A norm which is symmetric enough is induced by an inner product?

Question

$\newcommand{\<}{\langle} \newcommand{\>}{\rangle} $

It is a fact that for every norm $\| \|$ on a finite dimensional (real) vector space, its isometry group $\text{ISO}(|| \cdot ||)$ is contained in some isometry group of a suitable inner product. (see this question).

Now assume we have a norm $\| \|$ such that $\text{ISO}(|| \cdot ||)=\text{ISO}(\<,\>)$ for some inner product.

Is $\| \|$ necessarily induced by an inner product?

Update:

The answer is yes. The key fact is the transitivity of the isometry group. Actually, as pointed out in this question in MO, the following statement is true:

Let $X$ be finite-dimensional normed space whose isometry group acts transitively on the unit sphere (i.e, for every two unit-norm vectors $x,y∈X$ there exist a linear isometry from $(X, \| \|)$ to itself that sends $x$ to y). Then X is a Euclidean space (i.e., the norm comes from a scalar product).

The proof is decomposed of two steps:

(1) showing there exists an inner product $\< ,\>$ whose isometry group $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$. The basic idea is this:

$\text{ISO}(|| \cdot ||)$ is compact, hence it admits an invariant probability measure, hence (by an averaging argument) there exists a Euclidean structure preserved by it. (The details can be found here).

(2) The transitivity of $\text{ISO}(|| \cdot ||)$ together with (1) impliy that the original norm is proportional to that Euclidean norm.

Answer 1

I think the answer is yes and the corresponding inner product is proportionnal to $\langle \cdot , \cdot \rangle$. Let's denote $E$ the vector space and $S$ the sphere associated to the inner product $\langle \cdot , \cdot \rangle$, i.e. the set $S=\lbrace x \in E \; | \; \langle x , x \rangle =1 \rbrace$.

The key for the demonstation is the fact that $ISO(\langle \cdot , \cdot \rangle)$ acts transitively on $S$. For the following, we will fix an element $x_0$ of $S$.

Let $x\in E$ be any element of $E$. Defining $y=\dfrac{x}{\sqrt{\langle x , x \rangle}}$, we have $y \in S$. Thus, by transitivity, there exists $u \in ISO(\langle \cdot , \cdot \rangle)$ such that $y=u(x_0)$. Then, we have : $$\begin{array}{rcl} \|x\|^2 & = & \| \sqrt{\langle x , x \rangle} y \|^2 \\ & = & \langle x , x \rangle \|y\|^2 \\ & = & \langle x , x \rangle \|u(x_0)\|^2 \\ & = & \langle x , x \rangle \|x_0\|^2 \; \; \; \; \text{as} \; u\in ISO(\langle \cdot , \cdot \rangle)=ISO(\| \cdot \|)\\ \end{array}$$

Thus, defining $\lambda=\|x_0\|^2$, we have, for all $x \in E$ : $$ \boxed{\| x \|^2 = \lambda \langle x , x \rangle}$$ which ends the demonstration.