370 CHAPTER 13. LEBESGUE MEASURE

Now by assumption, hi (x) = xi on ∂B(0,R) and so one can integrate by parts and write

I′ (λ ) =−∑i

∫B(0,R)

∑j

cof(Dpλ (x))i j, j (hi (x)− xi)dx = 0.

Therefore, I (λ ) equals a constant. However, |h(x)|2 = R2 so ∑pi=1 hi (x)hi (x) = R2 and so,

differentiating with respect to j,

2p

∑i=1

hi, j (x)hi (x) = 0 so Dh(x)T h(x) = 0

and so, since h(x) ̸= 0,Dh(x)T is not invertible. Hence det(Dh(x)) = 0 and so I (1) = 0 ̸=I (0) = mp (B(0,R)). This is a contradiction.

The following is the Brouwer fixed point theorem for C2 maps.

Lemma 13.11.4 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. Then define

g(x) = h(x)+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 retraction onto ∂B(0,R) contraryto Lemma 13.11.3. Thus t (x) is the nonnegative solution to

|h(x)|2 +2(

h(x) ,x−h(x)|x−h(x)|

)t + t2 = R2 (13.11.28)

then by the quadratic formula,

t (x) =−(

h(x) ,x−h(x)|x−h(x)|

)+

√(h(x) ,

x−h(x)|x−h(x)|

)2

+(

R2−|h(x)|2)

Is x→t (x) C2? If what is under the radical is positive, then there is no problem becauses→√

s is smooth for s> 0. In fact, this is the case here. The inside of the radical is positive