what characterizes a characteristic function of a probability measure in separable Hilbert spaces? As we all know on real line $\mathbb{R}$, the following is valid 

A $\mathbb{C}$-valued function $\varphi$ is a characteristic function of a probability measure on $\mathbb{R}$ if and only if $\varphi$ is a continuous, positive definite function such that $\varphi(0)=1$.

But on separable Hilbert space, Theorem 2.13 in Stochastic Equations in infinite Dimensions (by Da Prato) says that

A $\mathbb{C}$-valued function on $H$ $\varphi$ is a characteristic function of a probability measure on $H$ if and only if 
(i) $\varphi$ is a continuous, positive definite function such that $\varphi(0)=1$.
(ii) $\forall \varepsilon >0, $ $\exists$ a nonnegative nuclear operator $S$ such that
  $$
1-\mathrm{Re}\ \varphi(\lambda) \leq \varepsilon, \mathrm{for\  all}\  \lambda \text{ satisfying }\langle S\lambda,\lambda \rangle \leq1 
$$

What confused me is that why (ii) condition is added in the infinite case? Is there any intuitive or intelligible explanation for it? 
 A: A good way to see why (ii) is needed in the infinite dimensional case is to find an example where (i) holds but not (ii). (in the finite dimensional case, (ii) is a consequence of (i)). Also, it could be good to check (ii) in the case of a Gaussian random variable. 
Define $\phi(x):=\exp(-\lVert x\rVert^2_H)$: it satisfies (i). Now we shall see that (ii) does not hold with $\varepsilon:=1/2$. Suppose that there exists a non-negative nuclear operator $S$ such that if $\langle Sx,x\rangle\leqslant 1$, then 
$\phi(x)\geqslant 1/2$. Notice that $S$ has the form 
$$S(x)=\sum_{j=1}^{+\infty}\lambda_j\langle x,a_j\rangle b_j$$
where $\lambda_j\geqslant 0$, $\sum_{j=1}^{+\infty}\lambda_j<+\infty$, 
$a_j,b_j\in H$ and $M:=\sup_j\lVert a_j\rVert+\lVert b_j\rVert$ is finite. 
Fix an integer $N\geqslant 1$. If $x$ belongs to the orthogonal of the vector space generated by $a_j,j\leqslant N$, then the condition $\langle Sx,x\rangle$ will be satisfied once $M^2\lVert x\rVert^2\sum_{j\geqslant N}\lambda_j\leqslant 1$. If we pick $x$ such that the equality holds, then we should have 
$$\exp\left(-\frac 1{M^2\sum_{j\geqslant N}\lambda_j}\right)\geqslant \frac 12.$$
If we choose $N$ large enough we get a contradiction.
The problem is that for infinite dimensional spaces $\langle Sx,x\rangle$ may be smaller than $1$ for $x$ having a large norm. 
