Let $H:R^n \to R$ be $C^2(R^n\setminus\{0\})$ if $H(x) \neq 0$ and $H(0)=0$, strictly convex function and homogeneous of degree 2. Define $G(y)= \max_{x \in R^n} ( \langle x,y \rangle - H(x))$.
So, $G(y)$ is well defined, this Is, for fixed $y$, $ x \mapsto \langle x,y \rangle - H(x)$ has a unique maximum.
I know that this followns from the strictly convex hypothesis and the fact that for some $c>0$ we have \begin{equation} \frac{1}{c} \langle y,y \rangle \leq H(y) \leq c \langle y,y \rangle, \; \forall y\neq 0. \end{equation}
And that, if $x_0$ is a maximum so $x_0$ must obey $ y= \nabla H (x_0)$.
Thank you for the help.
Yes, a unique maximum is achieved, and only strict convexity is necessary for this.
Fix $y \in \Bbb{R}^n$ and suppose $x_1, x_2$ both maximise the problem. That is, $$\langle x_1, y \rangle - H(x_1) = \langle x_2, y \rangle - H(x_2) = G(y).$$ Let $x = \frac{x_1 + x_2}{2}$. Then, $$\langle x, y \rangle - H(x) \le G(y) = \langle x_1, y \rangle - H(x_1).$$ By strict convexity, we must have $$H(x) < \frac{H(x_1) + H(x_2)}{2},$$ hence \begin{align*} \langle x_1, y \rangle - H(x_1) &\ge \langle x, y \rangle - H(x) \\ &> \frac{\langle x_1, y \rangle + \langle x_2, y \rangle}{2} - \frac{H(x_1) + H(x_2)}{2} \\ &= \frac{(\langle x_1, y\rangle - H(x_1)) + (\langle x_2, y\rangle - H(x_2))}{2} \\ &= \frac{G(y) + G(y)}{2} = G(y). \end{align*} This contradicts the definition of $G(y)$.