Equivalence between norms induced by a scalar product 
Given a vector space $V$ over $\mathbb R$, all norms induced by a positive scalar product on $V$ are equivalent. True or false? (The scalar product can of course change.)

I don't know whether this question can be easily answered. I do suspect this is false, but all examples I know of non equivalent norms in (infinte-dimensional) vector spaces (i.e. in $C^0([a,b])$) are not induced by a scalar product.
Any suggestions of solution/techniques/counterexamples?
Thank you in advance.
 A: On the real  $\ell^2 (\mathbb N)$ take the usual inner product 
$$
\langle x,y\rangle=\sum_n x_ny_n,
$$
and 
$$
[x,y]=\sum_n\frac1n\,x_ny_n.
$$
A: Suppose that $V$ is a Hilbert space with respect to two norms $\|\cdot\|_1$, $\|\cdot\|_2$ induced by inner products. Then the norms are equivalent iff there exists a bounded invertible selfadjoint $A$ on $(V,\langle\cdot,\cdot\rangle_1)$ such that $\langle x,y\rangle_2 = \langle Ax,y\rangle_1$. You can prove this using the Riesz Representation Theorem. The selfadjoint property follows from the symmetry of the inner products, and the invertibility follows by applying the result after swapping the two inner products.
For a given infinite-dimensional Hilbert space $V$ with inner product $\langle \cdot,\cdot\rangle_1$, it is always possible to construct a bounded selfadjoint operator $A$ such that $\mathcal{N}(A)=\{0\}$, but which is not continuously invertible; that leads to an inequivalent norm $\langle x,y\rangle_2 = \langle Ax,y\rangle_1$. You can't do that for finite-dimensional Hilbert spaces $V$.
If the two norms are not complete, but are equivalent, then you can extend to the completions and obtain a related result.
