MATH 519 Homework
Fall 2013
26. (Problem 3.6 from the text) Show that the integral equation
Z
1 a
1
f (x) = 1 +
f (y) dy for − a ≤ x ≤ a
π −a 1 + (x − y)2
has a unique bounded, continuous solution for any 0 < a < ∞. Prove that this solution
is nonnegative. (Suggestions: make use of the complete metric space
C+ ([−a, a]) = {f ∈ C([−a, a]) : f (x) ≥ 0 for all x}
Estimating the integral operator using the maximum of the kernel won’t be quite good
enough here – see the proof of Theorem 3.3 in the text.)
27. From the proof of the contraction mapping it is clear that the smaller L is, the
faster the sequence xn converges to the fixed point x. With this in mind, explain why
Newton’s method
f (xn )
xn+1 = xn − 0
f (xn )
is in general a very rapidly convergent method for approximating roots of f : R → R,
as long as the initial guess is close enough.
28. Let X be a complete metric space and T : X → X satisfy d(T (x), T (y)) < d(x, y)
for all x, y ∈ X. Show that T can have at most one fixed point,
but may have none.
√
2
(Suggestion: for an example of non-existence look at T (x) = x + 1 on X = [0, ∞), or
see Exercise 3.1 in the text.)
29. Let
Z
F = {f ∈ C([0, 1]) : |f (x) − f (y)| ≤ |x − y| for all x, y,
1
f (x) dx = 0}
0
Show that F is compact in C([0, 1]). (Suggestion: to prove that F is uniformly bounded,
justify and use the fact that if f ∈ F then f (x) = 0 for some x ∈ [0, 1].)
30. Let S denote the linear Volterra type integral operator
Z x
K(x, y)u(y) dy
Su(x) =
a
where the kernel K is continuous and satisfies |K(x, y)| ≤ M for a ≤ y ≤ x.
a) Show that
|S n u(x)| ≤
M n (x − a)n
max |u(y)| x > a n = 1, 2, . . .
a≤y≤x
n!
b) Deduce from this that for any b > a, there exists an integer n such that S n is a
contraction on C([a, b]).
c) Show that for any f ∈ C([a, b]) the second kind Volterra integral equation
Z x
K(x, y)u(y) dy = u(x) + f (x) a < x < b
a
has a unique solution u ∈ C([a, b]).