When inner products are equivalent and preserve orthogonality? Let $H=(H,(\cdot, \cdot))$ be a Hilbert space (over $\mathbb{R}$) and $L: D(L) \subset H \longrightarrow H$, with $\overline{D(L)}=H$, a linear, self-adjoint and closed operator (not necessarily bounded). Consider the function $a: D(L) \times D(L) \longrightarrow \mathbb{R}$ defined by
$$a(u,v):=(L(u),v),\; \forall \; u,v \in D(L).$$
Question 1. The function  $a: D(L) \times D(L) \longrightarrow \mathbb{R}$ defines a inner product in $D(L)$?
I think so, since the linearity of the operator $ L $ and the properties of the inner product $ (\cdot, \cdot) $ of $H$.
Question 2. If $a$ defines an inner product in $ D(L) $, then $ a $ is an inner product equivalent to $ (\cdot, \cdot) $, that is, there exists $b,c \in \mathbb{R}$ such that $0<b\leq c$ and
$$b\cdot a(f,f) \leq (f,f) \leq c\cdot a(f,f), \;  \forall \; f \in D(L)?$$
And more, if $a(u,v)=0$, for some $u,v \in D(L)$ then also $(u,v)=0$?
Here I don't know what to think anymore, since $ L $ is not necessarily bounded.
 A: No and no.
For $a$ to be an inner product you need that $\langle Lu,u\rangle\geq0$ for all $u$ (so you need $L$ to be positive, not just selfadjoint) and you also need that $\langle Lu,u\rangle=0$ implies $u=0$; with $L$ positive, this requires $L$ to be injective. In summary, for $a$ to be an inner product you need $L$ to be positive and injective.
Even when $a$ is an inner product, it is not necessarily equivalent to the original one, even if $L$ is bounded. For instance let $H$ be any separable Hilbert space with orthonormal basis $\{e_n\}$ and let $L$ be the linear operator induced by $Le_n=\tfrac1n\,e_n$ (actually any positive compact operator would do). Then
$$
\langle Le_n,e_n\rangle=\frac1n,\ \ \ \langle e_n,e_n\rangle=1,
$$
showing the $a$ and $\langle\cdot,\cdot\rangle$ not equivalent. Actually, $a$ and the original inner product are equivalent precisely when $L$ is bounded and invertible (with bounded inverse).
As for preserving orthogonality, if $\langle Le_n,e_m\rangle=\langle e_n,e_m\rangle$ for all $n,m$ then $\langle Lu,v\rangle=\langle u,v\rangle$ for all $u,v$, which implies that $L=I$.
