TITLE: Equivalent norms imply equivalent inner products?
QUESTION [6 upvotes]: Let $H$ be Hilbert and let it have two innner products $(\cdot,\cdot)_1$ and $(\cdot,\cdot)_2$.
If the norms $|\cdot|_1$ and $|\cdot|_2$ are equivalent, does this ever imply: there exist constants $C_i$ such that
$$C_1(a,b)_2 \leq (a,b)_1 \leq C_2(a,b)_2$$
(i.e. the inner products are equivalent in some sense)?
under what conditions on $H$ does this happen?

REPLY [3 votes]: You have such constants that
$$C_1\langle a\mid b\rangle_2 \leqslant \langle a \mid b\rangle_1 \leqslant C_2\langle a\mid b\rangle_2$$
for all $a,\, b \in H$ if and only if the two norms are constant multiples of each other.
Firstly, such an inequality can only hold if $\langle\mid\rangle_2$-orthogonality implies $\langle\mid\rangle_1$-orthogonality.
Assume that to be the case.
Then let $(e_\nu)$ be a $\langle\mid\rangle_2$-ONB.
For $\nu \neq \mu$ (if $\dim H = 1$, there is nothing to show), we have $\langle e_\nu + e_\mu \mid e_\nu - e_\mu\rangle_2 = \lVert e_\nu\rVert_2^2 - \lVert e_\mu\rVert_2^2 = 0$, and therefore
$$\langle e_\nu + e_\mu \mid e_\nu - e_\mu\rangle_1 = \lVert e_\nu\rVert_1^2 - \lVert e_\mu\rVert_1^2 = 0,$$
i.e. $\lVert e_\nu \rVert_1$ is constant, and thus, up to scaling by a factor independent of $\nu$, $(e_\nu)$ is also an ONB for $\langle\mid\rangle_1$.