Proving that the Cauchy kernel $(P_t)_{t > 0}$ is a convolution semigroup

Question

Let's consider the Cauchy kernel $P_t(x)$ on $\mathbb R$, given by

$$P_t(x) = \frac t{\pi(x^2 + t^2)}; \qquad x \in \mathbb R, t > 0$$

I now want to show that $(P_t)_{t > 0}$ is a convolution semigroup, i.e. that it satisfies the property $$P_{s {\color{red} +} t} = P_s * P_t \tag{1}$$ for all $t, s > 0$, where $*$ is the convolution of the two functions.

I'm given the hint that I should consider the function $h_t(\xi) = \frac 1{\sqrt{2 \pi}} e^{- t|\xi|}$, that I should show $\check h_t(x) := \widehat h_t(x) = P_t(x)$ (where $\hat{\space}$ denotes the Fourier transformation), and that I should apply the inversion theorem [for a function $f :\mathbb R \to \mathbb R$ with $f, \hat f \in L^1(\mathbb R^n)$, we have $f(x) = (\hat f)^\check{\space}(x) = \frac 1{\sqrt{2 \pi}} \int_{\mathbb R} \hat f(\xi) e^{i x \xi} d \xi$ for almost all $x \in \mathbb R$].

Now I started using the hint and by calculating the Fourier transform $h_t(\xi)$, arriving (if I didn't make a mistake) at

$$\widehat h_t(- \xi) = \frac 1{2 \pi} \int_{\mathbb R} e^{i x \xi - t |x|} d x$$

And I'm a bit lost where to go from here. I'm not really able to see how this expression for $\hat h_t(-\xi)$ is equal to $P_t(\xi)$. And assuming I'd have shown that, how can I use that to deduce the formula $(1)$? I'm guessing that the $P_t, \hat P_t$ would be my $f, \hat f$ in the inversion theorem, but how can I get from that to the convolution of two such functions $P_t, P_s$?

Answer 1

No! You actually want to show $P_s*P_t=P_{s+t}$, not $P_{st}$. (Also, although this is less important, $P_s$ is the Poisson kernel, not the Cauchy kernel.)

Evaluate that integral by writing it as $\int_{-\infty}^0+\int_0^\infty$; in one integral you have $|t|=-t$ and in the other you have $|t|=t$; they both become integrals you can do by elementary calculus.

Now since these functions are even the $L^1$ inversion theorem shows that $$\widehat P_t=h_t,$$ and it follows that $$\widehat{P_{s+t}}=\widehat P_s\widehat P_t.$$

Recall that if $f,g\in L^1$ then $$\widehat{f*g}=\hat f\hat g.$$So you've shown that $$\widehat{P_{s+t}}=\widehat{P_s*P_t}.$$Since the Fourier transform is injective on $L^1$ you're done: $P_{s+t}=P_s*P_t$.

Answer 2

The fact that $\left\{f_t(x)=\frac{t}{\pi(x^2+t^2)}\right\}_{t\geq 0}$ is a semigroup with respect to the convolution can be proved without invoking explicit Fourier transforms. We have:

$$ \left(f_s * f_t\right)(x) = \frac{st}{\pi^2}\int_{-\infty}^{+\infty}\frac{du}{(u^2+t^2)((x-u)^2+s^2)} $$ where $$\small\operatorname*{Res}_{u=ti}\left(\frac{1}{(u^2+t^2)((x-u)^2+s^2)}\right)+ \operatorname*{Res}_{u=x+si}\left(\frac{1}{(u^2+t^2)((x-u)^2+s^2)}\right)=-\frac{i (s+t)}{2 s t \left((s+t)^2+x^2\right)}$$ immediately gives $f_s * f_t = f_{s\color{red}{+}t}$.

For a proof through the Lagrange-Brahmagupta-Fibonacci identity, have a look at this similar question.