22.1. BASIC THEORY 577
Now let λ = inf{∥x− y∥ : y ∈ K} and let yn be a minimizing sequence. This means{yn} ⊆ K satisfies limn→∞ ∥x− yn∥= λ . By the parallelogram identity,
∥yn− x+ ym− x∥2 +∥yn− ym∥2 = 2(∥yn− x∥2 +∥ym− x∥2
)and so, since ∥yn− x+ ym− x∥2 = 4(∥ yn+ym
2 − x∥2),
∥yn− ym∥2 = 2(∥yn− x∥2 +∥ym− x∥2
)−4(∥yn + ym
2− x∥2)
≤ 2(∥yn− x∥2 +∥ym− x∥2
)−4λ
2
The right side converges as m,n→ 0 to 0. Therefore, {yn}∞
n=1 is a Cauchy sequence. SinceH is complete, yn→ y for some y ∈ H which must be in K because K is closed. Therefore
∥x− y∥= limn→∞∥x− yn∥= λ .
Let Px = y. ■
Corollary 22.1.9 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 (22.6)
for all y ∈ K.
Before proving this, consider what it says in the case where the Hilbert space is Rn.
Ky θ
xz
Condition 22.6 says the angle θ , shown in the diagram, is always obtuse. Rememberfrom calculus, the sign of x ·y is the same as the sign of the cosine of the included anglebetween x and y. Thus, in finite dimensions, the conclusion of this corollary says thatz = Px exactly when the angle of the indicated angle is obtuse. Surely the picture suggeststhis is reasonable.
The inequality 22.6 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 in this 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