Topics In Analysis

14.3 Something Which Is Not A Retract

The next lemma is a fundamental result which will be used to develop the Brouwer degree. It will also be used to give a short proof of the Brouwer fixed point theorem in the exercises. This major fixed point theorem is probably the most fundamental theorem in nonlinear analysis. The proof outlined in the exercises is from [?].

Proof: From the cofactor expansion theorem,

∑n det(Dg ) = gi,j cof (Dg)ij i=1

and so

∂det(Dg-) ∂gi,j = cof (Dg )ij (14.3.2)

(14.3.2)

which shows the last claim of the lemma. Also

∑ δkjdet(Dg ) = gi,k(cof (Dg ))ij (14.3.3) i

(14.3.3)

because if k≠j this is just the cofactor expansion of the determinant of a matrix in which the k^th and j^th columns are equal. Differentiate 14.3.3 with respect to x_j and sum on j. This yields

∑ ∑ ∑ δkj∂-(detDg-)gr,sj = gi,kj(cof (Dg )) + gi,kcof (Dg ) . r,s,j ∂gr,s ij ij ij ij,j

Hence, using δ_kj = 0 if j≠k and 14.3.2,

∑ ∑ ∑ (cof (Dg ))rs gr,sk = gr,ks(cof (Dg))rs + gi,kcof (Dg)ij,j. rs rs ij

Subtracting the first sum on the right from both sides and using the equality of mixed partials,

∑ ( ∑ ) gi,k( (cof (Dg ))ij,j) = 0. i j

If det

≠0 so that is invertible, this shows ∑ _jij,j = 0. If det = 0, let

g = g + εI k k

where ε_k → 0 and det

≡ det≠0. Then

∑ (cof (Dg )) = lim ∑ (cof (Dgk )) = 0 ■ j ij,j k→∞ j ij,j

Proof: Suppose such an h exists. Let λ ∈

and let p_λ ≡ x + λ. This function, p_λ is a homotopy of the identity map and the retraction, h. Let

∫ I(λ) ≡ -----det(Dp λ(x))dx. B(0,R)

Then using the dominated convergence theorem,

Now by assumption, h_i = x_i on ∂B

(0,R)

and so one can integrate by parts and write

∫ I′(λ) = − ∑ ∑ cof (Dp (x)) (h (x)− x )dx = 0. i B(0,R) j λ ij,j i i

Therefore, I

equals a constant. However,

( ) I (0) = mn B-(0,R-) > 0

but

∫ ∫ I(1) = det(Dh (x))dmn = # (y)dmn = 0 B(0,R) ∂B(0,R)

because from polar coordinates or other elementary reasoning, m_n

= 0. ■

The last formula uses the change of variables formula for functions which are not one to one. In this formula, #

equals the number of x such that h = y. To see this is so in case you have not seen this, note that h is C¹ and so the inverse function theorem from advanced calculus applies. Thus

Thus h is locally one to one on the two open sets ,. Now use inverse function theorem and change of variables for one to one h to verify that both of these integrals equal 0. You cover with countably many balls on which h is one to one and then use change of variables for each of these integrals over intersected with this ball.

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

Proof: Suppose the lemma is not true. Then for all x,

≠0. Then define

g (x) = h(x)+ -x-− h-(x)t(x) |x − h (x)|

where t

is nonnegative and is chosen such that g ∈ ∂B. This mapping is illustrated in the following picture.

If x → t

is C² near B

(0,R )

, it will follow g is a C² retraction onto ∂B contrary to Lemma 14.3.3. Thus t is the nonnegative solution to

( ) H (x,t) ≡ |h(x)|2 +2 h (x ), x-−-h(x) t+ t2 = R2 (14.3.4) |x − h(x)|

(14.3.4)

Then

( ) Ht (x,t) = 2 h (x), x-−-h(x)- + 2t. |x − h(x)|

If this is nonzero for all x near B

, it follows from the implicit function theorem that t is a C² function of x. Then from 14.3.4

and so If   < R, this is nonzero. If = R, then it is still nonzero unless

(h(x),x − h(x)) = 0.

But this cannot happen because the angle between h

and x − h cannot be π∕2. Alternatively, if the above equals zero, you would need

(h (x ),x) = |h(x)|2 = R2

which cannot happen unless x = h

which is assumed not to happen. Therefore, x → t is C² near B

(0,R )

and so g given above contradicts Lemma 14.3.3. ■

Then the Brouwer fixed point theorem is as follows.

Proof: You can extend f to assume it is defined on all of ℝ^p, f

⊆B

(0,R)

, the convex hull of B

(0,R )

. Then letting be a mollifier, let f_n ≡ f ∗ ψ_n. Thus

|∫ | ∫ ∫ |fn(x)| = || f (t)ψn(x − t)dt||≤ |f (t)|ψn (x− t)dt ≤ R ψn (x− t)dt = R | ℝp | ℝp ℝp

and so the restriction of f_n to B

is C². Therefore, there exists x_n ∈B

(0,R )

such that f_n = x_n. The functions f_n converge uniformly to f on B

(0,R)

.

provided n is large enough, this for every x ∈B

(0,R)

, this by uniform continuity of f on B

(0,R + 1)

. There exists a subsequence, still called which converges to x ∈B

(0,R )

. Then using the uniform convergence of f_n to f,

f (x) = lnim→∞ f (xn) = nl→im∞ fn (xn ) = nli→m∞ xn = x ■