On Convolution: Show that $g_a *g_b = g_{\min(a,b)}$

Question

For $a>0$, I have been given following the functions $$f_a(x)=\frac{a}{π(x^2+a^2)}$$ and $$g_a(x)=\frac{\sin(ax)}{π x}~~x\neq0,\qquad g_a(0)= \frac{a}{π}. $$

Question Show that,

$$f_a *f_b = f_{a+b}$$ and $$g_a *g_b = g_{\min(a,b)}$$

I was able to prove that $f_a *f_b = f_{a+b}$ through Fourier Transform.

Can anyone help to show that $g_a *g_b = g_{\min(a,b)}$ or any hint?

Answer 1

You can use the Fourier transform too: $$\mathcal F(g_a*g_b)=\mathcal F(g_a)\mathcal F(g_b).$$ Up to normalisation factors, $\mathcal F(g_a)$ is the indicator function of the interval $[-a,a]$, so it is apparent that $$\mathcal F(g_a)\mathcal F(g_b)=\mathcal F(g_{\min(a,b)})$$ etc.