has the form
u
(
x
) =
A
+
Bx
for constants
A
and
B
that are determined by the boundary conditions. In order for
u
to be in
C
2
b
[0
,
1], we must have
u
0
(0) = 0; since
u
0
(
x
) =
B
, we have
B
= 0. Now
u
∈
C
2
b
[0
,
1] also requires
u
(1) = 0, and since
u
(1) =
A
, we conclude that
A
= 0 too, meaning that
u
(
x
) =
A
+
Bx
= 0 for
all
x
∈
[0
,
1]. Thus, the only element of the null space is the zero function, that is, the null space
is trivial,
N
(
L
b
) =
{
0
}
.
(c) [
GRADERS
: please count (
L
b
u, u
)
≥
0 for 5 points, and count
λ
≥
0 for 4 points.]
We wish to show that (
L
b
u, u
)
≥
0 for all
∈
C
2
b
[0
,
1]. Using the first integration by parts from
part (a), we have
(
L
b
u, u
) =

[
u
0
(
x
)
u
(
x
)]
1
0
+
Z
1
0
u
0
(
x
)
u
0
(
x
)
dx
=
Z
1
0
(
u
0
(
x
))
2
dx.
Thus, (
L
b
u, u
) is the integral of a nonnegative function, so it is nonnegative.
This statement implies that all eigenvalues are nonnegative, since
λ
(
u, u
) = (
λu, u
) = (
L
b
u, u
)
≥
0
,
and we know that (
u, u
)
>
0 for all nonzero
u
due to the positivity of the inner product.
(Though not asked for in the problem statement, one can show that
u
0
(
x
)
6
= 0 for some
x
∈
[0
,
1]:
otherwise
u
would be a constant, and the only constant that satisfies the boundary conditions is
u
(
x
) = 0, which is prohibited from consideration in the problem statement. Thus, we can actually
say (
L
b
u, u
)
>
0 for all nonzero
u
.)
(d) Suppose that
L
b
u
=
λu
. The general solution to the equivalent differential equation

d
2
u
dx
2
(
x
) =
λu
(
x
)
has the form
u
(
x
) =
A
sin(
√
λx
) +
B
cos(
√
λx
)
.
Now we must find values of the constants
A
,
B
, and
√
λ
that will satisfy the boundary conditions.