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 Incoming links: Cauchy convolution semigroup

Transform pair. In the paper's symmetric normalization, the inverse Fourier transform is

g (x) = \frac{1}{2 π} \int_{- \infty}^{\infty} g (k) e^{ik x} d k .

Split the transform of

e^{- c ∣ x ∣}

at zero, with

c > 0

F [e^{- c ∣ x ∣}] (k) = \frac{1}{2 π} [\int_{- \infty}^{0} e^{(c - ik) x} d x + \int_{0}^{\infty} e^{- (c + ik) x} d x] = \frac{1}{2 π} (\frac{1}{c - ik} + \frac{1}{c + ik}) = \frac{2}{π} \frac{c}{c ^{2} + k ^{2}} .

The inverse formula applied to this result gives

e^{- c ∣ x ∣} = \frac{c}{π} \int_{- \infty}^{\infty} \frac{e ^{ik x}}{c ^{2} + k ^{2}} d k .

Because the kernel is even, replacing the transform variable by

x

and the argument by

k

yields

F [\frac{1}{x ^{2} + c ^{2}}] (k) = \frac{2 π}{2 c} e^{- c ∣ k ∣} .

The factors of

2 π

depend on the normalization and will also enter the convolution theorem.

Convolution theorem. For integrable functions define the convolution

(f * g) (x) = \int_{- \infty}^{\infty} f (x - y) g (y) d y .

Substitute this into the Fourier transform, interchange the two absolutely integrable factors using Fubini's theorem, and set

z = x - y

f * g (k) = \frac{1}{2 π} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} f (x - y) g (y) e^{- ik x} d y d x = \frac{1}{2 π} (\int f (z) e^{- ik z} d z) (\int g (y) e^{- ik y} d y) .

Therefore

f * g (k) = 2 π f (k) g (k) .

Integral equation. Put

f_{a} (x) = 1/ (x^{2} + a^{2})

. Since this is even, the left side is

f_{a} * g

. The transformed equation is

\frac{π}{a} e^{- a ∣ k ∣} g (k) = \frac{2 π}{2 b} e^{- b ∣ k ∣},

g (k) = \frac{a}{b 2 π} e^{- (b - a) ∣ k ∣} .

The assumption

b > a

makes the new scale

d = b - a

positive. Inverting with the previously established transform pair gives

g (x) = \frac{a ( b - a )}{πb [ x ^{2} + ( b - a ) ^{2} ]} .

The Cauchy convolution semigroup gives a quick verification. With

K_{c} (x) = c / [π (x^{2} + c^{2})]

, the formula is

g = (a / b) K_{b - a}

and

f_{a} = (π / a) K_{a}

. Hence

f_{a} * g = (π / b) K_{b} = 1/ (x^{2} + b^{2})

, as required. The solution is unique among integrable functions because the multiplier

f_{a} (k)

never vanishes.