How can prove the following sentence?
a correspondence having a convex graph is convex valued as well.
In order to explain the functions which used in this proof context,
This comes from Kakutani’s theorem:
Let$ f : S \to S $ be a correspondence. If S is nonempty, compact, and convex, and if f is nonempty-valued, convex-valued, and has a closed graph, then f has a fixed point. When we say that f is nonempty-valued and convex-valued, we mean that $f(s) $ is a nonempty convex set for every $s \in S$.
Please look atthe lecture note, page 3
Also please look at the book, page 229 “ Every correspondence with convex graph is also convex valued, but the converse is false.
I want to prove this first part. Please share your ideas with me. Thank you.
Let $f: X\to\!\!\!\to Y$ be a correspondence with convex graph.
Let $x\in X$ and $y_1, y_2\in f(x)$, i.e., $(x, f(y_1)), (x, f(y_2))\in \text{Gr}f$.
Let $\lambda\in (0,1)$.
Since $(x, f(y_1)), (x, f(y_2))\in \text{Gr}(f)$ and $f$ has convex graph, $$\lambda(x, f(y_1)) + (1-\lambda)(x, f(y_2))\in \text{Gr}(f).$$ Let us check the LHS. $$\begin{aligned} \lambda(x, f(y_1)) + (1-\lambda)(x, f(y_2)) &=(\lambda x, \lambda f(y_1)) +((1-\lambda)x, (1-\lambda)f(y_2))\\ &=(\lambda x+(1-\lambda)x,\, \lambda f(y_1) + (1-\lambda) f(y_2))\\ &=(x,\, \lambda f(y_1) + (1-\lambda) f(y_2)). \end{aligned}$$ So, $(x, \lambda f(y_1) + (1-\lambda) f(y_2))\in\text{Gr(f)}$.
That means $\lambda f(y_1) + (1-\lambda) f(y_2)\in f(x)$.
Hence, $f(x)$ is convex-valued. $\quad\checkmark$