2.12. EXERCISES 77

22. Let K be a nonempty closed and convex set in an inner product space (X , |·|) whichis complete. For example, Fn or any other finite dimensional inner product space.Let y /∈ K and let λ = inf{|y− x| : x ∈ K} . Let {xn} be a minimizing sequence. Thatis λ = limn→∞ |y− xn| Explain why such a minimizing sequence exists. Next explainthe following using the parallelogram identity in the above problem as follows.∣∣∣∣y− xn + xm

2

∣∣∣∣2 = ∣∣∣ y2 − xn

2+

y2− xm

2

∣∣∣2=−

∣∣∣ y2− xn

2−( y

2− xm

2

)∣∣∣2 + 12|y− xn|2 +

12|y− xm|2

Hence ∣∣∣∣xm− xn

2

∣∣∣∣2 = −∣∣∣∣y− xn + xm

2

∣∣∣∣2 + 12|y− xn|2 +

12|y− xm|2

≤ −λ2 +

12|y− xn|2 +

12|y− xm|2

Next explain why the right hand side converges to 0 as m,n→ ∞. Thus {xn} is aCauchy sequence and converges to some x ∈ X . Explain why x ∈ K and |x− y|= λ .Thus there exists a closest point in K to y. Next show that there is only one closestpoint. Hint: To do this, suppose there are two x1,x2 and consider x1+x2

2 using theparallelogram law to show that this average works better than either of the two pointswhich is a contradiction unless they are really the same point. This theorem is ofenormous significance.

23. Let K be a closed convex nonempty set in a complete inner product space (H, |·|)(Hilbert space) and let y ∈ H. Denote the closest point to y by Px. Show that Px ischaracterized as being the solution to the following variational inequality

Re(z−Px,y−Px)≤ 0

for all z ∈ K. Hint: Let x ∈ K. Then, due to convexity, a generic thing in K is of theform x+ t (z− x) , t ∈ [0,1] for every z ∈ K. Then

|x+ t (z− x)− y|2 = |x− y|2 + t2 |z− x|2− t2Re(z− x,y− x)

If x = Py, then the minimum value of this on the left occurs when t = 0. Functiondefined on [0,1] has its minimum at t = 0. What does it say about the derivativeof this function at t = 0? Next consider the case that for some x the inequalityRe(z− x,y− x)≤ 0. Explain why this shows x = Py.

24. Using Problem 23 and Problem 22 show the projection map, P onto a closed con-vex subset of a complete inner product space is Lipschitz continuous with Lipschitzconstant 1. That is |Px−Py| ≤ |x− y| .

25. Suppose S is an uncountable set and suppose f (s) is a positive number for each s∈ S.Also let Ŝ denote a finite subset of S. Show that

sup

{∑s∈Ŝ

f (s) : Ŝ⊆ S

}= ∞

2.12.22.23.24.25.EXERCISES 77Let K be a nonempty closed and convex set in an inner product space (X,|-|) whichis complete. For example, F” or any other finite dimensional inner product space.Let y ¢ K and let A = inf {|y —x|: x € K}. Let {x,} be a minimizing sequence. Thatis A = limps. |y —X,| Explain why such a minimizing sequence exists. Next explainthe following using the parallelogram identity in the above problem as follows.— atin! _|y yam?2 ~“1I2 2°2 2=—|2-%_(2-*)/ 1 eal, 2B 2 (a7 a) | Fy oral + gly amlHence2 2Xm —X, Xn, +x, 1 15 n _ | _ SS m +5 lanl? +5 ly aml?< -A?4 5 ly —2xn|? + ly — Xml?Next explain why the right hand side converges to 0 as m,n — oo. Thus {x,} is aCauchy sequence and converges to some x € X. Explain why x € K and |x—y| =A.Thus there exists a closest point in K to y. Next show that there is only one closestpoint. Hint: To do this, suppose there are two x;,x2 and consider ape using theparallelogram law to show that this average works better than either of the two pointswhich is a contradiction unless they are really the same point. This theorem is ofenormous significance.Let K be a closed convex nonempty set in a complete inner product space (H,|-|)(Hilbert space) and let y € H. Denote the closest point to y by Px. Show that Px ischaracterized as being the solution to the following variational inequalityRe(z— Px, y— Px) <0for all z © K. Hint: Let x € K. Then, due to convexity, a generic thing in K is of theform x+t(z—x),t € [0,1] for every z © K. Thenbe +t (z—x)—y = ley? +0? |g — x]? —12Re(z—x,y—x)If x = Py, then the minimum value of this on the left occurs when t = 0. Functiondefined on [0,1] has its minimum at t = 0. What does it say about the derivativeof this function at t = 0? Next consider the case that for some x the inequalityRe (z—x,y—x) <0. Explain why this shows x = Py.Using Problem 23 and Problem 22 show the projection map, P onto a closed con-vex subset of a complete inner product space is Lipschitz continuous with Lipschitzconstant 1. That is |Px — Py| < |x—y].Suppose S is an uncountable set and suppose f (s) is a positive number for each s € S.Also let S denote a finite subset of S. Show thatwp { Esto :Ses} ==se$