Kenneth Kuttler

11.5. FOURIER TRANSFORMS IN Rn 273

Theorem 11.4.1 Let g be a measurable function defined on (0,∞) which has expo-nential growth,|g(t)| ≤Ceηt for some real η and is Holder continuous from the right andleft as in 11.2 and 11.3. For Re(s)> η ,L g(s)≡

∫∞

0 e−sug(u)du. Then for any γ > η , andt > 0,

limR→∞

12π

∫ R

−Re(γ+iy)tL g(γ + iy)dy =

g(t+)+g(t−)2

(11.6)

In case of t = 0, you would only assume the Holder continuity from the right and the aboveresult would be g(0+)/2.

Proof: This follows from plugging in the formula for the Laplace transform of g andthen using the above. Thus

12π

∫ R

−Re(γ+iy)tL g(γ + iy)dy =

12π

∫ R

−Re(γ+iy)t

∫∞

0e−(γ+iy)ug(u)dudy =

12π

∫ R

−Reγteiyt

∫∞

0e−(γ+iy)ug(u)dudy

= eγt 12π

∫ R

−Reiyt∫

∞

0e−iyue−γug(u)dudy

Let ĝ(u) = 0 for all u ≤ 0 so this equals eγt 12π

∫ R−R eiyt ∫ ∞

−∞e−iyue−γuĝ(u)dudy. Now apply

Theorem 11.3.6 which said that for g in L1 having the Holder condition at t,

limR→∞

12π

∫ R

−Reiyt∫

∞

−∞

e−iyug(u)dudy =g(t+)+g(t−)

to conclude that if t > 0,

limR→∞

12π

∫ R

−Re(γ+iy)tL g(γ + iy)dy

= eγt limR→∞

12π

∫ R

−Reiyt∫

∞

−∞

e−iyue−γuĝ(u)dudy

= eγt ĝ(t+)e−γt++ ĝ(t−)e−γt−

g(t+)+g(t−)2

If t = 0 you would have ĝ(0−) = 0 so you would end up finding 12 g(0+). ■

In particular, this shows that if L g(s) =L h(s) for all s large enough, both g,h havingexponential growth, then these must be equal except for jumps and in fact, at any pointwhere they are both Holder continuous from right and left, the mid point of their jumpsis the same. This gives an alternate proof of Corollary 11.2.6 in the case of points ofcontinuity.

11.5 Fourier Transforms in Rn

In this section is a general treatment of Fourier transforms. It turns out you can take theFourier transform of almost anything you like. First is a definition of a very specialized setof functions. Here the measure space will be (Rn,mn,Fn) , mn Lebesgue measure on Rn.

First is the definition of a polynomial in many variables.

11.5. FOURIER TRANSFORMS IN R" 273Theorem 11.4.1 Lez g be a measurable function defined on (0,%) which has expo-g(t)| <Ce™ for some real n and is Holder continuous from the right andleft as in 11.2 and 11.3. For Re(s) >, 2g (s) = fo e “g (u) du. Then for any y >, andt>0,1 + (t+) +8 (t—)— (y+iy)t — SVT) Ts)iim 5) le L2(y+iy) dy 5} (11.6)In case of t =0, you would only assume the Holder continuity from the right and the aboveresult would be g (0+) /2.Proof: This follows from plugging in the formula for the Laplace transform of g andthen using the above. Thus1elYtiy)t . _x | L2(y+iy) dyR co R co=| ero | eon ae/ ete | e (Yi) 9 (4) dudyR JO 20 JR Jo27 Jart [ et [ee Ue Mg (u) dudyLet @(u) = 0 for all u < 0 so this equals ett [Re edt [=e “e-™ 8 (u) dudy. Now applyTheorem 11.3.6 which said that for g in L' having the Holder condition at f,1 fk oy fe g (t+) +8 (t—). tyt —lyu _jim. 20 I, ° —oo eng (u) dudy 2to conclude that if t > 0,1plim — I, elt" Lo (y+ iy) dyR00 270= e” lim x/, em [ie ~M eM 8 (u) dudyRoo 270en dlttje ™ +8t)e™ _ gltt) ta (t-)2 2 :If t = 0 you would have g(0—) = 0 so you would end up finding 5¢ (0+). llIn particular, this shows that if “g(s) = £h/(s) for all s large enough, both g,/ havingexponential growth, then these must be equal except for jumps and in fact, at any pointwhere they are both Holder continuous from right and left, the mid point of their jumpsis the same. This gives an alternate proof of Corollary 11.2.6 in the case of points ofcontinuity.11.5 Fourier Transforms in R”In this section is a general treatment of Fourier transforms. It turns out you can take theFourier transform of almost anything you like. First is a definition of a very specialized setof functions. Here the measure space will be (R", my, -Fn) , m, Lebesgue measure on R".First is the definition of a polynomial in many variables.