[proofplan]
We prove the equivalence of the Stampacchia variational inequality $A(u) \circ (v - u) \ge 0$ and the Minty variational inequality $A(v) \circ (v - u) \ge 0$ for a monotone, hemicontinuous operator on a closed convex subset of a Hilbert space. The forward direction adds the monotonicity inequality to the Stampacchia inequality. The reverse direction substitutes convex combinations $v_t = u + t(w - u)$ into the Minty inequality, divides by $t > 0$, and passes to the limit $t \to 0^+$ using hemicontinuity.
[/proofplan]
[step:Derive the Minty inequality from the Stampacchia inequality using monotonicity]
Assume $A(u) \circ (v - u) \ge 0$ for all $v \in K$. Let $v \in K$ be arbitrary. By monotonicity of $A$:
\begin{align*}
(A(v) - A(u)) \circ (v - u) &\ge 0.
\end{align*}
Expanding by linearity of the dual pairing in the first argument:
\begin{align*}
A(v) \circ (v - u) - A(u) \circ (v - u) &\ge 0.
\end{align*}
Since $A(u) \circ (v - u) \ge 0$ by hypothesis, adding this non-negative quantity to the left-hand side preserves the inequality:
\begin{align*}
A(v) \circ (v - u) &\ge A(u) \circ (v - u) \ge 0.
\end{align*}
Since $v \in K$ was arbitrary, the Minty inequality holds.
[guided]
The forward direction is direct. We want to show $A(v) \circ (v - u) \ge 0$ for every $v \in K$, given that $A(u) \circ (v - u) \ge 0$ for every $v \in K$.
Fix an arbitrary $v \in K$. Monotonicity of $A$ means $(A(x) - A(y)) \circ (x - y) \ge 0$ for all $x, y \in K$. Applying this with $x = v$ and $y = u$:
\begin{align*}
(A(v) - A(u)) \circ (v - u) &\ge 0.
\end{align*}
Expanding the dual pairing (which is linear in the first argument since elements of $V^*$ are linear functionals):
\begin{align*}
A(v) \circ (v - u) &\ge A(u) \circ (v - u).
\end{align*}
The right-hand side is $\ge 0$ by the Stampacchia hypothesis. Chaining the two inequalities:
\begin{align*}
A(v) \circ (v - u) &\ge A(u) \circ (v - u) \ge 0.
\end{align*}
Since $v \in K$ was arbitrary, the Minty variational inequality holds for all $v \in K$.
[/guided]
[/step]
[step:Derive the Stampacchia inequality from the Minty inequality using convex combinations and hemicontinuity]
Assume $A(v) \circ (v - u) \ge 0$ for all $v \in K$. Let $w \in K$ be arbitrary. Since $K$ is convex and $u, w \in K$, the convex combination
\begin{align*}
v_t &= u + t(w - u) \in K, \quad t \in (0, 1],
\end{align*}
belongs to $K$. Substituting $v = v_t$ into the Minty inequality:
\begin{align*}
A(v_t) \circ (v_t - u) &\ge 0.
\end{align*}
Since $v_t - u = t(w - u)$, linearity of the dual pairing in the second argument gives
\begin{align*}
t \cdot [A(u + t(w - u)) \circ (w - u)] &\ge 0.
\end{align*}
Dividing by $t > 0$:
\begin{align*}
A(u + t(w - u)) \circ (w - u) &\ge 0.
\end{align*}
By hemicontinuity, the map $t \mapsto A(u + t(w - u)) \circ (w - u)$ is continuous on $[0, 1]$. Taking $t \to 0^+$:
\begin{align*}
A(u) \circ (w - u) &\ge 0.
\end{align*}
Since $w \in K$ was arbitrary, the Stampacchia inequality holds.
[guided]
The reverse direction is more subtle. We cannot directly pass from $A(v) \circ (v - u) \ge 0$ to $A(u) \circ (v - u) \ge 0$ because we lack strong continuity of $A$. Instead, we use a convex interpolation argument combined with hemicontinuity.
Fix an arbitrary $w \in K$. We want to show $A(u) \circ (w - u) \ge 0$. The idea is to test the Minty inequality at points $v_t$ that approach $u$ along the segment from $u$ to $w$.
Since $K$ is convex and $u, w \in K$, the point $v_t = u + t(w - u) = (1-t)u + tw$ lies in $K$ for every $t \in [0, 1]$. Substituting $v = v_t$ into the Minty inequality:
\begin{align*}
A(v_t) \circ (v_t - u) &\ge 0.
\end{align*}
Computing $v_t - u = t(w - u)$ and using linearity of the dual pairing in the second argument (for fixed $f \in V^*$, $f \circ (\lambda x) = \lambda (f \circ x)$ for $\lambda \in \mathbb{R}$):
\begin{align*}
t \cdot [A(u + t(w - u)) \circ (w - u)] &\ge 0.
\end{align*}
Since $t \in (0, 1]$ is strictly positive, dividing by $t$ preserves the inequality:
\begin{align*}
A(u + t(w - u)) \circ (w - u) &\ge 0, \quad \forall t \in (0, 1].
\end{align*}
Now we take $t \to 0^+$. The hypothesis of hemicontinuity states that the map $h: [0,1] \to \mathbb{R}$ defined by $h(t) = A(u + t(v - u)) \circ w'$ is continuous for all $u, v, w' \in K$. Applied with $v = w$ and $w' = w - u$, the function $t \mapsto A(u + t(w - u)) \circ (w - u)$ is continuous on $[0, 1]$. Since continuous functions preserve limits and the inequality $\ge 0$ is preserved under limits:
\begin{align*}
A(u) \circ (w - u) &= \lim_{t \to 0^+} A(u + t(w - u)) \circ (w - u) \ge 0.
\end{align*}
Since $w \in K$ was arbitrary, the Stampacchia variational inequality holds for all test vectors in $K$.
[/guided]
[/step]
