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?
 A: 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.
A: In general, given two equivalent (i.e. defining the same topology) metrics $d_1,d_2$ on a given space $X$, we have $(X,d_1)$ separable if and only if $(X,d_2)$ is separable. This is simply because, having the same topology, $(X,d_1)$ and $(X,d_2)$ have the same dense subsets.
Two norms on a normed vector space $E$ define the same topology if and only if there exist constants $A,B>0$ such that $A\|x\|_1\leq \|x\|_2\leq B\|x\|_1$ for every $x\in E$. So there is no ambiguity in saying such norms are "equivalent". 
If $H$ is equipped with two inner products $(\cdot,\cdot)_1$ and $(\cdot,\cdot)_2$, and if the resulting norms are equivalent, it follows that $H$ is separable for the first norm if and only if it is separable for the second one. You can simply keep the same countable dense set from one to the other.
For your other question, note that there does not exist a single orthonormal basis in a given Hilbert space. And if $\{e_n\}$ is an orthonormal basis for $(\cdot,\cdot)_1$, it need not be an orthonormal basis for $(\cdot,\cdot)_2$. A trivial counterexample would be to take $(\cdot,\cdot)_2=5(\cdot,\cdot)_1$. But it need not even be orthogonal. It is easy to construct counterexamples in dimension $2$ already.
To understand all this (renorming), it is interesting to note that a new inner product $(\cdot,\cdot)_2$ is norm equivalent to the old one if and only if it is given by
$$
(x,y)_2=(Tx,Ty)_1
$$
for some invertible bounded operator operator on $(H,(\cdot,\cdot)_1 )$.
If such a $T$ is given, it is easy to see that the formula above defines an inner product on $H$. And $\|x_n\|_2=\|Tx_n\|_1$ tends to $0$ if and only if $\|x_n\|_1$ tends to $0$ by invertibility and boundedness of $T$. So the two norms are equivalent.
Conversely, assume $(x,y)_2$ yields an equivalent norm. Then $H$ is a separable Hilbert space for the new inner product as well. So take an orthonormal basis $\{f_n\}$ for the new inner product and consider the linear map defined on the original basis $\{e_n\}$ by $Te_n=f_n$. Then $T$ is a unitary operator from $(H,\|\cdot\|_1)$ onto $(H,\|\cdot\|_2)$. By norm equivalence, it follows that $T$ in bounded invertible on $(H,\|\cdot\|_1)$. And it is easily seen that $(x,y)_2=(Tx,Ty)_1$ for every $x,y\in H$.
With the latter construction, it is clear that any orthonormal basis $\{e_n\}$ is an orthonormal basis for the new inner product if and only if $T$ is a unitary operator.
