5.13. EXERCISES 151

6. ↑If A⊆ (X ,d) is totally bounded, show that Ā the closure of A is also totally bounded.In the above problem, explain why F̄ the closure of F is compact. This uses thebig theorem on compactness. Try and do this on your own, but if you get stuck,it is in the section on Arzela Ascoli theorem. When you have done this problem,you have proved the important part of the Arzela Ascoli theorem in the special casewhere the functions are defined on an interval. You can use this to prove one ofthe most important results in the theory of differential equations. This theorem is areally profound result because it gives compactness in a normed linear space whichis not finite dimensional. Thus this is a non trivial generalization of the Heine Boreltheorem.

7. Let (X ,∥·∥) be a normed linear space. A set A is said to be convex if whenever x,y ∈A the line segment determined by these points given by tx+(1− t)y for t ∈ [0,1] isalso in A. Show that every open or closed ball is convex. Remember a closed ballis D(x,r)≡ {x̂ : ∥x̂−x∥ ≤ r} while the open ball is B(x,r)≡ {x̂ : ∥x̂−x∥< r}.This should work just as easily in any normed linear space with any norm.

8. Let K be a nonempty closed and convex set in an inner product space (X , |·|) which iscomplete. For example, Fn or any other finite dimensional inner product space. Lety /∈ K and let λ = inf{|y− x| : x ∈ K} . Let {xn} be a minimizing sequence. That isλ = 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+xm2

∣∣2 + 12 |y− xn|2 + 1

2 |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.

9. 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−Py,y−Py)≤ 0

for all z ∈ K. That is, show that x = Py if and only if Re(z− x,y− x) ≤ 0 for allz ∈ K. Hint: Let x ∈ K. Then, due to convexity, a generic thing in K is of the formx+ 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)