Computing and proving existence of convolution.

Question

Let $$f(x)= \sum_{k=0}^{\infty}\frac{\lambda^k}{k!}e^{-\lambda}\chi_{[k,k+1[}(x)$$ and $$g(x)=\sum_{j=0}^{\infty}\frac{\mu^j}{j!}e^{-\mu}\chi_{[j,j+1[}(x)$$ Show that exits $\forall x \in \mathbb{R}$ the convolution $f*g(x)$ and compute $f*g(n)$ for $n \in \mathbb{N}$

I have tried to prove for example that $f(x) \in \mathcal{L}_2$ to prove that the convolution exists at all points but I have had problems, I understand that I must prove that such function is well defined but I can not see how, and with the calculation of the convolution I have not been able to, any suggestion or help to complete the problem would be greatly appreciated.

Answer 1

Notice that $f,g \in L^1(\mathbb R)$ and $\Vert f \Vert_1 = \Vert g \Vert_1 = 1$. Also $$0 \le f(x),g(x) \le 1$$ for all $x \in \mathbb R$. Therefore, applying dominated convergence theorem we get

$$\begin{aligned} 0 \le (f * g) (x) &= \int_{-\infty}^\infty \left(\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}e^{-\lambda}\chi_{[k,k+1[}(x-t)\right)\left(\sum_{j=0}^{\infty}\frac{\mu^j}{j!}e^{-\mu}\chi_{[j,j+1[}(t)\right) \ dt\\ &=\sum_{j=0}^{\infty} \frac{\mu^j}{j!}e^{-\mu}\int_{-\infty}^\infty \left(\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}e^{-\lambda}\chi_{[k,k+1[}(x-t)\right)\chi_{[j,j+1[}(t)\ dt\\ &=\sum_{j=0}^{\infty} \frac{\mu^j}{j!}e^{-\mu}\int_{j}^{j+1} \left(\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}e^{-\lambda}\chi_{[k,k+1[}(x-t)\right)\ dt\\ &=\sum_{j=0}^{\infty} \frac{\mu^j}{j!}e^{-\mu}\int_{x-j-1}^{x-j} \left(\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}e^{-\lambda}\chi_{[k,k+1[}(u)\right)\ du\\ &\le \sum_{j=0}^{\infty} \frac{\mu^j}{j!}e^{-\mu}\int_{-\infty}^{\infty} \left(\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}e^{-\lambda}\chi_{[k,k+1[}(u)\right)\ du = 1 \end{aligned}$$ Which proves that $(f * g)(x)$ is well defined for all $x \in \mathbb R$. Now for $n \in \mathbb N$, we get applying above equations:

$$\begin{aligned} (f * g) (n) &= \sum_{j=0}^{\infty} \frac{\mu^j}{j!}e^{-\mu}\int_{n-j-1}^{n-j} \left(\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}e^{-\lambda}\chi_{[k,k+1[}(u)\right)\ du\ &=\sum_{j=0}^{n-1}\frac{\mu^j}{j!}e^{-\mu}\frac{\lambda^{n-j-1}}{(n-j-1)!}e^{-\lambda}\\ &= e^{-(\mu+\lambda)}\sum_{j=0}^{n-1}\frac{\mu^j}{j!}\frac{\lambda^{n-j-1}}{(n-j-1)!}\\ &=\frac{(\mu + \lambda)^{n-1}}{(n-1)!}e^{-(\mu+\lambda)} \end{aligned}$$

Answer 2

Hint: Take the Laplace transforms of the two factors. Each Laplace transform can be summed explicitly. The product of the two Laplace transforms belongs to the same family as that of the original factors.