Fixed sign typo in Fenchel-Moreau.
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Bokuan Li
2026-06-24 21:55:34 -04:00
parent 81e4ef6ffe
commit c897b3022e

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@@ -190,7 +190,7 @@
-\dpn{y, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha &< f(y) -\dpn{y, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha &< f(y)
\end{align*} \end{align*}
so $(-\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha) \le f$ and so $(-\mu^{-1}\phi, -\dpn{x, \mu^{-1}\phi}{\lambda} - \alpha) \le f$ and
\[ \[
f^{**}(x) \ge -\dpn{x, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha \ge \alpha f^{**}(x) \ge -\dpn{x, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha \ge \alpha
\] \]
@@ -202,21 +202,19 @@
\gamma = \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} < \dpn{x, \phi}{\lambda} \gamma = \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} < \dpn{x, \phi}{\lambda}
\] \]
For each $t > 0$, let $\Phi_t = \phi_0 + t\phi$ and $\Gamma_t = \gamma_0 - t\gamma$, then for each $y \in \bracs{f < \infty}$, For each $t > 0$, let $\Phi_t = \phi_0 + t\phi$ and $\Gamma_t = t\gamma + \gamma_0$, then for each $y \in \bracs{f < \infty}$,
\[ \[
\dpn{y, \Phi_t}{\lambda} + \Gamma_t \le f(y) + t\dpn{y, \phi}{\lambda} - t\gamma \le f(y) \dpn{y, \Phi_t}{\lambda} - \Gamma_t \le f(y) + t\dpn{y, \phi}{\lambda} - t\gamma \le f(y)
\] \]
so $(\Phi_t, \Gamma_t) \le f$. By (1), so $(\Phi_t, \Gamma_t) \le f$. By (1),
\[ \[
f^{**}(x) \ge \dpn{x, \Phi_t}{\lambda} + \Gamma_t = \dpn{x, \phi_0}{\lambda} + \gamma_0 + t\underbrace{(\dpn{x, \phi}{\lambda} - \gamma)}_{> 0} f^{**}(x) \ge \dpn{x, \Phi_t}{\lambda} - \Gamma_t = \dpn{x, \phi_0}{\lambda} + \gamma_0 + t\underbrace{(\dpn{x, \phi}{\lambda} - \gamma)}_{> 0}
\] \]
As the above holds for all $t > 0$, $f^{**}(x) = \infty \ge \alpha$. Since $f^{**}(x) \ge \alpha$ for all $(x, \alpha) \in E \times \real \setminus A$, $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$. As the above holds for all $t > 0$, $f^{**}(x) = \infty \ge \alpha$. Since $f^{**}(x) \ge \alpha$ for all $(x, \alpha) \in E \times \real \setminus A$, $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$.
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, $\text{epi}(f^{**}) \supset \ol{\text{Conv}}(\text{epi}(f))$. 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 $\text{epi}(f^{**}) \supset \ol{\text{Conv}}(\text{epi}(f))$.
\end{proof} \end{proof}