Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 27J / iv / Solution

Assume ${\psi }_{+}(c)=e^{-Ac}$ with $A\ge 0$. Part (ii) givesInsert this into (iii). If $A\ne a$, the convolution becomesEquality for all $c>0$ forces $q^2=(a+A)(a-A)$, hence $A^2=a^2-q^2=\lambda (2q+\lambda )$. The case $A=a$ would instead give $e^{-ac}$ times a nonzero term linear in $c$, so is impossible. Thus