Topological Degree: An Introduction

Abstract

In this chapter, we construct the Brouwer topological degree and extended it for compact perturbations of the identity in a Banach space, namely, the Leray–Schauder degree. Some topological consequences are presented. Moreover, we give applications to some boundary value problems.

Appendices

Appendix

Many important results in analysis have been used in our construction of the Brouwer degree. Most of them are very well known and require no presentation here, for example, the divergence theorem, the change of variables theorem, or the Stone–Weierstrass theorem. But others, still very important, are less famous and deserve a few paragraphs. The first one is usually referred as a lemma, although it is very strong and beautiful. We shall only prove the case that has been used in this work (as mentioned, a proof for the general case can be found, for example, in [87]).

Theorem 5.8.

(Sard’s lemma, particular case) Let $U \subset {\mathbb{R}}^{n}$ be open and $f: U \rightarrow {\mathbb{R}}^{n}$ be a C ² function. Then the set $\mathcal{C}\subset {\mathbb{R}}^{n}$ of critical values of f has zero Lebesgue measure.

Proof.

By the σ-subadditivity of the Lebesgue measure, we may assume that U is a cube of side L and that f and its derivatives are bounded by some constant M. Divide U into N ⁿ cubes of side $\frac{L} {N}$, let x be a critical point of f, and denote by C _x the small cube of side $\frac{L} {N}$ that contains x. For y ∈ Cx, we may write

$$\displaystyle{f(y) - f(x) = Df(x)(y - x) + R(y),}$$

where $\vert R(y)\vert \leq \frac{M} {2} \vert y - x{\vert }^{2} \leq \frac{Mn} {2}{ \left ( \frac{L} {2N}\right )}^{2}$. Recall that x is a critical point, so the linear transform Df(x) maps the cube C _x into a ball of smaller dimension and radius $M\sqrt{n} \frac{L} {2N}$; hence, the set {f(y): y ∈ C _x } is contained in a cylinder centered at f(x) whose basis is $c{\left ( \frac{L} {N}\right )}^{n-1}$ for some c and whose height is $Mn{\left ( \frac{L} {2N}\right )}^{2}$. Next, observe that every critical value is an image of some critical point (and there are N ⁿ cubes), so it follows that

$$\displaystyle{\vert \mathcal{C}\vert \leq {N}^{n}c{\left ( \frac{L} {N}\right )}^{n-1}M\sqrt{n}{\left ( \frac{L} {2N}\right )}^{2} \leq \frac{C} {N}}$$

for some constant C. Letting N → ∞, we conclude that $\vert \mathcal{C}\vert = 0$. □ 

In addition, several extension theorems have been used in different applications. The first of them is very well known, and the proof may be found in all basic textbooks on metric spaces.

Theorem 5.9.

(Tietze) Let X be a metric space, let A ⊂ X be closed, and let $f: A \rightarrow \mathbb{R}$ be continuous. Then there exists a continuous function $F: X \rightarrow \mathbb{R}$ such that F| _A = f.

As an immediate consequence, the result is still valid for $f: A \rightarrow {\mathbb{R}}^{n}$; furthermore, the extension F can be chosen in such a way that F(A) ⊂ co(f(A)), where, as mentioned, “co” stands for the convex hull. But in many cases a more general result is needed for a function with values in an arbitrary Banach space. This generalization is due to Dugundji. Its proof was first published in [33] and can be found in [27], among other texts.

Theorem 5.10.

(Dugundji) Let X be a metric space, let A ⊂ X be closed, and let f: A → Y be continuous, where Y is a Banach space. Then there exists a continuous function F: X → co(f(A)) such that F| _A = f.

Remark 5.5.

In particular, if $\overline{f(A)}$ is compact, then its convex hull is compact, and hence F is a compact operator.

The preceding theorems have several useful variants. In particular, when f does not vanish, it might happen that 0 ∈ co(f(A)), but still, if the domain has a smaller dimension than the codomain, then there is enough “room” to construct a nonvanishing extension. More precisely:

Lemma 5.6.

Let $K \subset {\mathbb{R}}^{n}$ be compact and $f: K \rightarrow {\mathbb{R}}^{m}\setminus \{0\}$ continuous, with m > n. Then there exists a continuous extension $F: {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{m}\setminus \{0\}$ .

Proof.

First, observe that if $K = \overline{B_{R}(0)}$, then f can be extended by

$$\displaystyle{F(x) = \left \{\begin{array}{cl} f(x) &\vert x\vert \leq R, \\ f\left (R \frac{x} {\vert x\vert }\right )&\vert x\vert > R. \end{array} \right.}$$

Thus, for the general case it suffices to show that f can be extended to some closed ball $\overline{B}$ centered at 0.

Let r = min _x ∈ K  | f(x) | , and take $g \in {C}^{1}(\overline{B}, {\mathbb{R}}^{m})$ such that $\vert g(x) - f(x)\vert < \frac{r} {4}$ for x ∈ K. By Sard’s lemma, there exists y∉Im(g) such that $\vert y\vert < \frac{r} {4}$. Let

$$\displaystyle{\sigma (t) = \left \{\begin{array}{cl} \frac{2t} {r} &\qquad t \leq \frac{r} {2},\\ 1 &\qquad t > \frac{r} {2}, \end{array} \right.}$$

and define $\phi (x) = \frac{g(x)-y} {\sigma (g(x)-y)}$.

It follows by simple computation that $\vert \phi (x)\vert \geq \frac{r} {2}$ for all $x \in \overline{B}$; furthermore, if x ∈ K, then

$$\displaystyle{\vert g(x) - y\vert \geq \vert f(x)\vert -\vert g(x) - f(x)\vert -\vert y\vert > \frac{r} {2}.}$$

Hence ϕ(x) = g(x) − y and, in particular, $\vert \phi (x) - f(x)\vert \leq \vert f(x) - g(x)\vert + \vert y\vert < \frac{r} {2}$. Thus, co(Im(ϕ − f)) ⊂ B _r∕2(0), so by Dugundji’s theorem there exists $T: \overline{B} \rightarrow {\mathbb{R}}^{m}$ continuous extending ϕ − f such that $\vert T(x)\vert \leq \max _{z\in K}\vert \phi (z) - f(z)\vert < \frac{r} {2}$ for all $x \in \overline{B}$. Then $F: \overline{B} \rightarrow {\mathbb{R}}^{m}$ given by F(x): = ϕ(x) − T(x) is continuous and satisfies $\vert F(x)\vert \geq \vert \phi (x)\vert -\vert T(x)\vert \geq \frac{r} {2}-\vert T(x)\vert >0$ for all $x \in \overline{B}$. Clearly, F is an extension of f. □ 

As corollaries, we deduce the next two lemmas, which were used in our first proof of Borsuk’s theorem.

Lemma 5.7.

Let $V \subset {\mathbb{R}}^{n}$ be open bounded and symmetric such that $0\notin \overline{V }$ , let m > n, and let $f: \partial V \rightarrow {\mathbb{R}}^{m}\setminus \{0\}$ be continuous and odd. Then there exists a continuous odd extension $F: \overline{V } \rightarrow {\mathbb{R}}^{m}\setminus \{0\}$ .

Proof.

We shall proceed by induction. The case n = 1 is left as an exercise. Next, assume the lemma is true for n − 1, and let $V _{0} = V \cap {\mathbb{R}}^{n-1}$. Then $\partial V _{0} \subset \partial V \cap {\mathbb{R}}^{n-1}$ and $f\vert _{\partial V _{0}}$ can be extended to a nonvanishing odd continuous function (still denoted f) on $\overline{V _{0}}$. Now define ${S}^{+}:=\{ x \in {\mathbb{R}}^{n}: x_{n} \geq 0\}$, $K = \overline{V _{0}} \cup (\partial V \cap {S}^{+})$. Using the preceding lemma, f is extended to a nonvanishing continuous function on $\overline{V } \cap {S}^{+}$, which in turn can be extended to $\overline{V }$ by f(x): = −f(−x) for x∉S ⁺. □ 

Lemma 5.8.

Let $V \subset {\mathbb{R}}^{n}$ be open bounded and symmetric such that $0\notin \overline{V }$ , and let $f: \partial V \rightarrow {\mathbb{R}}^{n}\setminus \{0\}$ be continuous and odd. Given a hyperplane H passing through the origin, there exists a continuous odd extension $F: \overline{V } \rightarrow {\mathbb{R}}^{n}$ such that F(x) ≠ 0 for $x \in \overline{V } \cap H$ .

Proof.

Let $V _{0} = V \cap H$; then $\partial V _{0} \subset \partial V \cap {\mathbb{R}}^{n}$, and from the previous lemma, f can be extended to a nonvanishing continuous odd function on $\overline{V }_{0}$. By Tietze’s theorem, this function can be extended continuously to $\overline{V } \cap {S}^{+}$ and thereafter to $\overline{V }$, as before. □ 

Problems

Brouwer Degree

• 5.1. Let $f: \overline{B_{1}(0)} \rightarrow {\mathbb{R}}^{n}$ be continuous such that $f(\partial B_{1}(0)) \subsetneq \partial B_{1}(0)$. Prove that deg(f, B ₁(0), 0) = 0.

• 5.2. Let $\varOmega \subset {\mathbb{R}}^{n}$ be open and bounded, $f: \overline{\varOmega } \rightarrow {\mathbb{R}}^{n}$ continuous with 0∉f(∂ Ω), and $g: U \rightarrow V \subset {\mathbb{R}}^{n}$ a diffeomorphism, where U is an open connected neighborhood of $f(\overline{\varOmega })$. Prove that if 0∉g ∘ f(∂ Ω), then

  $$\displaystyle{deg(g \circ f,\varOmega,0) = sgn(Jg(y))deg(f,\varOmega,0)}$$

  for any y ∈ U. In particular, deg(−f, Ω, 0) = (−1)ⁿ deg(f, Ω, 0).

  Note: the previous equality is an elementary case of the well-known product formula (e.g., [30, 35, 72]).

• 5.3. Let $B =\{ z \in \mathbb{C}: \vert z\vert < 1\}$, and let $f: \overline{B} \rightarrow \mathbb{C}$ be a nonconstant analytic function such that f(∂ B) ⊂ ∂ B. Prove that

  $$\displaystyle{\sum _{n\geq 1}n{\left (\frac{\vert {f}^{(n)}(0)\vert } {n!} \right )}^{2} \in \mathbb{N}.}$$

Remark 5.6.

According to the well-known minimum modulus principle, it follows that f has at least one zero in B. Can you see any relation between the number of zeros and the previous sum?

• 5.4. Prove the following theorem (Rothe): Let $\varOmega \subset {\mathbb{R}}^{n}$ be open and bounded, and let $f: \overline{\varOmega } \rightarrow {\mathbb{R}}^{n}$ be continuous. Suppose there exists x ₀ ∈ Ω such that $f(x) - x_{0}\neq t(x - x_{0})$ for all t > 1 and all x ∈ ∂ Ω. Then f has a fixed point.

• 5.5. Prove the following generalization of the fundamental theorem of algebra. Let $P(z,\overline{z})$ be a polynomial of degree less than or equal to n. Then the equation ${z}^{n} = P(z,\overline{z})$ has at least one solution. Is it true that it has at most n solutions?

• 5.6.

  • 1. Let $V \subset {\mathbb{R}}^{n}$ be open, bounded, and symmetric, and let $\varphi: \overline{V } \rightarrow \mathbb{R}$ be continuous and odd. Prove that for any $\varepsilon > 0$ there exists $\psi: \overline{V } \rightarrow \mathbb{R}$ of class C ¹ and odd such that $\|\varphi -\psi \|_{\infty } <\varepsilon$.

    Hint: extend $\varphi$ to a continuous odd function in ${\mathbb{R}}^{n}$ with compact support and take an even smooth function $\phi: {\mathbb{R}}^{n} \rightarrow \mathbb{R}$ with compact support and $\int _{{\mathbb{R}}^{n}}\phi (x)\,dx = 1$. Prove that the convolution $\varphi {\ast}\phi _{\varepsilon }$, where $\phi _{\varepsilon }(x) {=\varepsilon }^{-n}\phi (\frac{x} {\varepsilon } )$, is odd and converges uniformly to $\varphi$. An alternative (and much simpler!) proof follows by taking an arbitrary C ¹ function $\tilde{\psi }$ such that $\|\varphi -tilde\psi \|_{\infty } <\varepsilon$ and then defining $\psi (x):= \frac{\tilde{\psi }(x)-\tilde{\psi }(-x)} {2}$.

  • 2. Let V be as before, and let $f: \overline{V } \rightarrow {\mathbb{R}}^{n}$ be continuous and odd. Prove that for any $\varepsilon > 0$ there exists $g: \overline{V } \rightarrow \mathbb{R}$ of class C ¹ and odd such that $\|f - g\|_{\infty } <\varepsilon$ and Dg(0) is invertible.

    Hint: using 1, approximate f by an odd C ¹ function $\tilde{f}$. Next, write $\tilde{f}(x) = D\tilde{f}(0)x + R(x)$ and use the fact that the set of invertible matrices is dense in ${\mathbb{R}}^{n\times n}$.

• 5.7. Let $f: {S}^{n} \rightarrow {S}^{n}$ be continuous with $n \in \mathbb{N}$ even. Prove that, for some x, either f(x) = x or f(x) = −x.

• 5.8. Let $G: [0,1] \times {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n}$ be continuous, sublinear, and locally Lipschitz in x, and let P be the Poincaré operator associated to the problem x′(t) + x(t) = G(t, x(t)). Prove that P is defined in ${\mathbb{R}}^{n}$ and $deg(I \pm P,B_{R}(0),0) = 1$ for any sufficiently large R.

• 5.9. Let $F: {\mathbb{R}}^{n} \rightarrow \mathbb{R}$ be an even C ² function. Prove that the antiperiodic problem

  $$\displaystyle{\left \{\begin{array}{c} x^{\prime}(t) = \nabla F(x(t)) + p(t)\\ x(0) + x(1) = 0 \end{array} \right.}$$

  has at least one solution for any continuous $p: [0,1] \rightarrow {\mathbb{R}}^{n}$. Compare with Exercise 4.7.

  Hint: because $F(x) -\frac{\vert x{\vert }^{2}} {2}$ is also an even C ² function, it suffices to prove that the equation x′(t) + x(t) = ∇F(x(t)) + p(t) has an antiperiodic solution. Multiply by x′(t) and integrate over [0, 1] to prove that any solution x satisfies $\|x^{\prime}\|_{{L}^{2}} \leq \| p\|_{{L}^{2}}$. On the other hand, write $x(t) = x(0) +\int _{ 0}^{t}x^{\prime}(s)\,ds = x(1) -\int _{t}^{1}x^{\prime}(s)\,ds$, and deduce that $2\vert x(t)\vert \leq \| x^{\prime}\|_{{L}^{2}} \leq \| p\|_{{L}^{2}}$ for all t. Thus, it may be assumed that ∇F is bounded, and the result follows from the previous exercise.

• 5.10. Let $F: {\mathbb{R}}^{n} \rightarrow \mathbb{R}$ be a C ¹ mapping such that

  $$\displaystyle{\lim _{\vert x\vert \rightarrow \infty }F(x) \rightarrow +\infty \qquad \mbox{ and }\quad \nabla F(x)\neq 0\,\;\mbox{ for $\vert x\vert \geq R$}.}$$

  Prove that deg(∇F, B _r (0), 0) = 1 for all r sufficiently large.

  Hint: consider the autonomous equation u′(t) + ∇F(u(t)) = 0 with initial value u(0) = u ₀, and prove that if u is a solution, then $F(u(t)) = F(u_{0}) -\int _{0}^{t}\vert \nabla F(u(s)){\vert }^{2}\,ds$. Moreover, deduce that u is defined for all t > 0 and, if | u ₀ | ≥ R, then F(u(t)) < F(u ₀) for all t > 0 and verify as in (5.11) that $deg(\nabla F,B_{r}(0),0) = deg(Id - P_{\lambda },B_{r}(0),0)$ for all r large enough and all λ > 0. Finally, prove the existence of λ, r > 0 such that if | u ₀ |  = r, then $\vert P_{\lambda }(u_{0})\vert < r$ and complete the proof.

• 5.11. Let $f: [0,1] \times {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n}$ be continuous and locally Lipschitz in u. Assume there exists a bounded open $\varOmega \subset {\mathbb{R}}^{n}$ containing 0 such that

  • 1. For all λ ∈ [0, 1] and all $v_{0} \in \overline{\varOmega }$, the solution $u_{\lambda,v_{0}}$ of the initial value problem

    $$\displaystyle{\left \{\begin{array}{cl} u^{\prime\prime}(t) =\lambda f(t,u(t),u^{\prime}(t))&t \in (0,1)\\ u(0) = 0, \quad u^{\prime}(0) = v_{ 0} & \end{array} \right.}$$

    is defined over [0, 1];

  • 2. If v ₀ ∈ ∂ Ω, then $u_{\lambda,v_{0}}(1)\neq 0$.

  Prove that the Dirichlet problem

  $$\displaystyle{\left \{\begin{array}{cl} u^{\prime\prime}(t) =\lambda f(t,u(t),u^{\prime}(t))&t \in (0,1)\\ u(0) = u(1) = 0 & \end{array} \right.}$$

  has at least one solution u with u′(0) ∈ Ω.

Leray–Schauder Degree

• 5.12. * Let $\hat{x},r \in {l}^{2}:=\{ (x_{n})_{n\in \mathbb{N}} \subset {\mathbb{R}}^{\mathbb{N}}:\sum _{ n=1}^{\infty }x_{n}^{2} < \infty \}$, and let $\mathcal{C}$ be the so-called Hilbert cube defined by

  $$\displaystyle{\mathcal{C}:=\{ x \in {l}^{2}: \vert x_{ i} -\hat{ x}_{i}\vert \leq \vert r_{i}\vert \quad \mbox{ for all }\,i\}.}$$

  Prove that if $F: \mathcal{C}\rightarrow {l}^{2}$ is continuous and

  $$\displaystyle{F_{i}(y) \leq 0 \leq F_{i}(z)}$$

  for every $y,z \in \mathcal{C}$ such that $y_{i} =\hat{ x}_{i} - r_{i}$, $z_{i} =\hat{ x}_{i} + r_{i}$, then F has at least one zero.

• 5.13.

  • 1. Give an example of a continuous retraction $r: \overline{B} \rightarrow \partial B$, where B is a ball of a Banach space E.

  • 2. Using the previous example, find $h: \overline{B} \times [0,1] \rightarrow E\setminus \{0\}$ continuous such that h(x, 0) = c and h(x, 1) = x for all x ∈ ∂ B. Does this contradict the homotopy invariance of the Leray–Schauder degree?

Remark 5.7.

Another example of such a homotopy was given in Exercise 4.5. According to the usual terminology, this asserts that the sphere of E is contractible.

• 5.14. Let E be a Banach space and let $K: E \rightarrow E$ be compact such that $\frac{Kx} {\|x\|} \rightarrow 0$ as ∥ x ∥ → ∞. Prove that the equation x = Kx + y has at least one solution for arbitrary y ∈ E.

• 5.15. Extend the Borsuk theorem and its consequences for an odd mapping F = I − K, where K is a compact operator defined in a symmetric subset of a Banach space. In particular, deduce an infinite-dimensional version of Theorem 5.6.

• 5.16. Prove the Fredholm alternative: if X is a Banach space and K: X → X is linear and compact, then for any μ ≠ 0 either μ is an eigenvalue of K or μ − K is an isomorphism. More generally, if K is compact and is homogeneous of degree 1, that is, K(rx) = rKx for all $r \in \mathbb{R}$ and Kx ≠ μ x for all x, then μ − K is surjective.

  Hint: it may be assumed without loss of generality that μ = 1. If F = I − K is injective, then prove that deg(I − K, B _R (0), 0) ≠ 0 for all R > 0.

• 5.17. Let K, L: E → E be compact operators, with L linear. Suppose that

  $$\displaystyle{\|Kx - Lx\| <\| x - Lx\|}$$

  for all x such that ∥ x ∥  = R. Prove that deg(I − K, B _R (0), 0) is odd.

• 5.18. Let $g: [0,1] \times {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n}$ be continuous with g(⋅ , 0) ≡ 0, and let

  $$\displaystyle{X:=\{ u \in C[0,1]: u(0) = u(1) = 0,\|u\|_{\infty } < r\}.}$$

  Prove that if u″ + g(⋅ , u) ≠ v″ + g(⋅ , v) for all $u,v \in X \cap {C}^{2}([0,1])$ such that u ≠ v, then there exists M > 0 such that the problem

  $$\displaystyle{u^{\prime\prime}(t) + g(t,u(t)) = p(t),\qquad u(0) = u(1) = 0}$$

  has a unique solution u ∈ X for any $p: [0,1] \rightarrow {\mathbb{R}}^{n}$ continuous with ∥ p ∥  _∞  < M. Furthermore, the mapping p↦u is continuous.

Remark 5.8.

In particular, taking g(t, u) = a(t)u, we retrieve a classical “injectivity implies surjectivity” result: if the problem u″(t) + a(t)u(t) = 0 admits only the trivial solution in X, then the nonhomogeneous problem u″(t) + a(t)u(t) = p(t) has a (unique) solution u ∈ X for all p, and the mapping p↦u is continuous. Observe that, due to linearity, it suffices to prove the existence of solutions for p in a neighborhood of 0. One may argue that, here, the involved mappings are C ¹, and hence the standard implicit function theorem can be used; however, the procedure would require verifying that the operator u″ + au is an isomorphism, and this is exactly what we want to prove. More generally, the reader might try proving the following variant of the Fredholm alternative: let X, Y be Banach spaces, and let L: D ⊂ X → Y be linear with compact inverse K: Y → X. For any T: X → Y linear and continuous, if L + T is injective, then it is surjective and (L + T)⁻¹ is continuous.

• 5.19.

  • 1. Extend Exercise 5.4 to the case of a compact operator $K: \overline{\varOmega } \rightarrow E$, where Ω ⊂ E is open and bounded.

  • 2. Deduce the existence of a fixed point for a compact operator $K: \overline{B_{1}(0)} \rightarrow E$ in the following cases:

    • $K(\partial B_{1}(0)) \subset \overline{B_{1}(0)}$;

    • $\|Kx - {x\|}^{2} \geq \| K{x\|}^{2} - 1$ for ∥ x ∥  = 1;

    • E is a Hilbert space and ⟨Kx, x⟩ ≤ 1 for ∥ x ∥  = 1.

• 5.20. Using the Leray–Schauder degree theory, prove the existence of solutions of the following problems:

  • 1.
    $$\displaystyle{x^{\prime}(t) = f(t,x(t)),\qquad x(t_{0}) = x_{0}}$$

    for $f: {\mathbb{R}}^{n+1} \rightarrow {\mathbb{R}}^{n}$ continuous.

  • 2.
    $$\displaystyle{u^{\prime\prime}(t) = f(t,u(t)),\qquad u(0) = u(1) = 0}$$

    for $f: [0,1] \times {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n}$ continuous and bounded.

• 5.21. * Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be continuous. Prove the following assertions:

  • 1. If there exists R ₀ > 0 such that

    $$\displaystyle{ g(-u) > 0 > g(u)\quad \mbox{ for all }\,u > R_{0}, }$$

    (5.16)

    then the periodic problem

    $$\displaystyle{u^{\prime\prime}(t) + g(u(t)) = p(t),\qquad u(0) = u(1),\quad u^{\prime}(0) = u^{\prime}(1)}$$

    has at least one solution for arbitrary p ∈ C([0, 1]) with $\overline{p} = 0$. Is the result true if inequalities (5.16) are reversed?

  • 2. If (5.16) or the reversed inequalities hold and c ∈ C([0, 1]) is such that c(t) ≠ 0 for all t, then the problem

    $$\displaystyle{u^{\prime\prime}(t) + c(t)u^{\prime}(t) + g(u(t)) = p(t),\qquad u(0) = u(1),\quad u^{\prime}(0) = u^{\prime}(1)}$$

    has at least one solution for arbitrary p ∈ C([0, 1]) with $\overline{p} = 0$.