Let g be a timelimted smooth function, say suppg⊂[−R,R]. Then the Fourier transform of g can be extended to a complex valued function Fg(z):=∫−∞+∞g(t)e−2πiztdt. Let z=x+iy, then Fg(z)=∫−RRg(t)e−2πitxe2πytdt. Moreover, ∂zˉFg(z)=∫−RRg(t)∂zˉe−2πitzdt=0 because e−2πitz is holomorphic. This implies Fg(z) is a holomorphic function on the whole complex plane, we call such function an entire function.
Another way to see that Fg is entire is to use power series. From complex analysis we know that ez=∑k≥0k!zn and the series converges unformly on compact sets of C, so e−2πizt=∑k≥0k!(−2πizt)k=∑k≥0(−2πi)ktkk!1zk. If g is supported on [−R,R], then ∫−RRg(t)e−2πitzdt=∫−RR∑k≥0k!(−2πiz)ktkg(t)dt≤sup∣g∣⋅∑k≥02Rk+1(−2πi)kk!zk. The Cauchy-Hadamard criterion implies that the series on the right hand side has convergence radius ∞ and defines an entire function. To justify the switch of integral and summation, let sn=∑k=0nk!(−2πiz)ktk, then ∣sn∣≤∑k≥0k!(2π∣t∣∣z∣)k=e2π∣z∣∣t∣ which is continuous and thus integrable on [−R,R] (as a function of t) and dominate ∣sn∣, then apply dominated convergence theorem.
We’ll consider the power series expansion in the proof of central limit theorem in later lecture.
(eqn:plcondition)
for every k≥0. Then U is the Fourier-Laplace theorem of a smooth function supported on [−R,R].
Proof. The idea to find u is Fourier inversion. Let U(x) be the restriction of U on the real line. Then the growth condition eqn:plcondition implies that U(x) is a Schwartz function. To see this, first the inequality implies U(x) is rapidly decreasing, then by Cauchy estimate, the derivative of a holomorphic function can be estimated by its value. Then u(t)=∫RxU(x)e2πixtdx is a Schwartz function. To show that U is the Fourier Laplace transform of u, we’ll show that u is supported in [−R,R]. Then Fu(z)=U(z) because they are entire functions but coincide on the real line.
We claim that u(t)=∫−∞+∞U(x)e2πixtdx=0 when ∣t∣>R. For r>0 and b∈R (note: we are not assuming b>0 even though the picture looks like b>0), let I1,I2,I3,I4 be path integrals as shown in the picture, they depend on r though not explicitly written. Since U(z) is entire function, the Cauchy integral formula (or Morera theorem) implies I1+I2+I3+I4=0. Note that limr→∞I1=u(t). First, we show that I2 and I4 go to zero when r→∞, so we get u(t)=−limr→∞I3.
When b>0, ∣I2∣≤∫0b∣U(r+iy)∣∣e2πit(r+iy)∣dy≤supy∈[0,b]∣U(r+ib)∣∫0be−2πtydy. The growth condition on ∣U(z)∣ implies supy∈[0,b]∣U(r+ib)∣≤Ck(1+r2+b2)ke2πR∣b∣. So ∣I2∣≤const⋅(1+r2+b2)ke2πR∣b∣→0 when r→∞. The same result holds when b<0 and for I4 by the same argument.
We have showed that for every b∈R, u(t)=−limr→∞I3=∫−∞∞U(x+ib)e2πt(x+ib)dx, so ∣u(t)∣≤C2e2πRb−2πtb∫−∞+∞(1+∣x∣)21dx for every b. When b>0, 2πRb−2πtb<0 when R−t<0, i.e. t>R, this implies u(t)=0 because the non-negative number which is less than any negative exponential is 0. Similarly, when b<0, we 2πRb−2πtb<0 when t<−R, which implies u(t)=0 when t<−R. The proof is complete.
The Paley-Wiener theorem can be generalized to distirbutions. But
Let T be a distribution. We say T is zero or zero mass at x if for some open neighbourhood U of x, (T,φ)=0 for every φ∈Cc∞(U). As the case of functions, the support of T is the set where T is not zero, which is automatically closed. The set E′ denote the distributions with compact support.
One important classes of distribution with compact support are compact supported measures.
Recall that a distribution is a coninuous linear functional on Cc∞(R). A natural question is the dual of C∞(R). For a distribution T, (T,f) may not be defined when f∈C∞(R) even when T is a continuous function. Not suprisingly this can be done when T has compact support.
Let T be a distribution supported on K, where K is a compact set. Let ψ∈Cc∞(R) which is ≡1 on K. For f∈C∞(R), we write f=ψf+(1−ψ)f. The (T,ψf) is defined because ψf has compact support, the other part (1−ψf), though may not have compact support, is gurenteed to support outside the support of T so it’s meaningful to assign zero. So we define (T,f):=(T,ψf). To show that it is well-defined, let ψ′∈Cc∞(R) satisfying also ψ′≡1 on K. Then (T,ψf)−(T,ψ′f)=(T,(ψ−ψ′)f)=0 because ψ−ψ′ is supported outside K. Moreover we see that ∣(T,f)∣=∣(T,ψf)∣≤CK∑0≤l≤ksupx∈K∣(fψ)(l)(x)∣≤CK′∑l≤ksupx∈K∣f(l)(x)∣ for every f∈C∞(R).
Note that the topology of C∞(R) is determined by the semi-norms ∥⋅∥l,K, and fn→f if and only if for every compact K and every l, ∥fn−f∥l,K→0. Then for T∈E′, ∣(T,fn)−(T,f)∣≤CKT⋅∑l≤k∥fn−f∥l,KT→0. If we denote by E the space C∞(R) together with its topology, then E′ are continuous linear functionals on E, which justifies the notation.
To verify the Cauchy-Riemann Fourier-Laplace transform of distribution we need to “differenciate under action of distributions”. These nice properties are consequence of continunity of T, as we’ll see.
Let ϕ(x,y)∈C∞(X×Y), where X,Y are open sets on R. Let T∈E′. Then
(1) (T(y),ϕ(x,y)) defines a smooth function on X.
(2) ∂x(k)(T(y),ϕ(x,y))=(T(y),∂xk(ϕ(x,y))).
Proof.
Let g(x)=(T(y),ϕ(x,y)). First we show that g(x) is continuous. Note that if ϕ(x,y)∈C∞(X,Y), then ϕ(x+h,y)→ϕ(x,y) in C∞(Y) when h→0. Since T is compact support, T is continuous on C∞(Y), this implies g(x+h)=(T(y),ϕ(x+h,y))→(T(y),ϕ(x,y))=g(x) when h→0.
Then we consider derivatives using this observation.
Taylor expansion of ϕ(x,y) near x implies
ϕ(x+h,y)=ϕ(x,y)+h∂xϕ(x,y)+2h2∂xxϕ(x+θh,y) for ∣θ∣<1. Then g(x+h)=g(x)+h(T,∂xϕ)+2h2(T,∂xxϕ(x+θh,y)).
∣hg(x+h)−g(x)−(T,∂xϕ)∣=2h∣(T(y),∂xxϕ(x+θh),y)∣≤2h(∣T(y),∂xxϕ(x,y)∣+ϵ)→0 when h→0. The ε comes continunity of (T,∂xx(x,y) when h is small.
Iterating the above argument shows that g is smooth and its derivative is given by g(k)(x)=(T(y),∂x(k)ϕ(x,y)).
The second bullet is a consequence of C∞ convergence of Riemann sum, still by a continunity argument. To verify this, we must show for every g∈Cc∞(R), (ϕ∗T,g)=(T,ϕ−∗g)=∫(T∗ϕ)(x)g(x)dx. Let Rh=∑k∈Zϕ−(x−kh)g(kh)h be the Riemann sum of ϕ−∗g. (T,Rh)=∑k∈Zhg(kh)(T,ϕ−(x−kh))=∑k∈Zhg(kh)(T∗ϕ)(kh). Since Rh→ϕ−∗g in Cc∞(R) and T is a distribution, the LHS →(T,ϕ−∗g)=(ϕ∗T,g) when h→0. But observe that the RHS, as a Riemann sum of ∫(T∗ϕ)(x)g(x)dx, goes to ∫(T∗ϕ)(x)g(x)dx. This finishes the proof. Note that actually the summation ∑k∈Z is actually a finite sum because ϕ,ψ,g are compactly supported.
Fourier-Laplace transform of distributions with compact support¶
Alternative definition of Fourier transform for compactly supported distributions¶
Recall that when we defined the Fourier transform for distributions, we tried a function (FTf,g)=∫y∫xf(x)e−2πixydxdy and applied Fubini theorem to conclude that (FTf,g)=(T,Fg) and defined the Fourier transform to be (FT,g)=(T,Fg). But what if we do not use Fubini theorem, and then (FTf,g)=((Tf(x),e−2πixy),g(y)), then try to define FT be the function y↦(T(x),e−2πixy)? Unfortunately this would not be successful for tempered distribution T because e−2πixy is not compacted supported and this function can be infinity. But if T has compact support, then T acts on all smooth functions and this approach works! Moreover, by the continunity argument, the function (T(x),e−2πixy) is a smooth function on y.
Definition. Let T be a distribution with compact support. The Fourier transform of T is a smooth function given by FT(x)=(T(y),e−2πixy).
Of course, we need to show the new definition of Fourier transform as a smooth function concide (in the sense of distribution) as the original definition. By the identity principle of distributions, it sufficies to check that their actions on Cc∞ are the same, i.e. ∫y(T(y),e−2πixy)g(x)dx=(T,Fg). This follows from the same argument of proving T∗ϕ=ϕ∗T by writing Fg=∫g(y)e−2πixy as a Riemann sum ∑k∈Zg(kh)e−2πixkhh.
The Fourier-Laplace transform of δa is Fδa(z)=(δa(t),e−2πizt)=e−2πiaz. In particular, F(2δa+δ−a)=cos(2πaz), F(−δa+δ−a)=2isin(2πaz).
Let χ[−a,a] be the step function. Then Fχ[−a,a](z)=(χ−a,a(t),e−2πitz)=∫−aae−2πitzdt=−2πiz1(e−2πiaz−e2πiaz)=−2πiz−2isin(2πaz)=πzsin(2πaz). This is an entire function with power series z1⋅(z−3!z3+5!z5+⋯)=∑n=0+∞(2n+1)!(−1)nz2n.