Let $(X, \|\cdot\|_X)$ is a finite dimensional real normed space and $X^*$ is the dual of $X$. And let $J\colon X\to 2^{X^*}$ be the duality mapping defined by $J(x) = \{f\in X^* : f (x)= \|x\|^2_{X} , \|f\|_{X^*} = \|x\|_{X}\}$. Then $X$ is Hilbert space if and only if $J(x)=\{x\}.$ I think this is folklore but I couldn't find any literature where this characterization is given, except this one "A Characterization of Hilbert Spaces" by Beata Radrianantoanina where this characterization is only mentioned without proof. Does anybody know the literature where this characterization of Hilbert spaces is given?
Here the equality $J(x)=\{x\}$ should be understand in the following way: Since $(X, \|\cdot\|_X)$ is a finite dimensional then there exists basis $e_1,e_2,\ldots e_n$ of $X$. For every $y\in X$ we can correspond $f_y\in X^*$ by the following way: $f_y(z) = y_1z_1+y_2z_2+\ldots +y_nz_n\in\mathbb R,\; \forall z\in X$ , where $y=y_1e_1+y_2e_2+\ldots y_ne_n$ and $z=z_1e_1+z_2e_2+\ldots z_ne_n$. The $J(x)=\{x\}$ actually means $J(x)=\{f_x\}$.
Let us assume that $J(x) = \{A \, x\}$, where $A \colon X \to X^*$ is linear. We set $$ (x,y)_A := \frac12 \, \big(\langle A \, x, y \rangle + \langle A \, y, x \rangle\big) . $$ This is bilinear and symmetric, and $$(x,x)_A = \langle A \, x, x\rangle = \|x\|^2.$$
Hence, $X$ is a Hilbert space.