Question on the proof of the uniqueness of Levy triplets in the Levy-Khintchine formula I am reading the proof on the uniqueness of the Levy triplets in the Levy-Khintchine formula from Ken Iti Sato's Levy Processes. However, there are two questions I cannot answer from the proof below.
First, why is the nonnegative-definite matrix $A$ uniquely determined by $\mu$ given that we have
$$s^{-2} \log \hat \mu(sz) \to -\frac{1}{2} \langle z, Az \rangle \; \text{as} \; s \to \infty?$$
Second, why is $\rho$ uniquely determined by $\psi$ given that we have (8.10), i.e. the Fourier transform of $\rho$ is given by $\int_c (\psi(z)-\psi(z+w))dw$?
I cannot think of a formal argument or other theorems guaranteeing the uniqueness from these forms. I would greatly appreciate any explanation.




 A: Concerning your $\textbf{first question}$: Let us assume the existence of another Lévy triplet $(A',\nu',\gamma')$. By repeating the steps that were done to get $$\lim_{s \to \infty} \frac{\log \widehat{\mu}(sz)}{s^{2}} = -\frac{1}{2} \langle Az,z \rangle$$ we will then get $$\lim_{s \to \infty} \frac{\log \widehat{\mu}(sz)}{s^{2}} = -\frac{1}{2} \langle A'z,z \rangle.$$ The left-hand side is independent of the Lévy triplets (since $\log \widehat{\mu}$ is already uniquely determined by $\widehat{\mu}$), hence we can conclude that the equation $$\begin{equation} -\frac{1}{2} \langle Az,z \rangle = -\frac{1}{2} \langle A'z,z \rangle \quad \text{or} \quad \langle Az,z \rangle = \langle A'z,z \rangle \tag{$1$} \end{equation}$$ holds for every $z \in \mathbb{R}^{d}$, especially for the standard unit vectors $e_{1}, \dots, e_{d}$. This leads us to $$\begin{equation} a_{i,i} = \langle Ae_{i},e_{i} \rangle = \langle A'e_{i},e_{i} \rangle = a'_{i,i} \tag{$2$} \end{equation}$$ for all $i$, assuring equal main diagonals of $A$ and $A'$. Moreover, $$\begin{align} \langle A(e_{i} + e_{j}),e_{i}+e_{j} \rangle &= \langle Ae_{i},e_{i} \rangle + \langle Ae_{i},e_{j} \rangle + \langle Ae_{j},e_{i} \rangle + \langle Ae_{j},e_{j} \rangle \\ &= a_{i,i} + a_{j,i} + a_{i,j} + a_{j,j} \\ &= a_{i,i} + 2a_{i,j} + a_{j,j} \end{align}$$ and $$\begin{align} \langle A'(e_{i} + e_{j}),e_{i}+e_{j} \rangle &= \langle A'e_{i},e_{i} \rangle + \langle A'e_{i},e_{j} \rangle + \langle A'e_{j},e_{i} \rangle + \langle A'e_{j},e_{j} \rangle \\ &= a'_{i,i} + a'_{j,i} + a'_{i,j} + a'_{j,j} \\ &= a'_{i,i} + 2a'_{i,j} + a'_{j,j} \end{align}$$ hold, as $A$ and $A'$ both are symmetric. Due to $(1)$, the left-hand sides are equal, resulting in $$a_{i,i} + 2a_{i,j} + a_{j,j} = a'_{i,i} + 2a'_{i,j} + a'_{j,j}.$$ Due to $(2)$, this is equivalent to $a_{i,j} = a'_{i,j}$ for all $i$ and $j$, proving that $A = A'$.
Concerning your $\textbf{second question}$: Because of the uniqueness of $\log \widehat{\mu}$ and $A$, the function $\psi$ is likewise uniquely determined by $\widehat{\mu}$. This means that the Fourier transform of $\rho$ is uniquely determined by $\widehat{\mu}$, as we can compute it with $\psi$. Knowing this, $\rho$ is also uniquely determined by $\widehat{\mu}$. This conclusion is possible, since $\rho$ is a finite measure. (Actually, $\rho$ can be treated as a probability measure and $\widehat{\rho}$ can be treated as its characteristic function in this context. With that, the step is maybe easier to understand.)
