# Self-duality in Banach spaces

I know few about Banach spaces, however I have came to a hypotheses. Before, let me give a definition:

Definition: A normed vector space X is trivial if it has the following two properties:

1. For every $z \in X$ there exists one and only one $f$ such that $f (z) = 1$ and $||f|| = \frac{1}{||z||}$.

2. For every $f \in X$ there exists one and only one $z$ such that $f (z) = 1$ and $||z|| = \frac{1}{||f||}$.

Let $(\cdot,\cdot) : X → X'$ denote a map (I like this notation) which assigns to each $x \in X$ some $(\cdot,x)\in X'$ such that $(x,x) = ||x||^2$ and $||(\cdot,x)|| = ||x||$. This map exists by Hann-Banach theorem and choice axiom. If $X$ has the property (1) above, the map $(\cdot,\cdot)$ is unique or well-defined.

My question is about the following hypotheses:

The map $(\cdot,\cdot)$ is a well-defined anti-linear isommetry between $X$ and $X'$ if and only if $X$ is trivial.

I know it is true for Hilbert spaces, where $(\cdot,\cdot)$ coincides with the usual inner product. Moreover, if we use the function $(\cdot,\cdot)$ to define orthogonality in general, we recover Birkhoff-James' orthogonality.

I have proved that (trivial spaces):

a) $(x,x)\ge 0$. $(x,x)=0$ if and only if $x=0$.

b) $(\alpha x+\beta y,z)=\alpha (x,z)+\beta (y,z)$.

c) $(x,\alpha y)=\overline{\alpha}(x,y)$.

But I have been not able to prove that $(z,x+y)=(z,x)+(z,y)$. If the hypotheses is true, trivial space would be reflexive.

Thank you.

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Your definition of "trivial" does not make sense unless you require $z\in X\setminus\{0\}$, $f\in X^*\setminus\{0\}$.
The hypothesis is not true. Choose $1<p<\infty$ with $p\neq 2$. Then $$(a_n)\mapsto (\sigma(\overline{a_n})|a_n|^{p-1} /\|a\|_p^{p-1})$$ is the map you denote by $x\mapsto (\cdot, x)$ if $X=\ell_p$. It is easy to see that this map is not anti-linear.