Kenneth Kuttler

12.2. BASIC THEORY OF HILBERT SPACES 305

Since ∥x−yn∥→ λ , this shows {yn−x} is a Cauchy sequence. Thus also {yn} is a Cauchysequence. Since H is complete, yn→ y for some y ∈ H which must be in K because K isclosed. Therefore ∥x− y∥= limn→∞ ∥x− yn∥= λ . Let Px = y. ■

Corollary 12.2.4 Let K be a closed, convex, nonempty subset of a Hilbert space, H,and let x ∈ H. Then for z ∈ K, z = Px if and only if

Re(x− z,y− z)≤ 0 (12.5)

for all y ∈ K.

Before proving this, consider what it says in the case where the Hilbert space is Rn.

Ky θ

Condition 12.5 says the angle, θ , shown in the di-agram is always obtuse. Remember from calculus,the sign of x ·y is the same as the sign of the co-sine of the included angle between x and y. Thus,in finite dimensions, the conclusion of this corollarysays that z = Px exactly when the angle of the in-dicated angle is obtuse. Surely the picture suggeststhis is reasonable.

The inequality 12.5 is an example of a variational inequality and this corollary charac-terizes the projection of x onto K as the solution of this variational inequality.

Proof of Corollary: Let z ∈ K and let y ∈ K also. Since K is convex, it follows that ift ∈ [0,1], z+ t(y− z) = (1− t)z+ ty ∈ K. Furthermore, every point of K can be written inthis way. (Let t = 1 and y ∈ K.) Therefore, z = Px if and only if for all y ∈ K and t ∈ [0,1],

∥x− (z+ t(y− z))∥2 = ∥(x− z)− t(y− z)∥2 ≥ ∥x− z∥2

for all t ∈ [0,1] and y ∈ K if and only if for all t ∈ [0,1] and y ∈ K

∥x− z∥2 + t2 ∥y− z∥2−2t Re(x− z,y− z)≥ ∥x− z∥2 (12.6)

If and only if for all t ∈ [0,1], t2 ∥y− z∥2− 2t Re(x− z,y− z) ≥ 0. Now this is equivalentto 12.6 holding for all t ∈ (0,1). Therefore, dividing by t ∈ (0,1) , 12.6 is equivalent tot ∥y− z∥2−2Re(x− z,y− z)≥ 0 for all t ∈ (0,1) which is equivalent to 12.5. ■

Corollary 12.2.5 Let K be a nonempty convex closed subset of a Hilbert space, H.Then the projection map, P is continuous. In fact, |Px−Py| ≤ |x− y| .

Proof: Let x,x′ ∈ H. Then by Corollary 12.2.4,

Re(x′−Px′,Px−Px′

)≤ 0, Re

(x−Px,Px′−Px

)≤ 0.

Hence

0 ≤ Re(x−Px,Px−Px′

)−Re

(x′−Px′,Px−Px′

)= Re

(x− x′,Px−Px′

)−∣∣Px−Px′

∣∣2and so |Px−Px′|2 ≤ |x− x′| |Px−Px′| .■

The next corollary is a more general form for the Brouwer fixed point theorem. Thiswas discussed in exercises and elsewhere earlier. However, here is a complete proof.

12.2. BASIC THEORY OF HILBERT SPACES 305Since ||x—y,|| + A, this shows {y, —x} is a Cauchy sequence. Thus also {y, } is a Cauchysequence. Since H is complete, y, — y for some y € H which must be in K because K isclosed. Therefore ||x — y|] = lim). ||¥ — yn || =A. Let Px = y.Corollary 12.2.4 Let K be a closed, convex, nonempty subset of a Hilbert space, H,and let x € H. Then for z € K, z = Px if and only ifRe(x—z,y—z) <0 (12.5)forally €K.Before proving this, consider what it says in the case where the Hilbert space is R”.Condition 12.5 says the angle, 8, shown in the di-agram is always obtuse. Remember from calculus,(e : y 0g the sign of x-y is the same as the sign of the co-K x sine of the included angle between x and y. Thus,NC Z in finite dimensions, the conclusion of this corollary— says that z = Px exactly when the angle of the in-dicated angle is obtuse. Surely the picture suggeststhis is reasonable.The inequality 12.5 is an example of a variational inequality and this corollary charac-terizes the projection of x onto K as the solution of this variational inequality.Proof of Corollary: Let z € K and let y € K also. Since K is convex, it follows that ift € [0,1], z+t(y—z) = (1 —1t)z+8ty € K. Furthermore, every point of K can be written inthis way. (Let t = 1 and y € K.) Therefore, z = Px if and only if for all y € K andt € [0,1],IIx — (<+t(y—2)) |? = ]@—z) —t(y—2) |? = lle zl?for all t € [0,1] and y € K if and only if for all t € [0,1] andy EKIIx 2]? +27 |ly —z|? —2#Re (x—z,y—z) > |lx—g |? (12.6)If and only if for all ¢ € [0,1], t? ||y —z||? — 2¢Re (x —z,y —z) > 0. Now this is equivalentto 12.6 holding for all t € (0,1). Therefore, dividing by ¢ € (0,1), 12.6 is equivalent tot ||y — ||” —2Re(x—z,y—z) > 0 for all t € (0,1) which is equivalent to 12.5. llCorollary 12.2.5 Let K be a nonempty convex closed subset of a Hilbert space, H.Then the projection map, P is continuous. In fact, |Px — Py| < |x—y].Proof: Let x,x’ € H. Then by Corollary 12.2.4,Re (x! — Px',Px— Px’) <0, Re (x — Px, Px’ — Px) <0.Hence=)IARe (x — Px, Px — Px’) —Re (x’ — Px’, Px— Px’)= Re(x—x’,Px— Px’) — |Px— Px’ |”and so |Px— Px'|? < |x —x'||Px — Px’ |.The next corollary is a more general form for the Brouwer fixed point theorem. Thiswas discussed in exercises and elsewhere earlier. However, here is a complete proof.