Define φ e r according to the formula φ u z 1 u x 2

Define φ : E → R according to the formula
φ ( u ) =
Z
1
0
(
u ( x )
2
- u ( x )
)
dx.
Prove there exists a function u ∈ E at which φ ( u ) attains an absolute maximum.
Solution. It is easy to see that E is closed and bounded in the norm metric. It is also easy (immediate) that it is
equicontinuous. Thus it is compact. It is also quite easy to prove that φ : E → R is continuous. Thus it attains a
maximum value.
This was, I assume, the proof that Pugh intended. However, a faster and quite elementary proof was discovered by
somebody and used by several people; though in every case with a bit of something not quite right suggesting that
none of the presenters was the real author. Or was she/he?

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
non-vanishing 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

credit for proving the result assuming Z is a singleton set. Though it isn’t quite necessary, I will prove the singleton
case first.