Adjusted Fenchel's inequality.

src/fa/convex/legendre.tex

@@ -69,24 +69,36 @@
     with equality if and only if $\phi \in \partial f(x)$. 
 \end{theorem}
 \begin{proof}
-    Let $x \in E$ and $\phi \in F$ with $f(x) < \infty$, then
+    Let $x \in E$ and $\phi \in F$. Assume without loss of generality that $f(x) < \infty$, then
     \[
         f(x) + f^*(\phi) \ge f(x) + \dpn{x, \phi}{\lambda} - f(x) = \dpn{x, \phi}{\lambda}
     \]
 
-    For the equivalence, 
+    Now let $x \in E$ and $\phi \in F$ such that $\dpn{x, \phi}{\lambda} = f(x) + f^*(\phi)$, then $f(x), f^*(\phi) < \infty$ and
     \begin{align*}
         f(x) + f^*(\phi) &= \dpn{x, \phi}{\lambda} \\
         f^*(\phi) &= \dpn{x, \phi}{\lambda} - f(x) 
     \end{align*}
 
-    if and only if for every $h \in E$, 
+    By definition, 
+    \[
+    \dpn{x, \phi}{\lambda} - f(x)  = f^*(\phi) = \sup_{h \in E}\dpn{x+h, \phi}{\lambda} - f(x+h)
+    \]
+
+    so for every $h \in E$, 
     \begin{align*}
         \dpn{x, \phi}{\lambda} - f(x) &\ge \dpn{x+h, \phi}{\lambda} - f(x+h) \\
         f(x + h) - f(x) &\ge \dpn{h, \phi}{\lambda} 
     \end{align*}
 
-    if and only if $y \in \partial f(x)$. 
+    Thus $\phi$ satisfies the subgradient inequality, and $\phi \in \partial f(x)$. 
+
+    On the other hand, if $f(x) < \infty$ and $\phi \in \partial f(x)$, then $f(x + h) - f(x) \ge \dpn{h, \phi}{\lambda}$ for all $h \in E$, and
+    \begin{align*}
+    f^*(\phi) &= \sup_{h \in E}\dpn{x+h, \phi}{\lambda} - f(x+h) = \dpn{x, \phi}{\lambda} - f(x) \\
+    f^*(\phi) + f(x) &=  \dpn{x, \phi}{\lambda}
+    \end{align*}
+    
 \end{proof}
 
 \begin{lemma}