By Theorem 3.6, \(\mu \) is Gaussian iff for every continuous linear form \(L \in E^*\), the characteristic function of \(\mu \) at \(L\) is

Every continuous linear form \(L \in E^*\) can be written as \(L(x) = \langle t, x \rangle \) for some \(t \in E\), hence we have that \(\mu \) is Gaussian iff for every \(t \in E\),

By Lemma 3.14, for every \(t \in E\),

By Lemma 3.13, \(\mu \) has finite first moment and \(\mu [\langle t, \cdot \rangle ] = \langle t, m_\mu \rangle \). By the same lemma, \(\mu \) has finite second moment and for any \(t\) we have \(\mathbb {V}_\mu [\langle t, \cdot \rangle ] = C'_\mu (t, t)\).

Lemma 3.15 states that the characteristic function of a Gaussian measure has the wanted form.

Suppose now that there exists \(m \in E\) and \(C\) positive semidefinite such that for all \(t \in E\), \(\hat{\mu }(t) = \exp \left(i \langle t, m \rangle - \frac{1}{2} C(t, t)\right)\).

We need to show that for all \(L \in E^*\), \(L_*\mu \) is a Gaussian measure on \(\mathbb {R}\). Such an \(L\) can be written as \(\langle u, \cdot \rangle \) for some \(u \in E\). Let then \(u \in E\). We compute the characteristic function of \(\langle u, \cdot \rangle _*\mu \) at \(x \in \mathbb {R}\) with Lemma 1.5:

This is the characteristic function of a Gaussian measure on \(\mathbb {R}\) with mean \(\langle u, m \rangle \) and variance \(C(u, u)\). By Theorem 1.4, \(\langle u, \cdot \rangle _*\mu \) is Gaussian, hence \(\mu \) is Gaussian.

By Lemma 3.15, we deduce that for any \(t \in E\) we have

In particular, for any \(t\) there exists \(n_t \in \mathbb {Z}\) such that

We deduce that \(n\) is a continuous map from \(E\) to \(\mathbb {Z}\), and thus must be constant because \(E\) is connected. By looking at the value at \(t = 0\), we deduce that for any \(t\), \(n_t = 0\). Looking at real and imaginary parts we obtain that for any \(t\),

We immediately deduce that \(m = m_\mu \). Moreover, because \(C\) and \(C'_\mu \) are symmetric, they are characterized by their values on the diagonal. Indeed, for any \(x, y\),

We deduce that \(C = C'_\mu \).

The forward direction is immediate.

For the converse direction, it is enough to show that \(\mu \) and \(\nu \) have the same characteristic function by Theorem 1.4. As they are both Gaussian, their characteristic functions only depend on their mean and covariance by Lemma 3.15. Thus they are equal.

As \(f\) is integrable, we can use Fubini theorem to obtain that

because the \(\mu _j\)s are probability measures.

Denote by \(\nu \) the standard Gaussian measure on \(\mathbb {R}\). This is a straightforward computation:

Since the standard Gaussian is a probability measure (hence finite), we can apply Lemma 3.16 that states that it suffices to show that the characteristic function has a particular form. That form is given by Lemma 3.22, taking \(m=0\) and \(C = \langle \cdot , \cdot \rangle \).

From Lemma 3.22, we know that for all \(t \in \mathbb {R}\),

As the identity is positive semidefinite, we deduce from Lemma 3.16 that \(\Sigma _\mu \) is the identity matrix.

The multivariate Gaussian measure is the pushforward of the standard Gaussian measure by an affine map, and is thus Gaussian by Lemma 3.8 and Lemma 3.7.

Since the multivariate Gaussian measure is a Gaussian measure, we can apply Lemma 3.15 to it. It suffices then to show that the mean and the covariance matrix of the multivariate Gaussian measure are equal to \(m\) and \(\Sigma \), respectively. This is given by Lemma 3.27 and Lemma 3.28.

Being a Gaussian process is defined in terms of the distribution of finite-dimensional random vectors. By Lemma 2.5, the random vector \((Y_{t_1}, \ldots , Y_{t_n})\) has the same distribution as the random vector \((X_{t_1}, \ldots , X_{t_n})\) for all \(t_1, \ldots , t_n \in T\).