Fix $H\in(\frac{1}{2},1)$, and let $\Omega:=C_0([0,T],\mathbb R^d)$ be the space of $\mathbb R^d$-valued continuous functions. There is a probability measure $P^H$ on $(\Omega,\mathcal B(\Omega))$, such that on the probability space $(\Omega,\mathcal B(\Omega),P^H)$ the coordinate process $B:\Omega\to\mathbb R^d$ defined as \begin{align*} B(t,\omega)=\omega(t),\quad \omega\in\Omega \end{align*} is a $d$-dimensional fBm.

Furthermore let \begin{align} \phi(s,t):=H(2H-1)|s-t|^{2H-2},\;s,t\in[0,T]. \end{align}

and considering the following space \begin{align*} \mathcal H_{\phi}:=\bigg\{f:[0,T]\to\mathbb R: |f|_{\phi}^2=\int_0^T\int_0^T f(s)f(t)\phi(s,t)dsdt<\infty\bigg\}. \end{align*} If $\mathcal H_{\phi}$ is equipped with the inner product \begin{align*} \langle f,g\rangle_{\phi}=\int_0^T\int_0^T f(s)g(t)\phi(s,t)dsdt, \end{align*} then it becomes a separable Hilbert space, moreover we can see that $\mathcal H_{\phi}$ equals the closure of $L^2([0,T])$ with respect to the inner product $\langle\cdot,\cdot\rangle_{\phi}$. For $f\in\mathcal H_{\phi}\cap C([0,T])$ we denote with $(\Phi f):[0,T]\to\mathbb R$ the following continuous map \begin{align*} &[0,T]\ni t\to \mathbb R,\\ &t\mapsto \int_0^T f(s)\phi(t,s)ds. \end{align*}

Now let $f\in\mathcal H_{\phi}\cap C([0,T])$ and consider the translation operator given by the action

\begin{align} \mathtt T_{f} X(\omega)=X(\omega+\Phi f(\cdot)) \end{align}

and let $f^K=\sum_{k=1}^K \langle f,e_k\rangle_{\phi}e_k$ where $\{e_k\}$ is a orthonormal basis for the separable Hilbert space $\mathcal H_{\phi}$.

I am trying to see whether $\mathtt T_{f^K} X(\omega)$ converges to $\mathtt T_{f} X(\omega)$ as $K\to\infty$ in $L^p(\Omega)$ for some $p$. My problem is that even if $f^K$ converges to $f$ in $\mathcal H_{\phi}$, the shift I am performing actually involves $\Phi f^K$ and $\Phi f$ and I haven't been able to show the convergence my no means.

My idea was to put \begin{align} &\mathbb E\left[|\mathtt T_{f^K} X(\omega)-\mathtt T_{f} X(\omega)|^p\right]\\ &=\mathbb E\left[|\mathtt T_{f^K} X(\omega)-\mathtt T_{f^K} \mathtt T_{(f^K)^{\perp}}X(\omega)|^p\right]\\ &\mathbb E\left[\mathtt T_{f^K} |X(\omega)- \mathtt T_{(f^K)^{\perp}}X(\omega)|^p\right]\\ \end{align} then use the fractional Girsanov theorem and then if would suffice to show that

$$\mathbb E\left[|X(\omega)- \mathtt T_{(f^K)^{\perp}}X(\omega)|^q\right]\to 0$$ for some $q>1$.

Fix $H\in(\frac{1}{2},1)$, and let $\Omega:=C_0([0,T],\mathbb R^d)$ be the space of $\mathbb R^d$-valued continuous functions. There is a probability measure $P^H$ on $(\Omega,\mathcal B(\Omega))$, such that on the probability space $(\Omega,\mathcal B(\Omega),P^H)$ the coordinate process $B:\Omega\to\mathbb R^d$ defined as \begin{align*} B(t,\omega)=\omega(t),\quad \omega\in\Omega \end{align*} is a $d$-dimensional fBm.

Furthermore let \begin{align} \phi(s,t):=H(2H-1)|s-t|^{2H-2},\;s,t\in[0,T]. \end{align}

and considering the following space \begin{align*} \mathcal H_{\phi}:=\bigg\{f:[0,T]\to\mathbb R: |f|_{\phi}^2=\int_0^T\int_0^T f(s)f(t)\phi(s,t)dsdt<\infty\bigg\}. \end{align*} If $\mathcal H_{\phi}$ is equipped with the inner product \begin{align*} \langle f,g\rangle_{\phi}=\int_0^T\int_0^T f(s)g(t)\phi(s,t)dsdt, \end{align*} then it becomes a separable Hilbert space, moreover we can see that $\mathcal H_{\phi}$ equals the closure of $L^2([0,T])$ with respect to the inner product $\langle\cdot,\cdot\rangle_{\phi}$. For $f\in\mathcal H_{\phi}\cap C([0,T])$ we denote with $(\Phi f):[0,T]\to\mathbb R$ the following continuous map \begin{align*} &[0,T]\ni t\to \mathbb R,\\ &t\mapsto \int_0^T f(s)\phi(t,s)ds. \end{align*}

Now let $f\in\mathcal H_{\phi}\cap C([0,T])$ and consider the translation operator given by the action

\begin{align} \mathtt T_{f} X(\omega)=X(\omega+\Phi f(\cdot)) \end{align}

and let $f^K=\sum_{k=1}^K \langle f,e_k\rangle_{\phi}e_k$ where $\{e_k\}$ is a orthonormal basis for the separable Hilbert space $\mathcal H_{\phi}$.

I am trying to see whether $\mathtt T_{f^K} X(\omega)$ converges to $\mathtt T_{f} X(\omega)$ as $K\to\infty$ in $L^p(\Omega)$ for some $p$. My problem is that even if $f^K$ converges to $f$ in $\mathcal H_{\phi}$, the shift I am performing actually involves $\Phi f^K$ and $\Phi f$ and I haven't been able to show the convergence my no means.

My idea was to put \begin{align} &\mathbb E\left[|\mathtt T_{f^K} X(\omega)-\mathtt T_{f} X(\omega)|^p\right]\\ &=\mathbb E\left[|\mathtt T_{f^K} X(\omega)-\mathtt T_{f^K} \mathtt T_{(f^K)^{\perp}}X(\omega)|^p\right]\\ &\mathbb E\left[\mathtt T_{f^K} |X(\omega)- \mathtt T_{(f^K)^{\perp}}X(\omega)|^p\right]\\ \end{align} then use the fractional Girsanov theorem and then if would suffice to show that

$$\mathbb E\left[|X(\omega)- \mathtt T_{(f^K)^{\perp}}X(\omega)|^q\right]\to 0$$ for some $q>1$.

My problem is that we are shifting $\omega$ by $\int_0^T (f^K)^{\perp}(s)\phi(s,\cdot)ds$ but I can't show that the latter converges to $0$ in some sense.