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Let $f: \mathbb{R}^{n} \rightarrow \overline {\mathbb{R}}$. Called conjugate in the sense of Young-Fenchel of $f$, the following function: $$f^{*}(x^{*})=Sup\lbrace \langle x,x^{*}\rangle -f(x) : x \in Dom(f)\rbrace$$ If $f: \mathbb{R}^{n} \rightarrow \overline {\mathbb{R}}$ be a lower semicontinuous function such that $\lim \limits_{\Vert x\Vert \rightarrow \infty}\dfrac{f(x)}{\Vert x\Vert} =\infty$ then $$\lim \limits_{\Vert x\Vert \rightarrow \infty}\dfrac{f^{**}(x)}{\Vert x\Vert} =\infty ?$$Thanks for any suggestions.

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up vote 1 down vote accepted

This is true. Let's introduce notation $m_f(r)=\inf\{f(x): \|x\| \ge r\}$ and $M_f(R)=\sup\{f(x): \|x\| \le R\}$. By the definition of conjugate function, $f^*(x^*)\le \sup_{r\ge 0} (r\|x^*\|-m_f(r))$, hence $$M_f(R)\le \sup_{r\ge 0}(Rr-m_f(r)).$$ On the other hand, $f^*(x^*)\ge \sup_{R\ge 0} (R\|x^*\|-M_f(R))$, hence $$m_{f^*}(r)/r\ge \sup_{R\ge 0} (R-M_f(R)/r).$$

The assumption $m_f(r)/r\to\infty$ implies $M_{f^*}(R)<\infty$ for all $R$, which in turn yields $m_{f^{**}}(r)/r \to \infty$ as $r\to\infty$. (If the latter is not clear, consider what happens when $r=M_{f^*}(10^{10})$ and $R=10^{10}$).

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