Лекции по функциональному анализу для начинающих специалистов по математической физике. Арсеньев А.А. - 255 стр.

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A L
2
(R
1
, dx)
A
f
(x) = (/π)
1/4
exp(x
2
/2).
( > 0) : f
Dom(A) , kf
k = 1 , kAf
k > const/ , 0.
R(λ , A)f(x) =
1
(λ x
2
)
f(x) , σ(A) = [0 , ).
A
f
n
(x) Af
n
(x) Gr(A),
L
2
(R
1
, dx) L
2
(R
1
, dx)
f
0
(x) ψ(x)
A
L
2
(R
1
, dx)
f
n
(x) f
0
(x) , x
2
f
n
(x) x
2
f
0
(x) = ψ(x) , n .
L
2
(R
1
, dx)
f
0
(x) L
2
(R
1
, dx) , x
2
f
0
(x) L
2
(R
1
, dx),
f
0
(x) Dom(A) , ψ(x) = Af
0
(x) f
0
(x) ψ(x) Gr(A).
A L
2
(R
1
, dx)
B = L
2
(R
1
, dx),
Dom(A) =
{f | f(x) L
2
(R
1
, dx) , f
0
(x) L
2
(R
1
, dx) , f
00
(x) L
2
(R
1
, dx)},
A : Dom(A) L
2
(R
1
, dx) , Af(x) =
d
2
dx
2
f(x).
Îáëàñòü îïðåäåëåíèÿ îïåðàòîðà A ïëîòíà â L2 (R1 , dx), òàê êàê îíà ñî-
äåðæèò âñå ôóíêöèè ñ êîìïàêòíûì íîñèòåëåì. Îïåðàòîð A íåîãðàíè÷åí.
Äåéñòâèòåëüíî, ïîëîæèì

                         f (x) = (/π)1/4 exp(−x2 /2).

Òîãäà

   ∀( > 0) : f ∈ Dom(A) , kf k = 1 , kAf k > const/ → ∞ ,  → 0.

Ëåãêî âèäåòü, ÷òî
                                    1
               R(λ , A)f (x) =             f (x) , σ(A) = [0 , ∞).
                                 (λ − x2 )
Îïåðàòîð A çàìêíóò. Äëÿ äîêàçàòåëüñòâà ýòîãî óòâåðæäåíèÿ ðàññìîò-
ðèì ïîñëåäîâàòåëüíîñòü

                           fn (x) ⊕ Afn (x) ∈ Gr(A),

êîòîðàÿ â òîïîëîãèè ïðÿìîé ñóììû L2 (R1 , dx) ⊕ L2 (R1 , dx) ñõîäèòñÿ
ê òî÷êå f0 (x) ⊕ ψ(x) è äîêàæåì, ÷òî ýòà òî÷êà ïðèíàäëåæèò ãðàôèêó
îïåðàòîðà A.
   Èç îïðåäåëåíèÿ íîðìû â ïðÿìîé ñóììå ïðîñòðàíñòâ ñëåäóåò, ÷òî â
ìåòðèêå ïðîñòðàíñòâà L2 (R1 , dx)

           fn (x) → f0 (x) , x2 fn (x) → x2 f0 (x) = ψ(x) , n → ∞.

Èç ïîëíîòû ïðîñòðàíñòâà L2 (R1 , dx) âûòåêàåò, ÷òî

                 f0 (x) ∈ L2 (R1 , dx) , x2 f0 (x) ∈ L2 (R1 , dx),

Ñëåäîâàòåëüíî,

        f0 (x) ∈ Dom(A) , ψ(x) = Af0 (x) è f0 (x) ⊕ ψ(x) ∈ Gr(A).

Çàìåòèì, ÷òî â ðàññìîòðåííîì ïðèìåðå îáëàñòü îïðåäåëåíèÿ îïåðàòîðà
A íå çàìêíóòà â ïðîñòðàíñòâå L2 (R1 , dx).
   Ðàññìîòðèì äðóãîé ïðèìåð. Ïîëîæèì

   B = L2 (R1 , dx),
   Dom(A) =
   {f | f (x) ∈ L2 (R1 , dx) , f 0 (x) ∈ L2 (R1 , dx) , f 00 (x) ∈ L2 (R1 , dx)},
                       2   1                       d2
   A : Dom(A) → L (R , dx) , Af (x) = − 2 f (x).
                                                  dx

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