|
Suppose that for $n \in \mathbb{N}$, $$f_n(x) = \begin{cases} -1, & -1 \leq x \leq - 1/n, \\ nx, & |x| \leq 1/n, \\ 1, & 1/n \leq x \leq 1. \end{cases}$$ a) Find the limit function for $\{f_n\}$. b) Determine whether $\{f_n\}$ converges uniformly on $[-1,1]$. |
|||||||||||||||||||
|
|
Have you observed that for almost all $\,x\in [-1,1]\,$ there exists a natural index from which on the function's value is $\,1\,$ ...what's this value that no doubt spoils the uniform convergence? |
|||||||||
|
|
a) $f_n(x) \rightarrow f(x)$, a.e where $$ F(x) = \left\{ \begin{array}{rcl} -1,& \mbox{se} & x < 0\\ 0, & \mbox{se} & x=0\\ 1, & \mbox{se} & x > 0 \end{array} \right. $$ In fact, $f_n(0) = 0 \ \forall n$ and if $x \neq 0$. Chosse $n_0$ such that $1/n_0 < |x|$. Then $$n \ge n_0 \Rightarrow |f_n(x) -1| = 0 < \epsilon $$ for any $\epsilon$. b) If $f_n$ converges uniformily, by a) it converges to $f$ above. But this is not true because for $\epsilon = 1/4, \forall n$ take $x = 1/2n$. Then $$ |f_n(x) - 1| = |n 1/2n - 1| = 1/2 > 1/4. $$ |
|||
|
|
