Corollary 21.11.26.8C4DFDD0ADE0 Let $F\colon (0,\infty )\rightarrow [-\infty ,\infty )$ be a convex function. If we define $G\colon \mathbb {R}\rightarrow (-\infty ,\infty ]$ by
\[ G(\tau )=\sup _{t>0} \Bigl ( t\tau -F(t) \Bigr ), \]
then $G$ is a convex function and
E918B6149A40
21.11.27
\begin{equation} F(t)=G^*(t),\quad \forall t>0. \end{equation}
Moreover,
3B0DC2BF53F0
21.11.28
\begin{equation} G(\tau )=\sup _{t\in \mathbb {R}_{>0}} \Bigl ( t\tau -F(t) \Bigr )=\sup _{t\in \mathbb {Q}_{>0}} \Bigl ( t\tau -F(t) \Bigr ). \end{equation}
Proof.
We distinguish two cases.
First suppose that $F(t)=-\infty $ for some $t>0$. Then $F(t)=-\infty $ for all $t>0$ by the convexity of $F$. Our assertions are clear in this case.
Next assume that $F(t)\neq -\infty $ for all $t>0$. In this case, theorem 21.11.17 guarantees that $F$ admits a closed proper extension $\tilde{F}\in \mathrm{Conv}(\mathbb {R})$ with
\[ \tilde{F}(t)=\infty ,\quad \forall t<0. \]
It follows from (21.11.23) that
\[ G(\tau )=\tilde{F}^*(\tau ),\quad \forall \tau \in \mathbb {R}. \]
Now theorem 21.11.24 implies (21.11.27). Finally (21.11.28) follows from the continuity of $F$.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.