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Problems set III

Problem 12\_June\_2015\_1

Let f and f_{n}(n=1,2,...) be functions from R \rightarrow R such that f_{n}(x_{n})\rightarrow f(x) whenever x_{n}\rightarrow x. Show that f is continuous.

Problem 12\_June\_2015\_2

Let f: \mathbb{R} \rightarrow \mathbb{R} be monotonically increasing. Suppose a<b, a < f(a) and f(b)<b. Show that there exists x\in (a,b) such that f(x)=x.

Problem 12\_June\_2015\_3

Let \Omega =\mathbb{N},\mathcal{F}= power set of \mathbb{N}, P\{n\}=2^{-n!} for n=2,3,.... [ n! stands for the product of first n positive integers]. Let P\{1\} be determined by the condition P(\Omega)=1. If X and Y are independent random variables on (\Omega ,\mathcal{F},P) show that at least one of them is almost surely constant.


6 thoughts on “Problems set III

  1. Solution for 2nd problem : Consider the non-empty set A=\{x : f(x)\textgreater x,x\in[a,b]\}. We claim that m=\sup\{A\} is such that f(m)=m. Suppose f(m)=m+\epsilon\textgreater  m. Clearly m+\epsilon<b. Let \delta\textgreater 0 be such that a<m+\delta<m+\epsilon. Since m+\delta\notin A\Rightarrow f(m+\delta)\leq m+\delta, which contradicts the increasing property since f(m)=m+\epsilon \textgreater  m+\delta. If f(m)=m-\epsilon\textless m, by the supremum property there exist \delta\textgreater 0 such that m-\epsilon\textless m-\delta\in A\Rightarrow f(m-\delta)\textgreater m-\delta\textgreater m-\epsilon which is again a contradiction to the increasing property.Hence f(m)=m and clearly m\in(a,b)

    p.s There has been some writing problem in my earlier comment so i am reposting it. Also it would be good if we have a preview option(by installing plugin) so that we can see our comment before posting.

  2. Consider the non-epmty set A=\{x : f(x)>x,x\in[a,b]\}$. We claim that m=sup\{A\} is such that f(m)=m. Suppose f(m)=m+\epsilon>m. Clearly a0 be such that
    a \leq m+m\delta+\delta. If f(m)=m-\epsilon0 such that m-\delta \geq m-\epsilon and m-\delta\in A \Rightarrow f(m-\delta) \geq ;m-\delta, which is again a contradiction since f(m)=m-\epsilon \geq ;m-\delta. Hence f(m)=m and clearly m\in [a,b]

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