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Spectral Theory in Hilbert Space

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4. OPERATORS 25<br />

Proposition 4.2. A projection P is orthogonal if and only if it<br />

satisfies P ∗ = P .<br />

Proof. If P ∗ = P and u ∈ M, v ∈ N, then 〈u, v〉 = 〈P u, v〉 =<br />

〈u, P ∗ v〉 = 〈u, P v〉 = 〈u, 0〉 = 0 so M and N are orthogonal. Conversely,<br />

suppose M and N orthogonal. For arbitrary u, v ∈ H we<br />

then have 〈P u, v〉 = 〈P u, P v〉 + 〈P u, (I − P )v〉 = 〈P u, P v〉 so that<br />

also 〈u, P v〉 = 〈P u, P v〉. Hence 〈P u, v〉 = 〈u, P v〉 holds generally, i.e.,<br />

P ∗ = P . <br />

An operator T for which T ∗ = T is called selfadjo<strong>in</strong>t. Hence an<br />

orthogonal projection is the same as a selfadjo<strong>in</strong>t projection. We will<br />

have much more to say about selfadjo<strong>in</strong>t operators <strong>in</strong> a more general<br />

context later. Another class of operators of great <strong>in</strong>terest are the unitary<br />

operators. This is an operator U : H1 → H2 for which U ∗ = U −1 .<br />

S<strong>in</strong>ce 〈Uu, Uv〉2 = 〈U ∗ Uu, v〉1 = 〈u, v〉1 the operator U preserves the<br />

scalar product; such an operator is called isometric. If U is isometric<br />

we have 〈u, v〉1 = 〈Uu, Uv〉2 = 〈U ∗ Uu, v〉1, so that U ∗ is a left<br />

<strong>in</strong>verse of U for any isometric operator. If dim H1 = dim H2 < ∞,<br />

then a left <strong>in</strong>verse of a l<strong>in</strong>ear operator is also a right <strong>in</strong>verse, so <strong>in</strong><br />

this case isometric and unitary (orthogonal <strong>in</strong> the case of a real space)<br />

are the same th<strong>in</strong>g. If dim H1 = dim H2 or both spaces are <strong>in</strong>f<strong>in</strong>itedimensional,<br />

however, this is not the case. For example, <strong>in</strong> the space<br />

ℓ 2 we may def<strong>in</strong>e U(x1, x2, . . . ) = (0, x1, x2, . . . ), which is obviously<br />

isometric (this is a so called shift operator), but the vector (1, 0, 0, . . . )<br />

is not the image of anyth<strong>in</strong>g, so the operator is not unitary. Its adjo<strong>in</strong>t<br />

is U ∗ (x1, x2, . . . ) = (x2, x3, . . . ), which is only a partial isometry,<br />

namely an isometry on the vectors orthogonal to (1, 0, 0, . . . ). See also<br />

Exercise 4.8.<br />

It is never possible to <strong>in</strong>terpret a differential operator as a bounded<br />

operator on some <strong>Hilbert</strong> space of functions. We therefore need to<br />

discuss unbounded operators as well. Similarly, we will need to discuss<br />

operators that are not def<strong>in</strong>ed on all of H. Thus we now consider a<br />

l<strong>in</strong>ear operator T : D(T ) → H2, where the doma<strong>in</strong> D(T ) of T is some<br />

l<strong>in</strong>ear subset of H1. T is not supposed bounded. Another such operator<br />

S is said to be an extension of T if D(T ) ⊂ D(S) and Su = T u for<br />

every u ∈ D(T ). We then write T ⊂ S. We must discuss the concept<br />

of adjo<strong>in</strong>t. The form u ↦→ 〈T u, v〉2 is, for fixed v ∈ H2, only def<strong>in</strong>ed for<br />

u ∈ D(T ), and though l<strong>in</strong>ear not necessarily bounded, so there may not<br />

be any v ∗ ∈ H1 such that 〈T u, v〉2 = 〈u, v ∗ 〉1 for all u ∈ D(T ). Even<br />

if there is, it may not be uniquely determ<strong>in</strong>ed, s<strong>in</strong>ce if w ∈ D(T ) ⊥ we<br />

could replace v ∗ by v ∗ + w with no change <strong>in</strong> 〈u, v ∗ 〉. We therefore<br />

make the basic assumption that D(T ) ⊥ = {0}, i.e., D(T ) is dense <strong>in</strong><br />

H1. T is then said to be densely def<strong>in</strong>ed 2 . In this case v ∗ ∈ H1 is<br />

ter 9.<br />

2 We will discuss the case of an operator which is not densely def<strong>in</strong>ed <strong>in</strong> Chap

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