More typo fixes.
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Bokuan Li
2026-06-24 22:09:17 -04:00
parent c81b77a721
commit 177ed32433

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@@ -202,14 +202,14 @@
\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 $\Gamma_t = t\gamma + \gamma_0$, then for each $y \in \bracs{f < \infty}$,
For each $t > 0$, let $\Phi_t = \phi_0 + t\phi$ and $\Gamma_t = \gamma_0 + t\gamma$, then for each $y \in \bracs{f < \infty}$,
\[
\dpn{y, \Phi_t}{\lambda} - \Gamma_t \le f(y) + t\dpn{y, \phi}{\lambda} - t\gamma \le f(y)
\dpn{y, \Phi_t}{\lambda} - \Gamma_t \le f(y) + t\underbrace{(\dpn{y, \phi}{\lambda} - \gamma)}_{\le 0} \le f(y)
\]
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 $(x, \alpha) \in E \times \real \setminus A$, $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$.