is a dense sub-Hilbert space invariant by bounded invertible operators Let $H$ be some complex Hilbert space and let $H_1$ be a dense sub-space in $H.$ Assume that $H_1$ is a Hilbert space with respect to a scaler product $\langle\cdot,\cdot \rangle_1$ which is different from that of $H.$ Let now $L \in \mathcal{B}(H)$ such that $L^{-1} \in \mathcal{B}(H).$ 
Is $L \in \mathcal{B}(H_1)$ ? In other words; is $L(H_1) \subset H_1$?
I think this is not true in general. I wonder how to finde a counterexample. 
Thank you in advance.

For a counterexample, let $H$ be the space $L^2(Q)$  of complex-valued square-integrable functions on the square $Q=[0,1]\times [0,1]$. Let 
$$H_1 = \left\{ f\in H : \int_{Q} |f(x,y)|^2  x^{-2}\,dx\,dy <\infty \right\}$$
which is a weighted $L^2$ space with the inner product 
$$\left<f, g \right> = \int_Q f(x, y)\overline{g(x, y)} x^{-2}\,dx\,dy$$
It is also dense in $H$ with respect  to $L^2(Q)$ norm, since any function $f\in H$ can be approximated by $f \chi_{\{x > \epsilon\}} $ where $\epsilon>0$ us small. 
The linear operator $Lf(x,y) = f(y,x)$ is a surjective isometry on $L^2(Q)$ but it does not map $H_1$ to itself; for example, the function $f(x,y)=x$ is mapped to $g(x,y)=y$ which is not in $H_1$.
