8.7. HARD TOPOLOGY THEOREMS 203

Lemma 8.7.4 If h ∈C2(

B(0,R))

and h : B(0,R)→ B(0,R), then h has a fixed point,

x such that h(x) = x.

Proof: Suppose the lemma is not true. Then for all x, |x−h(x)| ̸= 0 for all x ∈B(0,R).Then define

g(x) = h(x)+(x−h(x)) t (x)

where t (x) is nonnegative and is chosen such that g(x) ∈ ∂B(0,R) .This mapping is illustrated in the following picture.

h(x)x

g(x)

If x→ t (x) is C2 near B(0,R), it will follow g is a C2 retract onto ∂B(0,R) contrary toLemma 8.7.3. Thus t (x) is the nonnegative solution t to

|h(x)+(x−h(x)) t (x)|2 = |h(x)|2 +2(h(x) ,x−h(x)) t + t2 = R2 (8.15)

then by the quadratic formula,

t (x) =−(h(x) ,x−h(x))+√(h(x) ,x−h(x))2 +

(R2−|h(x)|2

)Is x→ t (x) a function in C2? If what is under the radical is positive, then this is so becauses→√

s is smooth for s> 0. In fact, this is the case here. The inside of the radical is positiveif R > |h(x)|. If |h(x)|= R, it is still positive because in this case, the angle between h(x)and x−h(x) cannot be π/2. This shows that x→ t (x) is the composition of C2 functionsand is therefore C2. Thus this g(x) is a C2 retract and by the above lemma, there isn’t one.■

Now it is easy to prove the Brouwer fixed point theorem. The following theorem is theBrouwer fixed point theorem for a ball.

Theorem 8.7.5 Let BR be the above closed ball and let f : BR→ BR be continuous.Then there exists x ∈ BR such that f(x) = x.

Proof: Let fk (x)≡ f(x)1+k−1 . Thus

∥fk− f∥ = maxx∈BR

{∣∣∣∣ f(x)1+(1/k)

− f(x)∣∣∣∣}= max

x∈BR

{∣∣∣∣ f(x)− f(x)(1+(1/k))1+(1/k)

∣∣∣∣}= max

x∈BR

{∣∣∣∣ f(x)(1/k)1+(1/k)

∣∣∣∣}≤ R1+ k

Letting ∥h∥ ≡max{|h(x)| : x ∈ BR} , It follows from the Weierstrass approximation theo-rem, that there exists a function whose components are polynomials gk such that

∥gk− fk∥<R

k+1