if we have $f: \mathbb{R}^n \longrightarrow \bar{\mathbb{R}}$ we define its conjugate as the function $f^*: \mathbb{R}^n \longrightarrow \bar{\mathbb{R}}$ given by $$f^*(u)=\sup_{x \in \mathbb{R}^n}\{u'x-f(x)\}$$
My question is if given $u, v \in \mathbb{R}^n$ there is some condition we can impose on $v $ such that $f^*(u+v)=f^*(u)$.
Thank you
Suppose $f < \infty$ everywhere. In order for $f^*(u) \ne \infty$, we must have $f(x) = u'x - g(x)$ where $g$ is bounded above. Then if $v \ne u$ $$f^*(v) = \sup_x\, v'x - f(x) = \sup_x\, (v'-u')x + g(x) = \infty$$ since the first term can be made as large as desired, while the second term is bounded.
So the only condition that can be put on $v$ to make $f^*(u+v) = f^*(u)$ for all maps $f$ is $v = 0$.