Are projection and norm enough to define an inner product?

Question

Given an inner product, one can define a projection and a norm. Can we do the opposite?

That is, suppose we have:

• a complex vector space V
• a norm $|V|^2 : V \rightarrow \mathbb{R}$ such that:
  • is posite definite
  • $|\alpha V|^2 = |\alpha|^2 |V|$ with $\alpha \in \mathbb{C}$
• a family of projections $\forall v \in V \; \exists \mathbf{P}_v : V \rightarrow V$ such that:
  • $\mathbf{P}_v ^2 = \mathbf{P}_v$
  • $\mathbf{P}_v (\alpha w_1 + \beta w_2) = \alpha \mathbf{P}_v (w_1) + \beta \mathbf{P}_v (w_2)$
  • $\forall w \in V$ $\mathbf{P}_v(w)=\alpha v$ with $\alpha \in \mathbb{C}$
  • $\mathbf{P}_v(v)=v$

Is the map $\langle \cdot , \cdot \rangle : V \times V \rightarrow \mathbb{C}$ defined such that $\mathbf{P}_v(w)=\frac{\langle v, w \rangle}{|v|^2} v$ an inner product?

$\langle \cdot , \cdot \rangle$ is positive definite. $\mathbf{P}_v(v) = v$. $\langle v, v \rangle = |v|^2$. The norm induced by the map is the original norm which is positive definite.

$\langle \cdot , \cdot \rangle$ is linear in the second argument because the projection is linear.

Conjugate symmetry is the only thing that I am missing. Is it an extra requirement, or can it be derived by the previous?

Answer 1

The family of projections with one-dimensional range is a red herring: It is equivalent to require a family of linear functionals (if you assume the axiom of choice). In the case of a normed space, it is equivalent to require a family of bounded projections/bounded linear functionals by Hahn-Banach (again, if you assume the axiom of choice).

Since you require nothing on the family of functionals/projections with respect to $v$, there are plenty of functionals/projections violating the conjugate symmetry.

A prominent family of examples are semi-inner products, see https://en.wikipedia.org/wiki/L-semi-inner_product (The examples given there are linear in the first argument instead of the second, but this is of course just a notational difference.)