Show that the covariance operator of a Gaussian measure is uniquely determined and nonnegative Remark: Feel free to assume $E=\mathbb R^d$, $d\in\mathbb N$, if it helps you to answer my question.
Let $E$ be a $\mathbb R$-Banach space, $R:E'\to E$ be symmetric ($\langle R\varphi_1,\varphi_2\rangle=\langle R\varphi_2,\varphi_1\rangle$ for all $\varphi_1,\varphi_2\in E'$) and $\rho$ be a probability measure on $\mathcal B(E)$ with characteristic function $$\hat\rho(\varphi)=e^{-\frac{\langle R\varphi,\:\varphi\rangle}2}\;\;\;\text{for all }\varphi\in E'\tag1.$$

How can we show that

*

*$R$ is nonnegative ($\langle R\varphi,\varphi\rangle\ge0$ for all $\varphi\in E'$);

*$R$ is uniquely determined?


Both claims are made in Proposition 2.4.4.1 of Structural Aspects in the Theory of Probability:

We know that there is a unique $f\in C(E',\mathbb C)$ with $f(0)=0$ and $$\hat\rho=e^f\tag2.$$ This function is denoted by $\log\hat\rho$ and called the distinguished logarithm of $\hat\rho$.
I would like to show the uniqueness claim in (2.) by the already known uniqueness of $\log\hat\rho$. We clearly should have $\log\hat\rho=-\frac{\langle R,\;\cdot\;\rangle}2$, but as you can see the book is claiming that $\log\hat\rho=-\langle R,\;\cdot\;\rangle$. Is this a typo?
In any case, I don't understand how the uniqueness immediately follows from this. And while the inequality shown for $\hat\rho$ is clear, how does it yield (1.)?
 A: I'll write $E^*$ instead of $E'$ (assuming it's the dual space). I'm assuming that by pairings $\langle R\varphi_1,\varphi_2\rangle$ we mean $\varphi_2 ( R\varphi_1)$.
Lemma. If $R,S: E^* \to E$ are such that for all $\varphi,\psi \in E^*$ we have $\langle R\varphi,\psi \rangle = \langle S\varphi,\psi \rangle$ then $R-S$ is a zero mapping. In other words, $R$ is uniquelly determined by $\langle R\varphi,\psi\rangle$ for any $\varphi,\psi \in E^*$.
Proof. Fixing $\varphi \in E^*$, we see that by linearity of $\psi \in E^*$ we get $\psi \big( R\varphi - S\varphi \big) = 0$. Since $\psi$ is arbitrary, by Hahn Banach theorem (the only element vanishing on every functional is a zero element) we get $0=R\varphi - S\varphi = (R-S)\varphi$ for any $\varphi \in E^*$. Since $\varphi$ is arbitrary, we see that $R-S$ is a zero mapping $\square$
(2). Hence as we see, it is sufficient to determine $\langle R\varphi,\psi\rangle$  for any $\psi,\varphi \in E^*$. Since $R$ is symmetric, values $\langle R\varphi,\psi \rangle$ are uniquelly determined by values of $\langle R(\varphi+\psi),\varphi+\psi\rangle$,$\langle R(\varphi-\psi),\varphi-\psi\rangle$, in other words by values $\langle R\eta,\eta \langle$ for any $\eta \in E^*$.
Note that $-2\log(\hat p(\eta)) = \langle R\eta,\eta \rangle$, so since $\hat p$ is itself unique, then so is $-2\log(\hat p)$, which means that $\eta \to \langle R\eta,\eta \rangle$ is unique, which by the above discussion and lemma, is sufficient to say that $R$ is itself unique.
(1). As for the non-negativeness, it's easier. Note that $\hat p(\varphi) \le 1$ for any $\varphi \in E^*$. Taking logarithm, we arrive at $\log(\hat p (\varphi)) \le 0$. By multiplying by $-2$, we get $ \langle R\varphi,\varphi \rangle = -2\ln(\hat p(\varphi)) \ge 0 $. Since $\varphi \in E^*$ was arbitrary, we can deduce the non-negativeness.
