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 $q>1$.