Here is the full proof; anything added or subtracted cost points.
Let
u
∈
E
. Then

u
(
x
)

=

u
(
x
)

u
(0)
 ≤ 
x

=
x
for 0
≤
x
≤
1. Thus
φ
(
u
)
≤
Z
1
0
fl
fl
u
(
x
)
2

u
(
x
)
fl
fl
dx
≤
Z
1
0
(
u
(
x
)
2
+

u
(
x
)

)
dx
≤
Z
1
0
(
x
2
+
x
)
dx
=
5
6
.
Thus
φ
(
u
)
≤
5
/
6 for all
u
∈
E
. Clearly,
u
0
(
x
) =

x
is in
E
and since
φ
(
u
0
) =
Z
1
0
(
x
2
+
x
)
dx
=
5
6
,
φ
attains its maximum at
u
0
∈
E
.
3. Suppose
f
: [
a, b
]
→
R
is continuous and satisfies
Z
b
a
x
n
f
(
x
)
dx
= 0
for
n
= 0
,
1
,
2
,
. . . .
Prove:
f
(
x
) = 0 for all
x
∈
[
a, b
].
Solution.
Let
² >
0 be given. By Weierstrass’ Theorem, there exists a polynomial
p
(
x
) =
∑
n
k
=0
a
k
x
k
such that

p
(
x
)

f
(
x
)

<
²
(
b

a
)
M
for all
x
∈
[
a, b
], where
M
=
k
f
k
+ 1, so
M >
0 and

f
(
x
)
 ≤
M
for all
x
∈
[
a, b
]. Now
Z
b
a
p
(
x
)
f
(
x
)
dx
=
n
X
k
=0
a
k
Z
b
a
x
k
f
(
x
)
dx
= 0
,
thus
Z
b
a
f
(
x
)
2
dx
=
Z
b
a
f
(
x
)
2
dx

Z
b
a
p
(
x
)
f
(
x
)
dx
=
Z
b
a
(
f
(
x
)

p
(
x
))
f
(
x
)
dx
≤
Z
b
a

f
(
x
)

p
(
x
)
 
f
(
x
)

dx
≤
(
b

a
)
·
²
(
b

a
)
M
M
=
².
Thus
R
b
a
f
(
x
)
2
dx < ²
; since
² >
0 is arbitrary this implies
R
b
a
f
(
x
)
2
dx
= 0. Since
f
, hence also
f
2
is continuous, and
f
2
≥
0, we conclude that
f
2
, hence also
f
, is 0.
4. Let
X
be a compact metric space and let
A
be a subalgebra of
C
(
X
) that separates points but does not satisfy the
nonvanishing property. That is, assume
A ⊂
C
(
X
) satisfies:
(a)
f, g
∈ A
implies
f
+
g, fg
∈ A
.
(b)
f
∈ A
,
c
∈
R
implies
cf
∈ A
.
(c) The set
Z
=
{
p
∈
X
:
f
(
p
) = 0
∀
f
∈ A}
is not empty.
(d) If
p, q
∈
X
,
p /
∈
Z
, there exists
f
∈ A
such that
f
(
p
)
6
=
f
(
q
).
Prove: If
f
∈
C
(
K
) is such that
f
(
p
) = 0 for all
p
∈
Z
, then for every
² >
0 there is
g
∈ A
such that

f
(
p
)

g
(
p
)

< ²
for all
p
∈
X
.
Solution.
I made a mistake in the statement of the exercise. The statement and the properties
Z
is supposed to
have are contradictory except in one case:
Z
is a singleton set (consists of a single point). In fact, if
Z
has more points,
then
A
cannot possibly separate points. If one assumes that
Z
is a singleton set the exercise has one small trick but
is otherwise not terribly hard. Just a little bit hard, perhaps. On the other hand, if one ignores the statement and
assumes that
A
satisfies (a)(d), things get a bit more involved. I will assume that properties (a)(d) hold, so that
A
does not separate points of
Z
if
Z
has more than one point, but separates points otherwise. However, I will give full
2