In the rest of the problem, we assume that $\lambda$ is a real distinct from 1 and we set $w = \frac{1}{\lambda - 1}$. We further set $\mathbf{f} = (1 + w)\delta - w\mathbf{1}$. Show that $\mathbf{f} * \mathbf{b} = \delta$.
In the rest of the problem, we assume that $\lambda$ is a real distinct from 1 and we set $w = \frac{1}{\lambda - 1}$. We further set $\mathbf{f} = (1 + w)\delta - w\mathbf{1}$.
Show that $\mathbf{f} * \mathbf{b} = \delta$.