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Suppose $F:\mathbb R^n\to \mathbb R$ is a convex continuous function.

Moreover, for any $x\in \mathbb R^n$, $$ \limsup_{\lambda\to\infty} \frac {|F(\lambda x)|}{\lambda}<\infty. $$ I would like to conclude that the Legendre-Fenchel transform $$ F^*(y) =\sup_{x\in\mathbb R^n} (\langle y,x\rangle-F(x)) $$ is continuous on its domain (which is a closed set).

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  • $\begingroup$ I might be mistaken here, but $F^*$ should be a convex function and convex functions are themselves continuous (on their domain). $\endgroup$
    – mlk
    Commented Dec 2, 2021 at 13:51
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    $\begingroup$ @mlk: convex functions don’t have to be continuous on the boundary of their domain. $\endgroup$
    – Anthony Quas
    Commented Dec 2, 2021 at 16:02
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    $\begingroup$ This does not seem to be true. If $F\equiv 0$, then the conjugate is the indicator of $\{0\}$ which is not continuous in $0$. $\endgroup$
    – Dirk
    Commented Dec 3, 2021 at 11:08
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    $\begingroup$ Maybe there is some confusion regarding "continuous on its domain". If the domain is not the full space, the conjugate will never be continuous on the domain, because it has to jump to $+\infty$ on the boundary. So are you asking if the conjugate has full domain? $\endgroup$
    – Dirk
    Commented Dec 3, 2021 at 11:09
  • $\begingroup$ How do you define the domain: the set of $y$ for which it is finite? $\endgroup$
    – Fedor Petrov
    Commented Dec 4, 2021 at 9:10

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