I am currently reading this paper and there is a particular step I am not getting.
First of all there is a process $Y$ defined by $$ Y_t = \int_0^t K(t,s)dW_s, $$ where $W$ is a standard Brownian motion and $K$ is some kernel (think of $K(t,s) = \alpha (t-s)^{\alpha -1}$ for now).
Let $C_K(t,t')= \int_0^{t \land t'}K(t,s)K(t',s)ds$ and $\theta_T(h):= \sup_{0\leq t,t'\leq T,|t-t'|\leq h} \{C_K(t,t) + C_K(t',t'),-2C_K(t,t')\}^{1/2}$. Now there are two propositions in the paper. The first one states that if $$ C_K(t,t) < \infty, \quad \text{for} \quad t > 0 $$ $$ \theta_T(0^+) = 0 \quad \text{and} \quad \int_{0^+} \sqrt{\ln(1/u)}d\theta_T(u) < \infty $$ Then $Y$ admits a continuous version.
The second proposition states that if (in addition to the above assumptions) $K(t-s) = \hat K(t-s)$ for $s \leq t$ and for all $\epsilon > 0$ we have $\int_0^\epsilon |\hat K| > 0$. Then for each $T \geq 0$, the law of $Y$ has full support in $C_0^T:= \{y \in C([0,T],\mathbb{R}): y(0) = 0\}$.
Later in the paper they conclude that $\mathbb{P}[Y_t \geq \lambda t -1, t \in [0,T] ]> 0$, with reference to the above (where $\lambda > 0$). I am not sure why this follows? Do we have (or have to assume) that $Y$ is a gaussian process? If so, why can we conclude that the probablility is strictly greater than 0 in the light of the above propositions? (This generalises the statement for Brownian motion, i.e. when $K \equiv 1$, where the statement is clear to me).