In Nualart's book "The Malliavin Calculus and related topics" ,
denotes by $\mathcal{E}$ the set of step functions on $[0,T]$ and says that $\mathcal{H}=\overline{(\mathcal{E},\langle\cdot,\cdot\rangle_H)}$ is a Hilbert space, where $\langle\mathbb{1}_{[0,t]},\mathbb{1}_{[0,s]}\rangle_H=R_H(t,s)=\frac{1}{2}(t^{2H}+s^{2H}-|t-s|^{2H})$ and $H\in(0,1)$.
What I've done so far:
If $(\mathcal{E},\langle\cdot,\cdot\rangle_H)$ is dense in $(L^2([0,T]),\langle\cdot,\cdot\rangle_H)$ then $\mathcal{H}$ is Hilbert. This could happen if the two norms induced by the inner products are equaivalent . That means, $\|\mathbb{1}_{[0,t]}\|_{L^2([0,T])}=\sqrt t$ is equivalent to $\|\mathbb{1}_{[0,t]}\|_H=t^{H}$. This is true after some calculations.
My question is: Is my thought right? And if not why $\mathcal{H}$ is Hilbert?