Let $\mathcal C$ be the class of continuous, nonnegative, not identically equal to zero, concave, positive homogeneous of first order functions from $\mathbb R^n_+ = \{ x \in \mathbb R^n \colon x \geq 0\}$ to $\mathbb R$. Define the transform $$ f^\times(y) = \inf_{x\colon f(x)>0} \left\{ \frac{x \cdot y}{f(x)} \right\}. $$ From this definition it follows that $f^\times \in \mathcal C$ provided $f \in \mathcal C$. I want to show that $f^{\times \times} = f$ for $f \in \mathcal C$. I tried to use a proof by contradiction.
From definition it follows that for any $x$, $y$ we have $f(x)f^\times(y) \leq x \cdot y$ and hence $f(x) \leq f^{\times \times}(x)$. Suppose now that there exists such $\tilde x$ that $f(\tilde x) < f^{\times\times}(\tilde x)$ and without loss of generality $\tilde x > 0$ (since functions are continuous). The set $F = \{ (x,t) \colon x \in \mathbb R^n_+, \; f(x) \geq t \geq 0$} is closed and convex and $(\tilde x,f^{\times \times}(\tilde x)) \not \in F$. Hence there are such $u \in \mathbb R^n$, $v$, $C_1$, $C_2 \in \mathbb R$ that $(u,v) \neq 0$ and $$ u \cdot \tilde x + v f^{\times \times}(\tilde x) > C_1 > C_2 > u \cdot x + v t, \quad \forall x \in \mathbb R^n_+, \; \forall t \geq 0, \; t \leq f(x). $$ From here we deduce that $v > 0$ and $u \in - \mathbb R^n_+$. But I don't see how to proceed. Maybe there is some other way of showing the required equality?