Hilbert space with two inner products; separability and orthonormal basis Let $H$ be a separable Hilbert space with inner product $(\cdot,\cdot)_H$. So it has an orthonormal basis $h_j$. (You can consider $H=L^2(\Omega)$).
Suppose I know that $(\cdot,\cdot)_G$  is an inner product on $H$ which is norm-equivalent with the norm $(\cdot,\cdot)_H$ generates.
Is $H$ separable with this latter inner product? I'm sure it is. But what is the orthonormal basis? It's not simply $h_j$ again, but is it related to $h_j$ somehow?

Recall that separability in a metric space is defined as the existence of a countable dense subset.  By hypothesis, the equivalence of the norms yields $$c \| h \|_G \le \|h\|_H \le C \|h\|_G$$ for some positive real constants $c, C$.
Now let $\{ f_i \}$ be a countable dense subset of $H$.  This means that for all $h \in H$ we have $\inf_i \| h - f_i \|_H = 0$.  By the inequality above this means that $\inf \| h - f_i \|_G = 0$.  Hence $H$ is still separable in the new metric space.
The orthonormal basis $\{h_j\}$ is still a basis under the new inner product.  Which follows since the set of finite linear combinations is a countable dense subset of $H$ under $(\cdot,\cdot)_H$ and still is under $(\cdot, \cdot)_G$ by the same argument as above.
You can always Gram-Schmidt the $h_j$ 's to get a new orthonormal basis for the new inner product.  Though, that isn't a very inspired answer.
