201A, Fall ’10, Thomases
Homework 5 Solutions
Mihaela Ifrim
1. If
f
is continuous on [0
,
1] and if
Z
1
0
f
(
x
)
x
n
dx
= 0
,
n
∈
N
*
,
then
f
(
x
) = 0 on [0
,
1].
Hint.
The integral of the product of
f
with any polynomial is zero.
Use the Weierstrass theorem to show that
R
1
0
f
2
(
x
)
dx
= 0
.
Proof.
Since
f
is continuous on the compact set [0
,
1], it is bounded,
i.e.,

f
(
x
)
 ≤
M
∀
x
. The complex conjugate
f
(
x
) is continuous also,
so by the Weierstrass Approximation Theorem, there is a sequence of
complex polynomials
P
n
, such that
P
n
→
f
uniformly on [0
,
1]. Since
f
is bounded, then
fP
n
→
f
f
=

f

2
, as
n
→ ∞
, uniformly as well.
More precisely, this is obvious from the following fact:
k
fP
n

f
f
k
∞
≤
M
k
P
n

f
k
∞
→
0
.
By problem no 6 from HW3, we can conclude that:
0 = lim
n
→∞
Z
1
0
fP
n
dx
=
Z
1
0
lim
n
→∞
fP
n
dx
=
Z
1
0

f

2
dx.
(1)
By the continuity of
f
on [0
,
1] and by 1, it follows that
f
≡
0 on [0
,
1].
2. Consider the following equation for an unknown function
f
: [0
,
1]
→
R
:
f
(
x
) =
g
(
x
) +
λ
Z
1
0
(
x

y
)
2
f
(
y
)
dy
+
1
2
sin(
f
(
x
))
.
(2)
Prove that there exists a number
λ
0
>
0 such that for all
λ
∈
[0
,λ
0
),
and all continuous functions
g
on [0
,
1], the equation (2) has a unique
continuous solution.