I have a question about closed operator. Operator $F:X \to Y$ is closed if $(x_n)$ is a sequence in $D(F)$ that is convergent in $X$ and the sequence $(F(x_n))$ is convergent in $Y$, then we have $\lim x_n \in D(F)$ and $F(\lim x_n)= \lim F(x_n)$. Now, how about F multivalued mapping? I mean $F:X \to 2^Y$. Can I use that definition? Thanks.
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One definition I have seen is that if $x_n \to x$ and $y_n \in F(x_n)$, then if $y^$ is an accumulation point of the sequence $y_n$, you have $y^ \in F(x)$. – copper.hat Nov 05 '14 at 08:13