Multivariable Advanced Calculus

248 THE LEBESGUE INTEGRAL FOR FUNCTIONS OF P VARIABLESNow by assumption, h i (x) = x i on ∂B (0, R) and so one can form iterated integralsand integrate by parts in each of the one dimensional integrals to obtainI ′ (λ) = − ∑ ∫∑cof (Dp λ (x)) ij,j(h i (x) − x i ) dx = 0.i B(0,R) jTherefore, I (λ) equals a constant. However,but∫I (1) =B(0,1)I (0) = m p (B (0, R)) > 0∫det (Dh (x)) dm p =∂B(0,1)# (y) dm p = 0because from polar coordinates or other elementary reasoning, m p (∂B (0, 1)) = 0. Thisproves the lemma. The following is the Brouwer fixed point theorem for C 2 maps.)Lemma 9.12.4 If h ∈ C(B 2 (0, R) and h : B (0, R) → B (0, R), then h has afixed point, x such that h (x) = x.Proof: Suppose the lemma is not true. Then for all x, |x − h (x)| ̸= 0. Then defineg (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 isillustrated in the following picture.xf(x)g(x)If x →t (x) is C 2 near B (0, R), it will follow g is a C 2 retraction onto ∂B (0, R)contrary to Lemma 9.12.3. Now t (x) is the nonnegative solution, t to()H (x, t) = |h (x)| 2 x − h (x)+ 2 h (x) ,t + t 2 = R 2 (9.22)|x − h (x)|Then()x − h (x)H t (x, t) = 2 h (x) ,+ 2t.|x − h (x)|If this is nonzero for all x near B (0, R), it follows from the implicit function theoremthat t is a C 2 function of x. From 9.22()x − h (x)2t = −2 h (x) ,|x − h (x)|√() 2x − h (x))± 4 h (x) ,− 4(|h (x)| 2 − R|x − h (x)|2