Serge Ballif
MATH 501 Homework 6
November 9, 2007
1.
Let
(
V,
h·
,
·i
)
be an inner product space. Prove that if
h
x, y
i
=
k
x
k · k
y
k
and
y
6
= 0
, then
x
=
ay
for some
a
∈
.
We follow the proof outlined in Folland. Define
a
=
h
x,y
i
h
x,y
i
and
z
=
ay
.
Claim 1.
h
x, z
i
=
h
z, x
i
=
h
x, y
i
and
k
z
k
=
k
y
k
.
Proof.
We use the fact that for
α
∈
,
α
α
=

α

2
.
h
x, z
i
=
h
x, ay
i
=
a
h
x, y
i
=
h
y, x
i
h
x, y
i
h
x, y
i
=
h
x, y
i
2
h
x, y
i
=
h
x, y
i
.
h
z, x
i
=
h
ay, x
i
=
a
h
y, x
i
=
h
x, y
i
h
x, y
i
h
y, x
i
=
h
x, y
i
2
h
x, y
i
=
h
x, y
i
.
h
z, z
i
=
h
ay, ay
i
=
a
a
h
y, y
i
=
h
x, y
i
h
x, y
i
h
y, x
i
h
x, y
i
h
y, y
i
=
h
y, y
i
.
Now we can use these facts to show that for
t
∈
0
≤ h
x

tz, x

tz
i
=
k
x
k
2
+ 2 Re
h
x,

tz
i
+
k 
tz
k
2
=
k
x
k
2

2
t
h
x, z
i
+
t
2
k
z
k
2
=
k
x
k
2

2
t
h
x, y
i
+
t
2
k
y
k
2
.
Since any quadratic function
ax
2
+
bx
+
c
has its extremum at
x
=

b
2
a
, our
quadratic function in
t
has an absolute minimum at
t
=
h
x,y
i
k
y
k
2
. Substituting this
value in for
t
yields
0
≤ k
x

tz
k
2
=
k
x
k
2

2
t
h
x, y
i
+
t
2
k
y
k
2
=
k
x
k
2

2
h
x, y
i
k
y
k
2
h
x, y
i
+
h
x, y
i
k
y
k
2
2
k
y
k
2
=
k
x
k
2

h
x, y
i
k
y
k
2
.
In this case
h
x, y
i ≤ k
x
kk
y
k
.
We get equality in this statement iff
x

tz
=
x

tay
= 0 iff
x
and
y
are linearly dependent.
2.
(a) Let
y
be a nonzero element of a Hilbert space
H
. Prove that, for every
x
∈
H
,
d
(
x,
{
y
}
⊥
) =
h
x, y
i
k
y
k
.
(b) Consider the space
L
2
([0
,
1])
and
E
=
{
f
∈
L
2
([0
,
1])

R
[0
,
1]
f
(
x
)
dm
= 0
}
. Determine
E
⊥
and find the distance
of the function
e
x
to
E
.
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