Лекции по функциональному анализу для начинающих специалистов по математической физике. Арсеньев А.А. - 441 стр.

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D
m
x
g
y
(ψ(x y)) = g
y
(D
m
x
ψ(x y)),
kg
y
(ψ(x y)) | (N , D(R
p
) , K
1
)k =
X
0≤|m|≤N
sup{|D
m
x
g
y
(ψ(x y))|x K
1
} =
X
0≤|m|≤N
sup{|g
y
(D
m
x
ψ(x y))|x K
1
}
φ
x
|g
y
(D
m
x
ψ(x y))| C(K
2
)kD
m
x
ψ(x y) | (M(K
2
) , D(R
q
) , K
2
)k =
C(K
2
)
X
0≤|n|≤M(K
2
)
sup{|D
n
y
D
m
x
ψ(x y)|y K
2
} < ,
(K R
p
, N) , (C(K , N) , K
0
R
p+q
, M(K , N) < ) :
kg
y
(ψ(x y)) | (N , D(R
p
) , K)k < C(K , N)kψ | M , D(R
p+q
) , K
0
k < .
(g D
0
(R
q
) , r D
0
(R
p
)) :
D(R
p+q
) 3 ψ(x y) 7→ r
x
(g
y
(ψ(x y))).
g D
?
(R
q
) r D
0
(R
p
)
r g : D(R
p+q
) 3 ψ(x y) 7→ r g(ψ) := r
x
(g
y
(ψ(x y))).
(g D
0
(R
q
) , r D
0
(R
p
) , ψ D(R
p+q
)) :
r
x
(g
y
(ψ(x y))) = g
y
(r
x
(ψ(x y))).
Èç ëåììû 6.2.6 ñëåäóåò, ÷òî

                     Dxm gy (ψ(x ⊕ y)) = gy (Dxm ψ(x ⊕ y)),
ïîýòîìó

                  kgy (ψ(x ⊕ y)) | (N , D(Rp ) , K1 )k =
                    X
                          sup{|Dxm gy (ψ(x ⊕ y))|x ∈ K1 } =
                  0≤|m|≤N
                    X
                            sup{|gy (Dxm ψ(x ⊕ y))|x ∈ K1 }
                  0≤|m|≤N

 ñèëó òåîðåìû 6.2.6 ñóùåñòâóþò òàêèå íå çàâèñÿùèå îò φ è ïîýòîìó íå
çàâèñÿùèå îò x êîíñòàíòû, ÷òî

   |gy (Dxm ψ(x ⊕ y))| ≤ C(K2 )kDxm ψ(x ⊕ y) | (M (K2 ) , D(Rq ) , K2 )k =
               X
   C(K2 )              sup{|Dyn Dxm ψ(x ⊕ y)|y ∈ K2 } < ∞,
          0≤|n|≤M (K2 )

Ñëåäîâàòåëüíî,

∀(K ∈ Rp , N ) , ∃(C(K , N ) , K0 ∈ Rp+q , M (K , N ) < ∞) :
kgy (ψ(x ⊕ y)) | (N , D(Rp ) , K)k < C(K , N )kψ | M , D(Rp+q ) , K0 k < ∞.
                                                                       (6.54)
Ëåììà äîêàçàíà.
  Èç ëåììû 6.2.8 ñëåäóåò, ÷òî êîððåêòíî îïðåäåëåíî ëèíåéíîå îòîáðà-
æåíèå

                  ∀(g ∈ D0 (Rq ) , r ∈ D0 (Rp )) :
                  D(Rp+q ) 3 ψ(x ⊕ y) 7→ rx (gy (ψ(x ⊕ y))).            (6.55)

Èç îöåíêè (6.54) ñëåäóåò, ÷òî îòáðàæåíèå (6.55) íåïðåðûâíî.
Îïðåäåëåíèå 6.2.8. Îïðåäåëåííîå ôîðìóëîé (6.55) îòîáðàæåíèå íàçû-
âàåòñÿ ïðÿìûì ïðîèçâåäåíèåì ðàñïðåäåëåíèé g ∈ D? (Rq ) è r ∈ D0 (Rp ):

       r ⊗ g : D(Rp+q ) 3 ψ(x ⊕ y) 7→ r ⊗ g(ψ) := rx (gy (ψ(x ⊕ y))).
   Àíàëîãîì òåîðåìû Ôóáèíè äëÿ ðàñïðåäåëåíèé ÿâëÿåòñÿ
Òåîðåìà 6.2.9. Ïðÿìîå ïðîèçâåäåíèå ðàñïðåäåëåíèé êîììóòàòèâíî:
                 ∀(g ∈ D0 (Rq ) , r ∈ D0 (Rp ) , ψ ∈ D(Rp+q )) :
                 rx (gy (ψ(x ⊕ y))) = gy (rx (ψ(x ⊕ y))).               (6.56)

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