8 I. Convex Sets at Large

m = 1, then u = u1 and u ∈ A since S ⊂ A. Suppose that m 1. Then αm 1

and we may write

u = (1 − αm)w + αmum, where w =

α1

1 − αm

u1 + . . . +

αm−1

1 − αm

um−1.

Now, w is a convex combination of u1,... , um−1 because

m−1

i=1

αi

1 − αm

=

1

1 − αm

m−1

i=1

αi =

1 − αm

1 − αm

= 1.

Therefore, by the induction hypothesis, we have w ∈ A. Since A is convex, [w, um] ⊂

A, so u ∈ A.

PROBLEMS.

1◦. Prove that conv

(

conv(S)

)

= conv(S) for any S ⊂ V .

2◦. Prove that if A ⊂ B, then conv(A) ⊂ conv(B).

3◦.

Prove that conv(A) ∪ conv(B) ⊂ conv(A ∪ B).

4. Let S ⊂ V be a set and let u, v ∈ V points such u / ∈ conv(S) and

v / ∈ conv(S). Prove that if u ∈ conv

(

S ∪ {v}

)be

and v ∈ conv

(that

S ∪ {u}

)

, then u = v.

5 (Gauss-Lucas Theorem). Let f(z) be a non-constant polynomial in one com-

plex variable z and let z1,... , zm be the roots of f (that is, the set of all solutions

to the equation f(z) = 0). Let us interpret a complex number z = x + iy as a

point (x, y) ∈

R2.

Prove that each root of the derivative f (z) lies in the convex

hull conv(z1, . . . , zm).

Hint: Without loss of generality we may suppose that f(z) = (z − z1) · · · (z −

zm). If w is a root of f (z), then

∑m

i=1 j=i

(w − zj) = 0, and, therefore,

∑m

i=1 j=i

(w − zj) = 0, where z is the complex conjugate of z. Multiply both

sides of the last identity by (w − z1) · · · (w − zn) and express w as a convex combi-

nation of z1,... , zm.

Next, we introduce two important classes of convex sets.

(2.2) Definitions. The convex hull of a finite set of points in

Rd

is called a

polytope.

Let c1,... , cm be vectors from

Rd

and let β1,... , βm be numbers. The set

P = x ∈

Rd

: ci,x ≤ βi for i = 1, . . . , m

is called a polyhedron (see Problem 2 of Section 1.3).