BANACH CONTRACTION MAPPING PRINCIPLE
1. Space of Continuous Functions on Sequentially Compact Spaces Theorem 1.1. Let f : M → N be a continuous function. If K is a sequentially compact subset of M , then f (K) is a sequentially compact subset of N.
Proof. Let (yn) be a sequence in f (K). Then we can choose a sequence (xn) in K such that yn = f (xn) for any n ≥ 1. Since K is sequentially compact, (xn) has a convergent subsequence (xnk) in K whose limit also belongs to K. Denote x = limk→∞xnk. Since ynk = f (xnk), by continuity of f,
k→∞lim ynk = lim
k→∞f (xnk) = f (x).
We see that (ynk) is convergent to f (x) ∈ f (K). Thus we prove that (yn) has a subsequence (ynk) converging to a point of f (K). Thus f (K) is sequentially compact. Corollary 1.1. Let K be any sequentially compact space and f : K → R be a continuous function. Then f is a bounded function on K and f (K) is a closed subset of R.
Proof. Since K is sequentially compact, f (K) is sequentially compact subset of R. By Bolzano-Weierstrass Theorem, f (K) is a bounded and closed subset of R. Proposition 1.1. The space C(K, R) of real valued continuous functions on K is a vector subspace of B(K, R) such that C(K, R) is a closed subset of B(K, R).
Proof. The previous corollary implies that C(K, R) is a subset of B(K, R). We leave it to the reader to check that C(K, R) is a vector subspace of B(K, R).
Let f be an adherent point of C(K, R). Choose a sequence (fn) in C(K, R) such that (fn) converges to f in B(K, R). To show that f belongs to C(K, R), it is equivalent to show that f is continuous.
For any > 0, we can choose N ∈ N so that kfN − f k∞< /3. Let x0 ∈ M. Since fN is continuous at x0, we can choose δ > 0 so that |fN(x) − fN(x0)| < /3 whenever d(x, x0) < δ.
For d(x, x0) < δ,
|f (x) − f (x0)| ≤ |f (x) − fN(x)| + |fN(x) − fN(x0)| + |fN(x0) − f (x0)|
≤ kfN− f k∞+
3+ kfN − f k∞
< 3 +
3 + 3 = .
Thus f is continuous at x0 for all x0 ∈ K. Thus f ∈ C(K, R). We also denote the restriction of k · k∞ to C(K, R) by k · k∞. Then
Corollary 1.2. The norm subspace (C(K, R), k · k∞) of (B(K, R), k · k∞) is a Banach space over R.
Proof. Since (B(K, R), k · k∞) is complete and C(K, R) is a closed subset of B(K, R). By Homework 7 (2), the normed subspace (C(K, R), k·k∞) is also complete. Hence (C(K, R), k·
k∞) is a Banach space over R.
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2 BANACH CONTRACTION MAPPING PRINCIPLE
2. Banach Contraction mapping principle
Let (M, d) and (N, ρ) be metric spaces. A function f : M → N is called a Lipschitz function if there exists C > 0 such that
ρ(f (x), f (y)) ≤ Cd(x, y), x, y ∈ M.
Furthermore, if C < 1, f is called a contraction mapping.
Lemma 2.1. A Lipschitz function f : M → N is continuous.
Proof. Let x0 be any point of M. For any > 0, we choose δ = /C. When d(x, x0) < δ, ρ(f (x), f (x0)) ≤ Cd(x, x0) < Cρ = .
Hence f is continuous at x0 for any x0 ∈ M. Therefore f is continuous. Definition 2.1. Let f : M → M be a function. We say that x is a fixed point of f if f (x) = x.
Theorem 2.1. Let f : M → M be a contraction mapping on a complete metric space.
Then f has a unique fixed point.
Proof. Since f is a contraction, there exists 0 < C < 1 so that d(f (x), f (x0)) ≤ Cd(x, x0).
Since any Lipschitz function is continuous, f is continuous. Choose x0 ∈ M and define a sequence (xn) by
xn= f (xn−1), n ≥ 1.
By the definition of (xn) and by the fact that f is a contraction,
d(xn+1, xn) = d(f (xn), f (xn−1)) ≤ Cd(xn, xn−1) for any n ≥ 1.
By homework 6’s technique, (xn) is a Cauchy sequence in M. Since (M, d) is complete, (xn) is convergent to a point x in M. By continuity, we find
x = lim
n→∞xn= lim
n→∞f (xn−1) = f (x).
We prove the existence of fixed points of f.
Suppose x, y are fixed points of f. If x 6= y, then d(x, y) > 0 and hence d(x, y) = d(f (x), f (y)) ≤ ρd(x, y) < d(x, y)
which is impossible. Hence the fixed point of f is unique. As an application, let us prove a special case for the existence of solution to the following ordinary differential equations. Let f : [0, 1] × R → R be a continuous function such that there exists 0 < C < 1 with
|f (t, y1) − f (t, y2)| ≤ C|y1− y2|
for any t ∈ [0, 1] and for any y1, y2 ∈ R. Let us recall the fundamental Theorem of calculus.
Theorem 2.2. (The fundamental Theorem of calculus I) Let f : [a, b] → R be a continuous function. For each a ≤ x ≤ b, define
F (x) = Z x
a
f (t)dt.
Then F is a C1 function such that F0(x) = f (x).
BANACH CONTRACTION MAPPING PRINCIPLE 3
Theorem 2.3. (Existence and Uniqueness of Solution of Ordinary Differential Equations) Let f : [0, 1] × R → R be as above. The initial value problem
(2.1) x0(t) = f (t, x(t)), x(0) = x0
has a unique solution in C[0, 1].
Proof. Integrating (2.1), we obtain an integral equation
(2.2) x(t) = x0+
Z t 0
f (s, x(s))ds.
Define T : C[0, 1] → C[0, 1] by
T (x)(t) = x0+ Z t
0
f (s, x(s))ds.
Then (2.2) has a unique solution if and only if T has a fixed point.
Let us prove that T is a contraction mapping. For x1, x2 ∈ C[0, 1], T (x1)(t) − T (x2)(t) =
Z t 0
(f (s, x1(s)) − f (s, x2(s)))ds.
For any t ∈ [0, 1],
|T (x1)(t) − T (x2)(t)| ≤ Z t
0
|f (s, x1(s)) − f (s, x2(s))|ds
≤ C Z t
0
|x1(s) − x2(s)|ds
≤ Ckx1− x2k∞t
≤ Ckx1− x2k∞. This implies that
kT (x1) − T (x2)k∞≤ Ckx1− x2k∞. In other words,
d∞(T (x1), T (x2)) ≤ Cd∞(x1, x2).
Hence T is a contraction mapping on a complete metric space (C[0, 1], d∞). By the con- traction mapping principle, T has a unique fixed point, said x ∈ C[0, 1] Since x is con- tinuous on [0, 1] and f : [0, 1] × R → R is continuous, the function g : [0, 1] → R de- fined by g(t) = f (t, x(t)) is continuous. By fundamental theorem of calculus, the function G : [0, 1] → R defined by G(s) =Rt
0g(s)ds is C1and G0(t) = g(t) = f (t, x(t)) with G(0) = 0.
Since x is a fixed point of T,
x(t) = T (x)(t) = x0+ Z t
0
f (s, x(s))ds = x0+ G(t), By fundamental Theorem of calculus, x is C1 and
x0(t) = G0(t) = f (t, x(t)).
Since x(0) = x0+ G(0) = x0. Thus x solves (2.1).
