PMA exercise 7.13

105 Views Asked by At

I don't understand why all of these steps of the hint to prove a). Is monotone convergence theorem enough to prove a? The sequence is bounded and monotone then $\exists N(\epsilon,x)$ s.t $|f_n(x)- f(x)|<\epsilon,\ \forall n\ge N(\epsilon,x)$.

enter image description here And why $f_n(x)$ does't converge for all $n$?

1

There are 1 best solutions below

0
On BEST ANSWER

It may help to consider a specific example. Let \begin{equation*} f_n(x) = \begin{cases} 0, & x<0\\ x^{n^{(-1)^{n+1}}}, & 0 \leq x < 1\\ 1, & 1 \leq x \end{cases} \end{equation*} so $f_1(x)=x$, $f_2(x)=x^{1/2}$, $f_3(x)= x^3$, $f_4(x) = x^{1/4}$, etc. on $[0,1]$.

Then each function is monotonically increasing. What can you say about subsequences?