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Spectral Theory in Hilbert Space

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66 10. BOUNDARY CONDITIONS<br />

uniformly to ũ, so that <br />

K |uj−u| 2 converges both to 0 and to <br />

K |ũ−u|2 .<br />

It follows that u = ũ a.e. <strong>in</strong> K. <br />

A bounded Hermitian form on a <strong>Hilbert</strong> space H is a map H × H ∋<br />

(u, v) ↦→ B(u, v) ∈ C such that |B(u, v)| ≤ Cuv for some constant<br />

C. It is clear that the boundedness of a Hermitian form is equivalent<br />

to it be<strong>in</strong>g cont<strong>in</strong>uous as a function of its arguments. The boundary<br />

form i(〈u, T1v〉 − 〈T1u, v〉) is a bounded Hermitian form on D1, i.e.,<br />

it is a Hermitian form <strong>in</strong> u, v and is bounded by u1v1, and by<br />

Lemma 10.8 the boundary form at a, i.e., u ′ (a)v(a) − u(a)v ′ (a), is also<br />

a bounded Hermitian form (bounded by 2CKu1v1 if a ∈ K). S<strong>in</strong>ce<br />

i(〈u, T1v〉 − 〈T1u, v〉) = −i lim<br />

x→b [u, v](x) + i[u, v](a)<br />

we see that i limx→b(u ′ (x)v(x) − u(x)v ′ (x)), the boundary form at b,<br />

is also a bounded Hermitian form on D1. S<strong>in</strong>ce the forms at a and b<br />

vanish if u is <strong>in</strong> the doma<strong>in</strong> of Tc, i.e., if u vanishes near a and b, it<br />

follows that they also vanish if u ∈ D0. In particular, if u ∈ D0, then<br />

u(a) = u ′ (a) = 0. Now T0 is the adjo<strong>in</strong>t of T1 so it follows that this<br />

is the only condition at a for an element of D1 to be <strong>in</strong> D0, s<strong>in</strong>ce this<br />

guarantees that the form at a vanishes. Of course, u ∈ D0 also requires<br />

that the form at b vanishes.<br />

Now let Ta be the closure of the restriction of T1 to those elements<br />

of D1 which vanish near b, and let Da be the doma<strong>in</strong> of Ta.<br />

Then Lemma 10.7 and the boundedness of the forms at a and b show<br />

that the boundary form at b vanishes on Da and that dim D1/D0 ≥<br />

dim Da/D0 ≥ 2. We obta<strong>in</strong> the follow<strong>in</strong>g theorem.<br />

Theorem 10.9. If the <strong>in</strong>terval I has one regular endpo<strong>in</strong>t a, then<br />

n+ = n− ≥ 1. If n+ = n− = 1, then the boundary form at the s<strong>in</strong>gular<br />

endpo<strong>in</strong>t vanishes on D1, and any selfadjo<strong>in</strong>t restriction of T1 is given<br />

by a boundary condition of the form (10.7) at a and no condition at all<br />

at b.<br />

Proof. If n+ = n− = 0 then T1 = T0 so that T1 is selfadjo<strong>in</strong>t,<br />

accord<strong>in</strong>g to Theorem 9.1. But then we can not have dim D1/D0 ≥ 2.<br />

If n+ = n− = 1, then 2 = dim D1/D0 ≥ dim Da/D0 ≥ 2 so that we<br />

must have D1 = Da. Thus the boundary form at the s<strong>in</strong>gular endpo<strong>in</strong>t<br />

vanishes on D1, and the boundary form at a vanishes precisely if we<br />

impose a boundary condition of the form (10.7). <br />

If n+ = n− = 2 we obta<strong>in</strong> a selfadjo<strong>in</strong>t restriction of T1 by impos<strong>in</strong>g<br />

two appropriate boundary conditions. One of them can be a condition<br />

of the form (10.7), and then a condition at the s<strong>in</strong>gular endpo<strong>in</strong>t also<br />

has to be imposed. There are also selfadjo<strong>in</strong>t restrictions obta<strong>in</strong>ed by<br />

impos<strong>in</strong>g coupled boundary conditions. See Exercise 10.3.<br />

Whether one obta<strong>in</strong>s deficiency <strong>in</strong>dices 1 or 2 when one endpo<strong>in</strong>t is<br />

regular clearly only depends on conditions near the s<strong>in</strong>gular endpo<strong>in</strong>t.

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