Numerical Methods By Kandasamy Thilagavathy Gunavathy Pdf Free DownloadQ: Uniform integrability and boundedness of set-valued functions Suppose $E$ is a complete, separable metric space and $F:E\to \mathbb{R}$ is a set-valued function. I want to show that if $F$ is bounded on bounded subsets of $E$, then $F$ is uniformly bounded. Suppose $F$ is unbounded on bounded subsets of $E$: For all $n\in \mathbb{N}$ and $\varepsilon > 0$, there exist $y_n \in E$ and $f_n\in F$ such that $|f_n(y_n)| > n$ and $d(y_n,y_m) n$. We may assume that $y_n\to \xi$ as $n\to \infty$ for some $\xi \in E$. Define $f:E\to \mathbb{R}$ by $$f(x) = \inf_{y\in E}d(x,y)+f_n(y)$$ where $d(x,y)$ denotes the distance between $x,y\in E$. Question: Is $f$ uniformly bounded? If not, how can I show this? A: The answer is no. For example, on $\mathbb{R}$, consider a non-uniformly bounded set-valued function $F$ with $F(x)=\{0\}$ for every $x$. *\* 0.16 0.47 0.00 0.00 0.78 0.00 0.00 1.00 0.00 0.00 0.00 0.96 0.00 DLD-S15 0.00