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I am stuck on the following exercise (this is one of the first exercises in Revuz and Yor, so the solution should be from first principles).

Let $f\in L^{2+\epsilon}_{\mathrm{loc}}(\mathbb R_+,dx)$ and put $Y_t = X(f\mathbf 1_{[0,t]})$, where $X$ is a gaussian measure with intensity $dx$ on $L^2(\mathbb R_+,dx)$. That is, $f\mapsto X(f)$ is a linear isometry from $L^2(\mathbb R_+,dx)$ to a centered Gaussian subspace of $L^2(\Omega,P)$ with $E[X(f)^2] = \|f\|^2_{L^2}$. Show $Y_t = X(f\mathbf 1_{[0,t]})$ admits a continuous modification if $f\in L^{2+\epsilon}_{\mathrm{loc}}$

My attempt: Evidently one wants to use Kolmogorov's continuity theorem. However, using that $X$ is a linear isometry, I get

$$E[|Y_{t+h}-Y_t|^2] = \int_t^{t+h}f^{2}(x)\,dx\,.$$

Since $f\in L^{2+\epsilon}_{\mathrm{loc}}$, I do not see how to remove the dependence on $h$, which prevents the application of Kolmogorov's continuity theorem. I am also not fully using the strength of the assumption that $f\in L^{2+\epsilon}_{\mathrm{loc}}$, but I can't quite see how to leverage it.

Another potential approach I think is to use the time reversal Brownian motion, i.e. $tB_{1/t}$ is a Brownian motion if $B_t$ is. However, I don't know how to make this compatible with $Y_t = X(f\mathbf 1_{[0,t]})$. I believe I need to define another linear isometry $\tilde X$ such that $\tilde X(\mathbf 1_{[0,t]}) = X(t\mathbf 1_{[0,1/t]})$, but I cannot figure out how to do this.

Christophe Leuridan
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Andrew
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1 Answers1

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Fix $p>0$. Set $C_p = \mathbb{E}[|N|^p]$, where $N$ is $\mathcal{N}(0,1)$. For every $t \ge 0$ and $h \ge 0$, $Y_{t+h}-Y_t$ is Gaussian, centered, with variance $\int_t^{t+h}f^{2}(x) \mathrm{d}x$, so $$E[|Y_{t+h}-Y_t|^p] = C_p\Big(\int_t^{t+h}f^{2}(x)\,dx\Big)^{p/2}.$$ But Hölder inequality yields $$\int_t^{t+h}f^{2}(x)\,dx \le \Big(\int_t^{t+h} 1 \,dx\Big)^{\epsilon/(2+\epsilon)} \Big(\int_t^{t+h}|f|^{2+\epsilon}(x)\,dx\Big)^{2/(2+\epsilon)} \le h^{\epsilon/(2+\epsilon)}\Vert f \Vert_{2+\epsilon}^2.$$ Hence $$E[|Y_{t+h}-Y_t|^p] \le C_p \times \Vert f \Vert_{2+\epsilon}^p \times h^{p\epsilon/(4+2\epsilon)}.$$ If one chooses $p>(4/\epsilon+2)$ so that $p\epsilon/(4+2\epsilon)>1$, Kolmogorov's continuity theorem applies.

Christophe Leuridan
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