Лекции по функциональному анализу для начинающих специалистов по математической физике. Арсеньев А.А. - 472 стр.

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Z
. . . dx
R
d
Z
exp(ax
2
)dx = (π/a)
d/2
,
Z
exp(ax
2
+ )dx =
Z
exp(a(x ξ/2a)
2
+ ξ
2
/4a)dx) = (π/a)
d/2
exp(ξ
2
/4a) ,
Z
exp(ax
2
+ ixξ)dx = (π/a)
d/2
exp(ξ
2
/4a),
exp(ξ
2
/4a) = (a/π)
d/2
Z
exp(ax
2
+ ixξ)dx ,
(4πt)
d/2
exp((x y)
2
/4t) = (2π)
d
Z
exp(tz
2
+ iz(x y))dz ,
(4πt)
d/2
Z
exp((x y)
2
/4t)f(y)dy = (2π)
d
Z
(
Z
exp(tz
2
+ iz(x y))f(y)dy)dz ,
(π)
d/2
Z
exp(z
2
)f(x + 2
tz)dz = (2π)
d
Z
exp(tz
2
+ ixz)F (f)(z)dz.
t +0
f(x) = (2π)
d
Z
exp(izx)F (f)(z)dz.
Z
f(x)
g(x)dx = (2π)
d
Z
f(x)
(
Z
exp(ixz)F (g)(z)dz)dx =
(2π)
d
Z
F (f)
(z)F (g)(z)dz.
6.5     Êîìåíòàðèè è ëèòåðàòóðíûå óêàçàíèÿ.

♠

6.5.1    Ïðåîáðàçîâàíèå Ôóðüå.

Ïðèâåäåì âûâîä îòíîñÿùèõñÿ ê ïðåîáðàçîâàíèþ Ôóðüå ôîðìóë. Íèæå
ñèìâîë                      Z
                              . . . dx

îçíà÷àåò èíòåãðàë ïî ïðîñòðàíñòâó Rd . Èìååì:
Z
   exp(−ax2 )dx = (π/a)d/2 ,
Z                      Z
   exp(−ax + xξ)dx = exp(−a(x − ξ/2a)2 + ξ 2 /4a)dx) = (π/a)d/2 exp(ξ 2 /4a) ,
           2



àíàëèòè÷åñêîå ïðîäîëæåíèå:
Z
   exp(−ax2 + ixξ)dx = (π/a)d/2 exp(−ξ 2 /4a),
                           Z
         2             d/2
exp(−ξ /4a) = (a/π)           exp(−ax2 + ixξ)dx ,
                                       Z
      −d/2              2           −d
(4πt)      exp(−(x − y) /4t) = (2π)       exp(−tz 2 + iz(x − y))dz ,
           Z                                     Z Z
      −d/2                  2                 −d
(4πt)         exp(−(x − y) /4t)f (y)dy = (2π)      ( exp(−tz 2 + iz(x − y))f (y)dy)dz ,

    −d/2
         Z
                   2
                              √            −d
                                              Z
(π)         exp(−z )f (x + 2 tz)dz = (2π)        exp(−tz 2 + ixz)F (f )(z)dz.

Ïåðåõîäÿ â ïîñëåäíåì íåðàâåíñòâå ê ïðåäåëó t → +0, ïîëó÷àåì ôîðìóëó
îáðàùåíèÿ:                    Z
                   f (x) = (2π)−d    exp(izx)F (f )(z)dz.

Îòñþäà ñëåäóåò ðàâåíñòâî Ïàðñåâàëÿ:
     Z                           Z         Z
             ∗                −d         ∗
        f (x) g(x)dx = (2π)         f (x) ( exp(−ixz)F (g)(z)dz)dx =
            Z
         −d
     (2π)      F (f )∗ (z)F (g)(z)dz.

    ♣

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