I have the following question regarding mappings on a Banach space $X$. If anyone has an idea or hint as to how to resolve this question it would appreciated.
Let $X$ be a Banach space, $X^{*}$ its dual space and $K \subset X$ a nonempty, closed, convex set. Assume that mapping $T: K \rightarrow X^{*}$ is an A-pseudomonotone and hemicontinuous mapping.
Definition: A mapping $T: K \rightarrow X^{*}$ is said to be A-pseudomonotone if for each $x,y \in K$ it follows that $$\langle T(y), x-y \rangle \geq 0 \text{ implies } \langle T(x), x-y \rangle \geq 0$$
and
Definition: A mapping $T: K \rightarrow X^{*}$ is said to be hemicontinuous, if the function $$t \mapsto \langle T(x+t(y-x)), y-x \rangle$$ is continuous at $0^{+}$, for all $x,y \in K$.
We now define the following two set-valued mappings $T_{1}, T_{2}$ by $$K \ni y \mapsto T_{1}(y) = \{ x \in K: \langle T(x), y-x \rangle \geq 0 \}$$ and $$K \ni y \mapsto T_{2}(y) = \{ x \in K: \langle T(y), y-x \rangle \geq 0 \}$$
Consider $[x,y]$ as a line segment joining points $x$ and $y$. How does it follow from the fact that $T: M \rightarrow X^{*}$ is A-pseudomonotone and hemicontinuous, and $M \subset X$ is closed and covex, that $$\bigcap_{z \in M}T_{1}(z) = \bigcap_{z \in M}T_{2}(z)$$, where $M = K \cap [x,y]$.
I can show the first inclusion $$\bigcap_{z \in M}T_{1}(z) \subset \bigcap_{z \in M}T_{2}(z)$$ quite easily. The second one I have the proposed answer below, what do you think? Thanks for any assistance.
First note that since $T: K \rightarrow X^{*}$ is A-pseudomonotone and hemicontinuous, it also follows that $T: M \rightarrow X^{*}$ is also A-pseudomonotone and hemicontinuous, where $M := K \cap [x,y]$. Note that since $K$ is assumed nonempty, convex and closed it follows that $M$ is also nonempty, closed and convex. We want to show that $$\bigcap_{z \in M }T_{2}(z) \subset \bigcap_{z \in M}T_{1}(z)$$ Let $\bar{x} \in \bigcap\limits_{z \in M} T_{2}(z)$. Therefore $$\langle T(z), z -\bar{x} \rangle \geq 0$$ for all $z \in M$. We want to show that $$\langle T(\bar{x}), z-\bar{x} \rangle \geq 0$$ for all $z \in M$. We have $\bar{x} + t(z - \bar{x}) = (1-t)\bar{x} + tz \in M$ since $M$ is convex. Using the assumed hemicontinuity we get $$\langle T(\bar{x} + t(z-\bar{x})),z-\bar{x}\rangle \rightarrow \langle T(\bar{x}), z-\bar{x} \rangle \text{ } \text{ as } t \rightarrow 0^{+}$$
Since $\langle T(\bar{x} + t(z-\bar{x})),z-\bar{x}\rangle \geq 0$ it follows that $\langle T(\bar{x}), z-\bar{x} \rangle \geq 0$. This gives the desired result $$\bigcap_{z \in M }T_{2}(z) \subset \bigcap_{z \in M}T_{1}(z)$$
$\square$