show/hide this revision's text 2 Changed "Cameron-Martin space" in last paragraph to "support of \mathbb P"

Suppose we already know that the process is continuous with probability one, so that the process takes values in the Banach space $X = C([0,1])$ with distribution $\mathbb P$. The covariance operator $C : X^* \to X$ is then a map from the dual space $X^*$ to $X$. The support of the Gaussian measure $\mathbb P$ is then the closure of the image of $X^*$ under $C$: $$\operatorname{supp} \mathbb P = \overline{ CX^* }.$$ (This is the main theorem of [Vakhania 1975])

Now let's construct the Cameron-Martin space. The operator $C$ defines an inner product on the dual space $X^*$ by $$\langle f, g \rangle = f(Cg)$$ for $f, g \in X^*$. The space $X^*$ isn't necessarily closed under the topology induced by the inner product, so let $H$ be the Hilbert space completion, and let $\iota : X^* \hookrightarrow H$ be the inclusion map. Define a map $\iota^* : H \hookrightarrow X$ first on the dense subspace $\iota X^* \subseteq H$ by $$\iota^*(\iota f) = Cf,$$ and extend continuously to all of $H$. Thus the covariance operator factors as $C = \iota^* \circ \iota$.

The Hilbert space $\iota^* H$ is a subspace of $X$, and is called the Cameron-Martin space of the process. Interpreting this in the context of the support of the Gaussian measure $\mathbb P$, we have $$\operatorname{supp} \mathbb P = \overline{\iota^* H},$$ so that the closure of the Cameron-Martin space (with respect to the original norm of $X$) is exactly the support of $\mathbb P$.

I go into these ideas in more detail in Section 2 of my preprint [LaGatta 2010].

Suppose your covariance operator is an integral operator with kernel $c(s,t)$, called the covariance function of the process. That is, if $\mu$ is a Radon measure on $[0,1]$, then $$(C\mu)(s) = \int_0^1 c(s,t) \, d\mu(t).$$

If we write $c_s(t) = c(s,t)$, then the Cameron-Martin space support of $\mathbb P$ is the closure of the span of the functions $c_s$ in $C([0,1])$. So to answer your question: if you know that the closure $\overline{\operatorname{span}\{c_s\}}$ is a space with regularity property $P$, then the process $\xi_t$ satisfies property $P$ with probability one.

show/hide this revision's text 1

Suppose we already know that the process is continuous with probability one, so that the process takes values in the Banach space $X = C([0,1])$ with distribution $\mathbb P$. The covariance operator $C : X^* \to X$ is then a map from the dual space $X^*$ to $X$. The support of the Gaussian measure $\mathbb P$ is then the closure of the image of $X^*$ under $C$: $$\operatorname{supp} \mathbb P = \overline{ CX^* }.$$ (This is the main theorem of [Vakhania 1975])

Now let's construct the Cameron-Martin space. The operator $C$ defines an inner product on the dual space $X^*$ by $$\langle f, g \rangle = f(Cg)$$ for $f, g \in X^*$. The space $X^*$ isn't necessarily closed under the topology induced by the inner product, so let $H$ be the Hilbert space completion, and let $\iota : X^* \hookrightarrow H$ be the inclusion map. Define a map $\iota^* : H \hookrightarrow X$ first on the dense subspace $\iota X^* \subseteq H$ by $$\iota^*(\iota f) = Cf,$$ and extend continuously to all of $H$. Thus the covariance operator factors as $C = \iota^* \circ \iota$.

The Hilbert space $\iota^* H$ is a subspace of $X$, and is called the Cameron-Martin space of the process. Interpreting this in the context of the support of the Gaussian measure $\mathbb P$, we have $$\operatorname{supp} \mathbb P = \overline{\iota^* H},$$ so that the closure of the Cameron-Martin space (with respect to the original norm of $X$) is exactly the support of $\mathbb P$.

I go into these ideas in more detail in Section 2 of my preprint [LaGatta 2010].

Suppose your covariance operator is an integral operator with kernel $c(s,t)$, called the covariance function of the process. That is, if $\mu$ is a Radon measure on $[0,1]$, then $$(C\mu)(s) = \int_0^1 c(s,t) \, d\mu(t).$$

If we write $c_s(t) = c(s,t)$, then the Cameron-Martin space is the closure of the span of the functions $c_s$ in $C([0,1])$. So to answer your question: if you know that the closure $\overline{\operatorname{span}\{c_s\}}$ is a space with regularity property $P$, then the process $\xi_t$ satisfies property $P$ with probability one.