Dual norm of the norm induced by inner product Given a positive definite matrix A, let $\langle x, y\rangle_A=x^\top Ay$. This inner product induces a norm $\|x\|_A^2=\langle x, x\rangle_A = x^\top A x$. My question is, what is the dual norm of $\|\cdot\|_A$?
The goal is to get something like a Holder's inequality:
$$
 |x^\top y|\leq \|x\|_A \|y\|_{A, *}
$$
Note the LHS is the inner product in Euclidean space. Thanks a lot for any suggestions.
 A: Let $A = L L^T$ be the Cholesky decomposition of $A$.
Note that 
$$
x^T A x = 1 \iff  \| y \|_2^2 = 1
$$
where $y = L^T x$.
The dual norm is 
\begin{align}
\| z \|_* &= \sup_{x^T Ax = 1} \langle z, x \rangle \\
&= \sup_{\|y\|_2 = 1} \langle z, L^{-T}y \rangle \\
&= \sup_{\|y\|_2 = 1} \langle L^{-1} z, y \rangle \\
&= \| L^{-1} z \|_2 \\
&= \sqrt{z^T L^{-T} L^{-1} z } \\
&= \sqrt{z^T A^{-1} z}\\
&= \|z\|_{A^{-1}}\quad .
\end{align}
A: It holds that 
$$
\lvert x^Ty\rvert^2 \le x^TAx\, y^TA^{-1}y,$$
(the right-hand side can be written $\langle x,x\rangle_A \langle y, y\rangle_{A^{-1}}$ in the notation of the OP). 
Proof: the left-hand side of the inequality is invariant under the transformation $x\mapsto Ox, y\mapsto Oy$, where $O$ is an orthogonal matrix, and since $A$ is symmetric it is orthogonally diagonalizable. So there is no loss of generality in assuming that $A=\text{diag}(\lambda_1,\lambda_2, \ldots, \lambda_n)$, and since $A$ is positive definite, $\lambda_j>0$ for all $j$. 
Now, setting 
$$
\xi_j=\sqrt{\lambda_j} x_j,\qquad \eta_j=\frac{1}{\sqrt{\lambda_j}}y_j, $$
we see that 
$$
|x^Ty|^2=|\xi^T\eta|^2\le \sum_{j=1}^n \xi_j^2\sum_{j=1}^n \eta_j^2=x^TAx\, y^TA^{-1}y,$$
and the proof is complete.
