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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

Eparoh
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  • So, by $u'$ you mean the transpose of $u$? And do you want a condition on $v$ that works for all $f$, or can the condition depend on $f$? – Paul Sinclair May 07 '20 at 23:54
  • Yes, $u'$ is the transpose of $u$ and I was looking for a general condition, independent of the $f$. – Eparoh May 08 '20 at 06:12

1 Answers1

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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 $v = 0$.

Paul Sinclair
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  • I think the first condition is not necessarely true if the domain of $f$ is not $\mathbb{R}^n$. For example, if $f$ is the indicator function of a compact set $C$ ($f$ is $0$ on $C$ and $+\infty$ outside of $C$) then its conjugate is the support function of $C$ and it is always real. – Eparoh May 11 '20 at 07:33
  • I see what you mean - When $g(x) = -\infty$, it doesn't matter how large $(v'-u')x$ is, the total value is still $-\infty$. – Paul Sinclair May 11 '20 at 13:36
  • However, that does not matter, since you want this to work for all $f$, not just infinite $f$. I've added an additional qualification on $f$. – Paul Sinclair May 11 '20 at 13:42