Measure, Integration & Real Analysis, 2021a

284 Chapter 10 Linear Maps on Hilbert Spaces
Because T ∗ ∈B(W, V), its adjoint (T ∗ ) ∗ : V → W is defined. Suppose f ∈ V.
Then
〈(T ∗ ) ∗ f , g〉 = 〈g, (T ∗ ) ∗ f 〉 = 〈T ∗ g, f 〉 = 〈 f , T ∗ g〉 = 〈Tf, g〉
for all g ∈ W. Thus (T ∗ ) ∗ f = Tf, and hence (T ∗ ) ∗ = T.
From 10.3, we see that ‖T ∗ ‖≤‖T‖. Applying this inequality with T replaced
by T ∗ we have
‖T ∗ ‖≤‖T‖ = ‖(T ∗ ) ∗ ‖≤‖T ∗ ‖.
Because the first and last terms above are the same, the first inequality must be an
equality. In other words, we have ‖T ∗ ‖ = ‖T‖.
Parts (a) and (b) of the next result show that if V and W are real Hilbert spaces,
then the function T ↦→ T ∗ from B(V, W) to B(W, V) is a linear map. However,
if V and W are nonzero complex Hilbert spaces, then T ↦→ T ∗ is not a linear map
because of the complex conjugate in (b).
10.12 properties of the adjoint
Suppose V, W, and U are Hilbert spaces. Then
(a) (S + T) ∗ = S ∗ + T ∗ for all S, T ∈B(V, W);
(b) (αT) ∗ = α T ∗ for all α ∈ F and all T ∈B(V, W);
(c) I ∗ = I, where I is the identity operator on V;
(d) (S ◦ T) ∗ = T ∗ ◦ S ∗ for all T ∈B(V, W) and S ∈B(W, U).
Proof
(a) The proof of (a) is left to the reader as an exercise.
(b) Suppose α ∈ F and T ∈B(V, W). Iff ∈ V and g ∈ W, then
〈 f , (αT) ∗ g〉 = 〈αTf, g〉 = α〈Tf, g〉 = α〈 f , T ∗ g〉 = 〈 f , α T ∗ g〉.
Thus (αT) ∗ g = α T ∗ g, as desired.
(c) If f , g ∈ V, then
Thus I ∗ g = g, as desired.
〈 f , I ∗ g〉 = 〈If, g〉 = 〈 f , g〉.
(d) Suppose T ∈B(V, W) and S ∈B(W, U). Iff ∈ V and g ∈ U, then
〈 f , (S ◦ T) ∗ g〉 = 〈(S ◦ T) f , g〉 = 〈S(Tf), g〉 = 〈Tf, S ∗ g〉 = 〈 f , T ∗ (S ∗ g)〉.
Thus (S ◦ T) ∗ g = T ∗ (S ∗ g)=(T ∗ ◦ S ∗ )(g). Hence (S ◦ T) ∗ = T ∗ ◦ S ∗ ,as
desired.
Measure, Integration & Real Analysis, by Sheldon Axler