bai tap diem bat dong

8/14/2019 Bai Tap Diem Bat Dong

1/3

1. Exercises-Fixed points of c.c. maps

(1) Let K be a bounded, open, convex subset of E. Let F : K E be

completely continuous and be such that F(K) K. Then F has a fixedpoint in K.

Solution 1:(Berton Earnshaw) This solution reduces the exercise to anapplication of the Schauder Fixed Point Theorem.

Choose c K and R > 0, such that the open ball of radius R centeredat c, BR(c) K. For each b BR(c) define

r(b) = {tb + (1 t)c : t 0},

the ray starting at c and containing b Since K is bounded, it follows that

r(b) K = ,

and equals exactly one point, since K is convex. Call this point p(b). Foreach x r(b), define

F(x) =

F(x), x K

F(p(b)), x / K.

Notice that F is continuous and that F(E) F(K), implying that F iscompletely continuous.

Put

A = co

F(K) K

.

Then A is a bounded, closed, and convex set and

F : A A,

hence by the Schauder Fixed Point Theorem, F has a fixed point in the setA, call it y. Notice that ify / K, then F(y) = F(p(b)), for some b BR(c),hence, by hypothesis (F(K) K) y K. Thus the fixed point y must liein K, but then F(y) = F(y).

Solution 2: Choose c K and make the change of variables

v = u c,

then the fixed point equation

u = F(u)

is equivalent to the fixed point equation

v = F(v + c) c =: G(v),

where G is a completely continuous mapping. Define the set

K := {v = u c : u K}.

This set is a bounded open neighborhood of 0 E and

u K v = u c K

and

u K v = u c K.

We furthermore conclude that

G(K) K.1


2/3

2

If we have that F has a fixed point u K, then the proof is complete andG has a fixed point in K. Thus, assume that G has no fixed points in K.Consider the family of mappings

g(t, v) := v tG(v), 0 t 1.

This is a family of completely continuous perturbations of the identity andfor v K, g(t, v) = 0, t [0, 1]. Thus by the homotopy invarianceprinciple of the Leray-Schauder degree, we have that

d(g(t, ), K, 0) = d(id, K, 0) = 1.

We therefore conclude that the equation

v G(v) = 0

has a solution in K.We remark that in the above proof the convexity of the set K may be

replaced by the weaker requirement that K be starlike with respect to apoint c K, i.e. that the ray emenating from c will intersect the boundaryof K in exactly one point.

(2) Let be a bounded open set in Ewith 0 . Let F : Ebe completelycontinuous and satisfy

x F(x)2 F(x)2 x2, x .

then F has a fixed point in .Solution: Let us assume that F has no fixed points in . Consider the

family of c.c. perturbations of the identity

f(t, x) := x tF(x), 0 t 1.

This family has no zeros on , for t = 0, 1. If, on the other hand

f(t, x) = 0,

for some t (0, 1), and some x , then

x = tF(x),

and the inequality in the exercise becomes

(1 t)2F(x)2 (1 t2)F(x)2.

But F(x) = 0, and thus

(1 t)2 1 t2,

i.e., t 1, contradicting that t (0, 1). We hence may conclude, by the

homotopy invariance principle of Leray-Schauder degree that

d(id, , 0) = 1 = d(id F, , 0).

Which implies that F has a fixed point in . Note a particular case, wherethe above condition hold is the following:

F(x) x, x .

As an example, where this condition holds, consider the following:


3/3

3

LetF : E E

be a completely continuous mapping such there exist nonnegative constantsa and b, a < 1, such that

F(x) ax + b.

Choose R >> 1, so that b (1 a)R. Then for x R

F(x) ax + b ax + (1 a)x = x.

Thus, for such R, we may choose = BR(0) and apply the above result.

bai tap diem bat dong

Documents