Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 24 / 2 / c / Solution

For fixed $u$, write $h(t)={\phi }_{X_t}(u)$. Independent increments and stationary increments giveThe preceding part gives continuity. Also $h$ never vanishes: if $h(t_0)=0$ with $t_0>0$, then $h(t_0/m{)}^m=0$ for every positive integer $m$, contradicting $h(t_0/m)\to 1$.

Here is a direct proof of the exponential form of Lévy characteristic functions. Choose $a>0$ small enough that $|h(t)-1|<1/2$ on $[0,a]$. The principal complex logarithm gives a continuous $L(t)=\log h(t)$ there, with $L(0)=0$. For $s,t\ge 0$ with $s+t\le a$, the multiplicative identity impliesThis difference is continuous on the connected triangle of allowed $(s,t)$ and equals zero at $(0,0)$, so it is identically zero. Thus $L$ satisfies the additive Cauchy functional equation locally. Subdivision gives $L(a/m)=L(a)/m$ and $L(ka/m)=kL(a)/m$; continuity then gives $L(t)=tL(a)/a$ for every $t\in [0,a]$.

Define $\eta (u)=L(a)/a$. For any $t\ge 0$, choose an integer $m$ with $t/m\le a$; thenThe coefficient is unique: if two coefficients give the same exponential for every $t\ge 0$, their derivatives at zero agree. In particular $\eta (0)=0$, and $\mathrm{Re}\eta (u)\le 0$ follows from $|h(t)|\le 1$. The characteristic exponent of a Lévy process has therefore been obtained from first principles, without invoking the Lévy–Khintchine formula.