Kenneth Kuttler

268 CHAPTER 11. FUNDAMENTAL TRANSFORMS

Now suppose only that | f (t)| ≤ Cert so f has exponential growth and that for all ssufficiently large, L ( f ) = 0. Does it follow that f = 0? Say this holds for all s≥ s0 wherealso s0 > r. Then consider f̂ (t)≡ e−s0t f (t) .

∣∣ f̂ (t)∣∣≤ e−s0tCert =Ce−(s0−r)t . Then if s> 0,∫∞

0e−st f̂ (t)dt =

∫∞

0e−ste−s0t f (t)dt =

∫∞

0e−(s+s0)t f (t)dt = 0

because s+ s0 is large enough for this to happen. It follows from Lemma 11.2.4 that f̂ = 0.But this implies that f = 0 also. This proves the following fundamental theorem.

Theorem 11.2.5 Suppose f has exponential growth and is continuous on [0,∞).Suppose also that for all s large enough, L ( f )(s) = 0. Then f = 0.

Now this will be extended to more general functions.

Corollary 11.2.6 Suppose | f (t)| , |g(t)| have exponential growth and are functions inL1 (0,∞). Then if L ( f )(s) = L (g)(s) for all s large enough, it follows that f = g a.e.

Proof: Say | f (t)| , |g(t)| ≤ Ceλ t and L ( f )(s) = L (g)(s) for all s > λ . Then bydefinition and Fubini’s theorem, for h = f − g, picking Borel measurable representativesfor f ,g, ∫

∞

(∫ t

0h(u)du

)e−stdt =

∫∞

0h(u)

∫∞

ue−stdtdu

=∫

∞

0h(u)

e−us =1sL (h)(s) = 0

Thus from Theorem 11.2.5,∫ t

0 h(u)du = 0. By fundamental theorem of calculus, h(t) = 0a.e. ■

11.3 Fourier TransformDefinition 11.3.1 The Fourier transform is defined as follows for f ∈ L1 (R) .

F f (t)≡ 1√2π

∫∞

−∞

e−itx f (x)dx

where here I am using the usual notation from calculus to denote the Lebesgue integral inwhich, to be more precise, you would put dm1 in place of dx. The inverse Fourier transformis defined the same way except you delete the minus sign in the complex exponential.

F−1 f (t)≡ 1√2π

∫∞

−∞

eitx f (x)dx

Does it deserve to be called the “inverse” Fourier transform? This question will beexplored somewhat below.

In studying the Fourier transform, I will use some improper integrals.

Definition 11.3.2 Define∫

∞

a f (t)dt ≡ limr→∞

∫ ra f (t)dt. This coincides with the

Lebesgue integral when f ∈ L1 (a,∞). However, situations will be considered below inwhich f is not in L1.

268 CHAPTER 11. FUNDAMENTAL TRANSFORMSNow suppose only that |f(t)| < Ce” so f has exponential growth and that for all ssufficiently large, Y (f) = 0. Does it follow that f = 0? Say this holds for all s > sq wherealso so > r. Then consider f (t) =e f (t). |f (t)| <e~°'Ce”™ = Ce~(%-")", Then if s > 0,° st P _ [Pst —sot _[ —(s+50)t _[e AOar= | oe rnar= | oH) F (1) dt =0because s + so is large enough for this to happen. It follows from Lemma 11.2.4 that f = 0.But this implies that f = 0 also. This proves the following fundamental theorem.Theorem 11.2.5 Suppose f has exponential growth and is continuous on [0,).Suppose also that for all s large enough, & (f)(s) =0. Then f =0.Now this will be extended to more general functions.Corollary 11.2.6 Suppose |f (t)| ,|g (¢)| have exponential growth and are functions inL! (0,00). Then if Z (f) (s) = Z (g) (s) for all s large enough, it follows that f = g a.e.Proof: Say |f (t)|,|g(t)| < Ce*? and Y(f) (s) = Y(g)(s) for all s > A. Then bydefinition and Fubini’s theorem, for h = f — g, picking Borel measurable representativesfor f,g,[ (fmm au) e “dt [hw [tara0 u[*h(u) we - ~£(h) (s) =0Thus from Theorem 11.2.5, {5 (u)du = 0. By fundamental theorem of calculus, h(t) = 0ae.11.3. Fourier TransformDefinition 11.3.1 The Fourier transform is defined as follows for f € L! (R).CO1~ V 27 J—cowhere here I am using the usual notation from calculus to denote the Lebesgue integral inwhich, to be more precise, you would put dm, in place of dx. The inverse Fourier transformis defined the same way except you delete the minus sign in the complex exponential.Ff (t) ef (x) dxco1~ V 27 J—coDoes it deserve to be called the “inverse” Fourier transform? This question will beexplored somewhat below.In studying the Fourier transform, I will use some improper integrals.F'f(t) el” f (x) dxDefinition 11.3.2 Define Jo f (t)dt = lim, J” f(t) dt. This coincides with theLebesgue integral when f € L'(a,~). However, situations will be considered below inwhich f is not in L!.