Bounded Linear Operators on a Hilbert Space

194 <strong>Bounded</strong> <strong>Linear</strong> <strong>Operators</strong> on a Hilbert Space
Example 8.15 Suppose that S and T are the right and left shift operators on the
sequence space ℓ 2 (N), defined by
S(x1, x2, x3, . . .) = (0, x1, x2, x3, . . .), T (x1, x2, x3, . . .) = (x2, x3, x4, . . .).
Then T = S ∗ , since
〈x, Sy〉 = x2y1 + x3y2 + x4y3 + . . . = 〈T x, y〉.
Example 8.16 Let K : L2 ([0, 1]) → L2 ([0, 1]) be an integral operator of the form
1
Kf(x) =
0
k(x, y)f(y) dy,
where k : [0, 1] × [0, 1] → C. Then the adjoint operator
K ∗ f(x) =
1
0
k(y, x)f(y) dy
is the integral operator with the complex conjugate, transpose kernel.
The adjoint plays a crucial role in studying the solvability of a linear equation
Ax = y, (8.11)
where A : H → H is a bounded linear operator. Let z ∈ H be any solution of the
homogeneous adjoint equation,
A ∗ z = 0.
We take the inner product of (8.11) with z. The inner product on the left-hand side
vanishes because
〈Ax, z〉 = 〈x, A ∗ z〉 = 0.
Hence, a necessary condition for a solution x of (8.11) to exist is that 〈y, z〉 = 0
for all z ∈ ker A ∗ , meaning that y ∈ (ker A ∗ ) ⊥ . This condition on y is not always
sufficient to guarantee the solvability of (8.11); the most we can say for general
bounded operators is the following result.
Theorem 8.17 If A : H → H is a bounded linear operator, then
ran A = (ker A ∗ ) ⊥ , ker A = (ran A ∗ ) ⊥ . (8.12)
Proof. If x ∈ ran A, there is a y ∈ H such that x = Ay. For any z ∈ ker A ∗ , we
then have
〈x, z〉 = 〈Ay, z〉 = 〈y, A ∗ z〉 = 0.