10.9. EXERCISES 281

hull for m > n+1, then there exist other nonnegative scalars{

t ′k}

summing to 1 suchthat v = ∑

m−1k=1 t ′kvk.Thus every vector in the convex hull of S can be obtained as a

convex combination of at most n+ 1 points of S. This incredible result is in Rudin[37]. Convexity is more a geometric property than a topological property. Hint:Consider L : Rm→V ×R defined by

L(a)≡

(m

∑k=1

akvk,m

∑k=1

ak

)

Explain why ker(L) ̸= {0} . This will involve observing that Rm has higher dimen-sion that V ×R. Thus L cannot be one to one because one to one functions takelinearly independent sets to linearly independent sets and you can’t have a linearlyindependent set with more than n+1 vectors in V×R. Next, letting a∈ ker(L)\{0}and λ ∈ R, note that λ a ∈ ker(L) . Thus for all λ ∈ R,

v =m

∑k=1

(tk +λak)vk.

Now vary λ till some tk+λak = 0 for some ak ̸= 0. You can assume each tk > 0 sinceotherwise, there is nothing to show. This is a really nice result because it can be usedto show that the convex hull of a compact set is also compact. Show this next. Thisis also Problem 22 but here it is again. This is because it is a really nice result.

7. Show that the usual norm in Fn given by

|x|= (x,x)1/2

satisfies the following identities, the first of them being the parallelogram identityand the second being the polarization identity.

|x+y|2 + |x−y|2 = 2 |x|2 +2 |y|2

Re(x,y) =14

(|x+y|2−|x−y|2

)Show that these identities hold in any inner product space, not just Fn.

8. 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. That is

λ = limn→∞|y− xn|

Explain why such a minimizing sequence exists. Next explain the following usingthe parallelogram identity in the above problem as follows.∣∣∣∣y− xn + xm

2

∣∣∣∣2 = ∣∣∣ y2 − xn

2+

y2− xm

2

∣∣∣2