Problem 1.
By CauchyShwartz inequality

1
0
f
(
t
) sin
πtdt
 ≤
1
0

f
(
t
) sin
πt

dt
≤
1
0

f
(
t
)

2
dt
1
0

sinπt

2
dt
=
1
√
2
1
0

f
(
t
)

2
dt
.
Also, equality holds if and only if
f
(
t
) =
C
sin
πt
. Indeed, consider functions
f
and
g
. Then we have
(
f
+
tg, f
+
tg
) =
f
2
+
t
2
g
2
+ 2
t
(
f, g
)
≥
0
.
Now, if we put
t
=

(
f, g
)
/ g
2
we get CauchyShwartz inequality.
So, for the
equality to be true we need (
f
+
tg, f
+
tg
) = 0, which means that
f
=

tg
for
some
t
.
Problem 2.
We know that (
f, g
) = 1
/
4(
f
+
g
2

f

g
2
). So, we have
(
f, g
) =
1
2
1

1

x

(
f
+
g
)
2
+ 3(
f
+
g
)
2
dx

1

1

x

(
f

g
)
2
+ 3(
f

g
)
2
dx
=
1

1

x

(
fg
) + 3(
f g
)
dx.
Now, apply CauchyShwartz inequality to functions
f
and
x
2
:

(
f, x
2
)

=

1

1

x

3
f
+ 6
xf dx
 ≤
f
x
2
=
1

1

x

f
2
+ 3

f

2
dx
1
/
2
1

1

x

3
+ 12
x
2
dx
1
/
2
=
1

1

x

f
2
+ 3

f

2
dx
1
/
2
5
√
3
.
Problem 3.
Since continuous functions are dense in
L
2
[0
,
1] and polynomials are dense in
C
[0
,
1] we conclude that polynomials are dense in
L
2
[0
,
1]. Since
L
2
[0
,
1] is sepa
rable, we can choose a countable dense set of polynomials. From this set we can
choose a linear independent set of polynomials and apply the GrammSchmidt pro
cess. In the process we use only addition and multiplication by a constant, so the
result will by also a set of polynomials.
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 Winter '08
 Zinchenko,M
 Continuous function, CauchyShwartz inequality

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