Fixed sign typo in Fenchel-Moreau.
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@@ -190,7 +190,7 @@
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-\dpn{y, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha &< f(y)
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-\dpn{y, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha &< f(y)
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\end{align*}
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\end{align*}
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so $(-\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha) \le f$ and
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so $(-\mu^{-1}\phi, -\dpn{x, \mu^{-1}\phi}{\lambda} - \alpha) \le f$ and
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\[
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\[
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f^{**}(x) \ge -\dpn{x, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha \ge \alpha
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f^{**}(x) \ge -\dpn{x, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha \ge \alpha
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\]
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\]
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@@ -202,21 +202,19 @@
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\gamma = \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} < \dpn{x, \phi}{\lambda}
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\gamma = \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} < \dpn{x, \phi}{\lambda}
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\]
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\]
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For each $t > 0$, let $\Phi_t = \phi_0 + t\phi$ and $\Gamma_t = \gamma_0 - t\gamma$, then for each $y \in \bracs{f < \infty}$,
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For each $t > 0$, let $\Phi_t = \phi_0 + t\phi$ and $\Gamma_t = t\gamma + \gamma_0$, then for each $y \in \bracs{f < \infty}$,
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\[
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\[
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\dpn{y, \Phi_t}{\lambda} + \Gamma_t \le f(y) + t\dpn{y, \phi}{\lambda} - t\gamma \le f(y)
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\dpn{y, \Phi_t}{\lambda} - \Gamma_t \le f(y) + t\dpn{y, \phi}{\lambda} - t\gamma \le f(y)
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\]
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\]
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so $(\Phi_t, \Gamma_t) \le f$. By (1),
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so $(\Phi_t, \Gamma_t) \le f$. By (1),
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\[
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\[
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f^{**}(x) \ge \dpn{x, \Phi_t}{\lambda} + \Gamma_t = \dpn{x, \phi_0}{\lambda} + \gamma_0 + t\underbrace{(\dpn{x, \phi}{\lambda} - \gamma)}_{> 0}
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f^{**}(x) \ge \dpn{x, \Phi_t}{\lambda} - \Gamma_t = \dpn{x, \phi_0}{\lambda} + \gamma_0 + t\underbrace{(\dpn{x, \phi}{\lambda} - \gamma)}_{> 0}
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\]
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\]
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As the above holds for all $t > 0$, $f^{**}(x) = \infty \ge \alpha$. Since $f^{**}(x) \ge \alpha$ for all $(x, \alpha) \in E \times \real \setminus A$, $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$.
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As the above holds for all $t > 0$, $f^{**}(x) = \infty \ge \alpha$. Since $f^{**}(x) \ge \alpha$ for all $(x, \alpha) \in E \times \real \setminus A$, $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$.
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On the other hand, $f^{**} \le f$ by \autoref{lemma:conjugate-function-gymnatics}, so $\text{epi}(f^{**}) \supset \text{epi}(f)$. Since $\text{epi}(f^{**})$ is closed and convex, $\text{epi}(f^{**}) \supset \ol{\text{Conv}}(\text{epi}(f))$.
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By \autoref{lemma:conjugate-function-gymnatics}, $f^{**}$ is lower semicontinuous and convex with so $f^{**} \le f$, so $\text{epi}(f^{**}) \supset \text{epi}(f)$ and $\text{epi}(f^{**}) \supset \ol{\text{Conv}}(\text{epi}(f))$.
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\end{proof}
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\end{proof}
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