Definition of Euclidean norm on dual space

Question

In the paper of Random Gradient-Free Minimization of Convex Functions, notation part shows as follows:

For a finite-dimensional space $E$, we denote by $E^*$ its dual space. The value of a linear function $s\in E^*$ at point $x \in E$ is denoted by $\langle s,x\rangle$. We endow the spaces $E$ and $E^*$ with Euclidean norms

$$\|x\| = \langle Bx, x\rangle^{1/2}, x \in E, \,\, \|s\| = \langle s, B^{-1}s\rangle^{1/2}, x \in E^*$$

where $B= B^*$ is a linear operator from $E$ to $E^*$. For any $u \in E$ we denote by $uu^*$ a linear operator from $E^*$ to $E$, which acts as follows:

$$uu^*(s) = u \cdot \langle s, u\rangle, \,\, s\in E^*.$$

1. Why $B$ is needed to define the Euclidean norm of $E$ and $E^*$?
2. In the definition of $uu^*(s)$, what is $u$ in the right hand side of definition $u\cdot\langle s, u\rangle$ ?

Answer 1

Let's say you have a finite-dimensional real inner product space $(E, g)$. Using the inner product $g$, you can define a linear map $B_g \colon E \rightarrow E^{*}$ by the formula

$$ B_g(u)(v) = \left<B_g(u), v \right> := g(u, v). $$

The map $B_g$ satisfies the following two properties:

1. (Symmetry) We have $\left< B_g(u), v \right> = \left< B_g(v), u \right>$ for all $u,v \in E$.
2. (Positivity) We have $\left< B_g(v), v \right> \geq 0$ for all $v \in E$ and $\left< B_g(v), v \right> = 0$ if and only if $v = 0$.

Going the other way around, given a finite dimensional real vector space $E$ and a linear map $B \colon E \rightarrow E^{*}$ which is symmetric and positive in the sense described above, you can define an inner product on $E$ by the formula $g_B(u,v) := \left< B(u), v \right>$ and, using the inner product $g_B$, you can then define a norm by the formula $$\|v\|_{B} := g_B(v,v)^{\frac{1}{2}} = \left< B(v), v \right>^{\frac{1}{2}}. $$

The constructions are mutually inverse (we have $g_{B_g} = g$ and $B_{g_B} = B$) and thus, a finite dimensional real inner product space $(E,g)$ (an Euclidean space) is the same as a finite dimensional vector space $E$ together with a symmetric and positive linear map $B \colon E \rightarrow E^{*}$.

Given an inner product $g$ on $E$, the map $B_{g}$ is an isomorphism and we can use it to define an inner product on $E^{*}$ by the formula

$$ \left< \psi, \phi \right> := g \left( B_{g}^{-1}(\psi), B_{g}^{-1}(\phi) \right) = \left< B_g(B_g^{-1}(\psi)), B_{g}^{-1}(\phi) \right> = \left< \psi, B_{g}^{-1}(\phi) \right> $$

which induces the norm

$$ \|\psi\| = \left< \psi, B_{g}^{-1}(\psi) \right>^{\frac{1}{2}} $$

on $E^{*}$, explaining the appearance of $B^{-1}$ in the formula for the norm on $E^{*}$.

Finally, the $u$ that appears in the formula is a vector in $E$. More explicitly,

$$ (uu^{*})(s) := s(u) \cdot u $$

where $s(u) \in \mathbb{R}$ is a scalar and $u$ is a vector. This is a rank-one (or rank zero) linear map whose image is (contained in) $\operatorname{span} \{ u \} $.