Remember, f is set-valued, so if ri−1=0, f(r)=[0,1]. In all other cases, f(r)=ri−12. f is a nonempty convex set-valued function, so all that’s left is to show the closed graph property. If the limiting value of ri−1 is something other than 0, the closed graph property holds, and if the limiting value of ri−1 is 0, the closed graph property holds because 0∈[0,1].
I agree that the multimap you described is Kakutani and gives the correct fair set, but in the OP it says that if ri−1=0 then f(r)=ri, not f(r)=[0,1]. Maybe I am missing something about the notation?
It looks legitimate, actually.
Remember, f is set-valued, so if ri−1=0, f(r)=[0,1]. In all other cases, f(r)=ri−12. f is a nonempty convex set-valued function, so all that’s left is to show the closed graph property. If the limiting value of ri−1 is something other than 0, the closed graph property holds, and if the limiting value of ri−1 is 0, the closed graph property holds because 0∈[0,1].
Hi Alex!
I agree that the multimap you described is Kakutani and gives the correct fair set, but in the OP it says that if ri−1=0 then f(r)=ri, not f(r)=[0,1]. Maybe I am missing something about the notation?