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Solutions: Problem Set 7
Math 201A, Fall 2006
Problem 1. Let rn = xn+1/xn be the ratio of successive terms in the
Fibonacci sequence (xn) defined by xn+1 = xn + xnโ 1 with x 0 = x 1 = 1.
Prove that rn โ ฯ as n โ โ where ฯ = (1 +
5)/2 is the golden ratio.
Solution.
- Dividing the recursion relation for the Fibonacci sequence by xn, we
get
rn+1 = 1 +
rn
, r 1 =
- Let f (r) = 1 + 1/r. Then f
โฒ (r) = โ 1 /r
2 < 0 in r > 0. Hence
f : (0, โ) โ R is monotone decreasing. Since f (3/2) = 5/3 and
f (2) = 3/2, we see that f : [3/ 2 , 2] โ [3/ 2 , 2]. Moreover |f
โฒ (r)| โค 4 / 9
for 3/ 2 โค r โค 2, so by the mean value theorem f is a contraction on
[3/ 2 , 2]. The closed interval [3/ 2 , 2] is complete.
- By the contraction mapping theorem, f has a unique fixed point r โ
[3/ 2 , 2] and the iterates rn defined by rn+1 = f (rn) with r 1 โ [3/ 2 , 2]
converge to r as n โ โ.
- The fixed points r of f satisfy r = 1 + 1/r, or r
2 โ r โ 1 = 0. Hence
r =
The unique fixed point in [3/ 2 , 2] is ฯ, so rn โ ฯ as n โ โ.
Problem 2. Show that the mapping T : R โ R defined by
T x = 1 + log (1 + e
x )
satisfies |T x โ T y| < |x โ y| for all x, y โ R with x 6 = y, but T does not
have any fixed points. Why doesnโt this example contradict the contraction
mapping theorem?
Solution.
T
โฒ (x) =
e
x
1 + ex^
we have 0 < T
โฒ < 1. The mean value theorem implies that if x 6 = y
T x โ T y = T
โฒ (ฮพ)(x โ y)
for some ฮพ between x and y, so
|T x โ T y| < |x โ y|.
T x โ x = 1 + log
1 + e
โx
so T has no fixed point. (In fact, T maps any point a distance greater
than 1 away from itself.)
- As x, y โ โ, we have |T x โ T y|/|x โ y| โ 1, so there is no constant
c < 1 such that |T x โ T y| โค c|x โ y| for all x, y โ R. Thus, T does not
satisfy the hypothesis of the contraction mapping theorem.
Problem 4. Consider the following nonlinear integral equation:
f (x) โ
ฯ
0
f
2 (y)
1 + x
2
2
dy =
, 0 โค x โค 1. (1)
Prove that there is a unique continuous solution f : [0, 1] โ [0, 1] of this
equation.
Solution.
(T f )(x) =
ฯ
0
f
2 (y)
1 + x^2 + y^2
dy.
Then f is a solution of the integral equation if and only if f = T f ,
meaning that f is a fixed point of T.
B = {f โ C([0, 1]) | f : [0, 1] โ [0, 1]}.
Then B is a closed subset of the complete space C([0, 1]), since point-
wise convergence โ and hence uniform convergence โ preserves the
condition 0 โค f (x) โค 1, so it is complete.
- Suppose that k : [0, 1] ร [0, 1] โ R is continuous. If f : [0, 1] โ R is
square-integrable and g : [0, 1] โ R is given by
g(x) =
1
0
k(x, y)f
2 (y) dy,
then
|g(x 1 ) โ g(x 2 )| =
0
[k(x 1 , y) โ k(x 2 , y)] f
2 (y) dy
0
f
2 (y) dy
sup yโ[0,1]
|k(x 1 , y) โ k(x 2 , y)|.
Since k is continuous and [0, 1] is compact, it follows that g is continu-
ous. In particular, this is true if f is continuous.
- By the previous part, if f โ C([0, 1]) then T f โ C([0, 1]). Since the
integrand in (1) is nonnegative, we have T f โฅ 0 for every f โ C[0, 1].
(In fact, T f โฅ 3 /4.) If |f | โค 1, then
T f (x) โค
ฯ
0
1 + x^2 + y^2
dy
ฯ
0
1 + y^2
dy
It follows that T maps B into itself.
- We claim that T is a contraction on B. If f, g โ B then โf โโ, โgโโ โค 1
and
|T f (x) โ T g(x)| =
ฯ
0
f
2 (y) โ g
2 (y)
1 + x^2 + y^2
dy
โค โf
2 โ g
2 โโ
ฯ
0
1 + x
2
2
dy
โค (โf โโ + โgโโ) โf โ gโโ
ฯ
0
1 + y
2
dy
โf โ gโโ.
Hence, โT f โ T gโโ โค cโf โ gโโ with c = 1/2.
- By the contraction mapping theorem, T has a unique fixed point f โ B,
which proves the result.
