I am looking for a non-trivial set-valued map with the following properties: $K:X \rightrightarrows Y$ is a weakly closed set-valued cone-map, where for every $x\in X$, $K(x)$ is proper, closed, convex and pointed cone in $Y$ with $\operatorname{int} K(x) \neq \emptyset$.
Here, $X$ and $Y$ are real reflexive Banach spaces and we call the map $K$ weakly closed, if the graph of $K$ is a weakly closed set, that is, if for all sequences $\{x_n \} \subseteq X$ and $\{y_n\} \subseteq Y$ with $x_n \rightharpoonup x$, $y_n \rightharpoonup y$ and $y_n \in K(x_n)$, we have $y\in K(x)$. For $x\in X$, the set $K(x) \subseteq Y$ is called a cone if we have $\lambda K(x) \subseteq K(x)$ for every $\lambda \geq 0$. The cone $K(x)$ is called convex if $K(x) + K(x) \subseteq K(x)$, proper (sometimes non-trivial) if we have $K(x) \neq \{0\}$ and $K(x) \neq Y$ and pointed if $K(x) \cap K(x) = \{0\}$.
Some of my own thoughts: let us take $X=Y=\ell^2$ and put $$K(x):= \left\{ y\in \ell^2 \mid 0 \leq (1+x_1^2) \sum_{j=1}y_j < +\infty\right\},$$ where $x_1$ is the first number of the sequence $x=\{x_j\}$. Then we have that for every $x\in X$, $K(x)$ is proper, closed, convex and pointed cone (if I see this right), but I do not know how to show the weakly closedness of $K$.