Consider $\mathbb{R}^2$ with inner product $\langle x,y\rangle_A = x^T A y$, where $A$ is diagonal positive definite, i.e. $A = \begin{pmatrix} a & 0\\ 0 & b\end{pmatrix}\quad a,b >0$. To compute the dual norm of this norm, according to The Riesz representation theorem, we can compute $\|z\|^* = \underset{\|x\|_A = 1}{\mathrm{max}} z^T A x$, this gives $\|z\|^* = \|z\|_A = \sqrt{az_1^2 + bz_2^2}$.
My question is why cannot we compute the dual norm using $\|z\|^* = \underset{\|x\|_A = 1}{\mathrm{max}} z^T x$. In this case, the "dual norm" would be $\|z\|^* = \sqrt{\frac{z_1^2}{a}+\frac{z_2^2}{b}}$. Specifically, given an element $z$ in the dual space, what is wrong with defining $z(x) = z^Tx$ instead of $z(x) = z^T A x$ ?
Best Answer
Suppose you have some functional $\zeta(x) = f^T x$. Suppose $A>0$ is symmetric and define the $A$ inner product as above. Then $\zeta(x) = (A^{-1}f)^T Ax$, and so the Riesz representation of $\zeta$ with respect to the $A$ inner product is $A^{-1}f$.
Then $\|\zeta\| = \sup_{\|x\|_A \le 1} \zeta(x) = \sup_{\|x\|_A \le 1} \langle A^{-1} f, x\rangle_A = \|A^{-1} f\|_A = \sqrt{f^T A^{-1} f}$.