle b,b,b,ldots } converges to both {displaystyle x} and {displaystyle b.} Thus {displaystyle b=c,} which proves the theorem. {displaystyle blacksquare } Consequences Proposition — If {displaystyle C} is a closed vector subspace of a Hilbert space {displaystyle H} then[note 3] {displaystyle H=Coplus C^{bot }.} show Proof[3] Properties 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 {displaystyle Csubseteq H} and some {displaystyle xin H,} define a function {displaystyle d_{C,x}:Cto [0,infty )quad {text{ by }}cmapsto |x-c|.} A global minimum point of {displaystyle d_{C,x},} if one exists, is any point {displaystyle m} in {displaystyle ,operatorname {domain} d_{C,x}=C,} such that {displaystyle d_{C,x}(m),leq ,d_{C,x}(c)quad {text{ for all }}cin C,} in which case {displaystyle d_{C,x}(m)=|m-x|} is equal to the global minimum value of the function {displaystyle d_{C,x},} which is: {displaystyle inf _{cin C}d_{C,x}(c)=inf _{cin C}|x-c|.} Effects of translations and scalings When this global minimum point {displaystyle m} exists and is unique then denote it by {displaystyle min(C,x);} explicitly, the defining properties of {displaystyle min(C,x)} (if it exists) are: {displaystyle min(C,x)in Cquad {text{ and }}quad left|x-min(C,x)right|leq |x-c|quad {text{ for all }}cin C.} The Hilbert projection theorem guarantees that this unique minimum point exists whenever {displaystyle C} is a non-empty closed and convex subset of a Hilbert space. However, such a minimum point can also exist in non-convex or non-closed subsets as well; for instance, just as long is {displaystyle C} is non-empty, if {displaystyle xin C} then {displaystyle min(C,x)=x.} If {displaystyle Csubseteq H} is a non-empty subset, {displaystyle s} is any scalar, and {displaystyle x,x_{0}in H} are any vectors then {displaystyle ,min left(sC+x_{0},sx+x_{0}right)=smin(C,x)+x_{0}} which implies: {displaystyle {begin{alignedat}{6}min &(sC,sx)&&=s&&min(C,x)\min &(-C,-x)&&=-&&min(C,x)\end{alignedat}}} {displaystyle {begin{alignedat}{6}min left(C+x_{0},x+x_{0}right)&=min(C,x)+x_{0}\min left(C-x_{0},x-x_{0}right)&=min(C,x)-x_{0}\end{alignedat}}} {displaystyle {begin{alignedat}{6}min &(C,-x){}&&=min(C+x,0)-x\min &(C,0);+;x;;;;&&=min(C+x,x)\min &(C-x,0){}&&=min(C,x)-x\end{alignedat}}} Examples The following counter-example demonstrates a continuous linear isomorphism {displaystyle A:Hto H} for which {displaystyle ,min(A(C),A(x))neq A(min(C,x)).} Endow {displaystyle H:=mathbb {R} ^{2}} with the dot product, let {displaystyle x_{0}:=(0,1),} and for every real {displaystyle sin mathbb {R} ,} let {displaystyle L_{s}:={(x,sx):xin mathbb {R} }} be the line of slope {displaystyle s} through the origin, where it is readily verified that {displaystyle min left(L_{s},x_{0}right)={frac {s}{1+s^{2}}}(1,s).} Pick a real number {displaystyle rneq 0} and define {displaystyle A:mathbb {R} ^{2}to mathbb {R} ^{2}} by {displaystyle A(x,y):=(rx,y)} (so this map scales the {displaystyle x-} coordinate by {displaystyle r} while leaving the {displaystyle y-} coordinate unchanged). Then {displaystyle A:mathbb {R} ^{2}to mathbb {R} ^{2}} is an invertible continuous linear operator that satisfies {displaystyle Aleft(L_{s}right)=L_{s/r}} and {displaystyle Aleft(x_{0}right)=x_{0},} so that {displaystyle ,min left(Aleft(L_{s}right),Aleft(x_{0}right)right)={frac {s}{r^{2}+s^{2}}}(1,s)} and {displaystyle Aleft(min left(L_{s},x_{0}right)right)={frac {s}{1+s^{2}}}left(r,sright).} Consequently, if {displaystyle C:=L_{s}} with {displaystyle sneq 0} and if {displaystyle (r,s)neq (pm 1,1)} then {displaystyle ,min(A(C),Aleft(x_{0}right))neq Aleft(min left(C,x_{0}right)right).} See also Orthogonal complement Orthogonal projection Orthogonality principle – Condition for optimality of Bayesian estimator Riesz representation theorem – Theorem about the dual of a Hilbert space Notes ^ Because the norm {displaystyle |cdot |:Hto mathbb {R} } is continuous, if {displaystyle lim _{nto infty }x_{n}} converges in {displaystyle H} then necessarily {displaystyle lim _{nto infty }left|x_{n}right|} converges in {displaystyle mathbb {R} .} But in general, the converse is not guaranteed. However, under this theorem's hypotheses, knowing that {displaystyle lim _{nto infty }left|c_{n}right|=d} in {displaystyle mathbb {R} } is sufficient to conclude that {displaystyle lim _{nto infty }c_{n}} converges in {displaystyle H.} ^ Explicitly, this means that given any {displaystyle epsilon >0} there exists some integer {displaystyle N>0} such that "the quantity" is {displaystyle ,leq epsilon } whenever {displaystyle m,ngeq N.} Here, "the quantity" refers to the inequality's right hand side {displaystyle 2left|c_{m}right|^{2}+2left|c_{n}right|^{2}-4d^{2}} and later in the proof, "the quantity" will also refer to {displaystyle left|c_{m}-c_{n}right|^{2}} and then {displaystyle left|c_{m}-c_{n}right|.} By definition of "Cauchy sequence," {displaystyle left(c_{n}right)_{n=1}^{infty }} is Cauchy in {displaystyle H} if and only if "the quantity" {displaystyle left|c_{m}-c_{n}right|} satisfies this aforementioned condition. ^ Technically, {displaystyle H=Koplus K^{bot }} means that the addition map {displaystyle Ktimes K^{bot }to H} defined by {displaystyle (k,p)mapsto k+p} is a surjective linear isomorphism and homeomorphism. See the article on complemented subspaces for more details. References ^ Petersen, Kaare. "The Matrix Cookbook" (PDF). Retrieved 9 January 2021. ^ Rudin 1991, pp. 306–309. ^ Rudin 1991, pp. 307−309. Bibliography Rudin, Walter (1987). Real and Complex Analysis (Third ed.). Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN 978-0-07-054236-5. OCLC 21163277. show vte Functional analysis (topics – glossary) show vte Hilbert spaces Categories: Convex analysisTheorems in functional analysis