Лекции по функциональному анализу для начинающих специалистов по математической физике. Арсеньев А.А. - 464 стр.

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МГУ | Москва

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f ∈ C
n , α
0
(R
d
)
∀(m , |m| ≤ n) : lim
|x|→∞
|D
m
x
f(x)| = 0.
f ∈ C
n , α
0
(R
d
)
∀( > 0) , ∃f
∈ S(R
d
) : kf − f
| C
n , α
0
(R
d
)k < ,
∀(m , |m| ≤ n , > 0) : lim sup
|x|→∞
|D
m
x
f(x)| ≤ .
f ∈ H
d/2+n+α
(R
d
) , 0 < α < 1,
f(x) ∈ C
n , α
0
(R
d
),
f C(d , α)
∀(f ∈ H
d/2+n+α
(R
d
)) : kf | C
n , α
0
(R
d
)k ≤ C(d , α)kf | H
d/2+n+α
(R
d
)k.
∀(m , |m| ≤ n , f ∈ S(R
d
)) : |D
m
x
f(x + y) − D
m
x
f(x)| =
(2π)
−d
Z
ξ
m
b
f(ξ)[exp(i(ξ , y)) −1] exp(i(ξ , x))dξ
≤
const.
Z
(1 + ξ
2
)
n/2
|
b
f(ξ)||exp(i(ξ , y)) − 1|dξ ≤
const.
Z
(1 + ξ
2
)
n+d/2+α
|
b
f(ξ)|
2
dξ
1/2
×
Z
|exp(i(ξ , y)) − 1|
2
(1 + ξ
2
)
−(d/2+α)
dξ
1/2
≤
const.kf | H
d/2+n+α
(R
d
)k
Z
|exp(i(ξ , y)) − 1|
2
|ξ|
−(d+2α)
dξ
1/2
=
C(d , α)kf | H
d/2+n+α
(R
d
)k|y|
α
.
   Îòìåòèì, ÷òî åñëè f ∈ C0n , α (Rd ), òî

                     ∀(m , |m| ≤ n) : lim |Dxm f (x)| = 0.
                                             |x|→∞


Äåéñòâèòåëüíî, åñëè f ∈ C0n , α (Rd ), òî

              ∀( > 0) , ∃f ∈ S(Rd ) : kf − f | C0n , α (Rd )k < ,

ïîýòîìó
                ∀(m , |m| ≤ n ,  > 0) : lim sup |Dxm f (x)| ≤ .
                                                  |x|→∞

Ñëåäóþùàÿ òåîðåìà íàçûâàåòñÿ òåîðåìîé âëîæåíèÿ Ñîáîëåâà (îí åå àâ-
òîð).
Òåîðåìà 6.4.4. Åñëè ðàñïðåäåëåíèå
                        f ∈ H d/2+n+α (Rd ) , 0 < α < 1,

òî îíî çàäàåòñÿ ôóíêöèåé

                                   f (x) ∈ C0n , α (Rd ),

ïðè÷åì ñóùåñòâóåò òàêàÿ íå çàâèñÿùàÿ îò                       f    êîíñòàíòà         C(d , α), ÷òî

   ∀(f ∈ H d/2+n+α (Rd )) : kf | C0n , α (Rd )k ≤ C(d , α)kf | H d/2+n+α (Rd )k.
                                                                           (6.121)

   Äîêàçàòåëüñòâî. Â ñèëó ôîðìóëû îáðàùåíèÿ ïðåîáðàçîâàíèÿ Ôóðüå
èìååì:

    ∀(m , |m| ≤ n , f ∈ S(Rd )) : |Dxm f (x + y) − Dxm f (x)| =
             Z
        −d
    (2π)       ξ m fb(ξ)[exp(i(ξ , y)) − 1] exp(i(ξ , x))dξ ≤
           Z
    const. (1 + ξ 2 )n/2 |fb(ξ)|| exp(i(ξ , y)) − 1|dξ ≤
           Z                               1/2
                       2 n+d/2+α b      2
    const.     (1 + ξ )          |f (ξ)| dξ      ×
     Z                                                     1/2
                               2          2 −(d/2+α)
          | exp(i(ξ , y)) − 1| (1 + ξ )                dξ          ≤
                                     Z                                                   1/2
                   d/2+n+α     d                                       2   −(d+2α)
    const.kf | H             (R )k        | exp(i(ξ , y)) − 1| |ξ|                   dξ          =

    C(d , α)kf | H d/2+n+α (Rd )k|y|α .                                                          (6.122)

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