ST the sequence $f_n$ where $f_n(x)=e^{-nx}$ is point wise but not uniformely convergent in $[0,\infty[$. Also show that the convergence is uniform in $[k,\infty[$, $k$ being a positive number.
I am able to show that point wise limit is $f(x)=0 \ \ \forall \ x$
Let $\epsilon>0$ be given then,
$|f_n(x)-f(x)|=|e^{-nx}-0|=e^{-nx}<\epsilon$
we can choose $m\in N$ such that $\large \large m>{1\over {\epsilon xlogx}} $
so as $x\rightarrow \infty \ \ \ m\rightarrow 0$ , but $m>0$ so not uniformely convergent. Is it good ?
How do i proceed further ?
Best Answer
The sequence $(f_n)$ is pointwise convergent on $[0,+\infty)$ to the function $f$ defined by
$$f(x)=\left\{\begin{array}\\1&\text{if}\ x=0\\ 0&\text{if}\ x>0\end{array}\right.$$ and since $f$ isn't continous then the convergence isn't uniform on $[0,+\infty)$.
For all $k>0$ we have $$\forall x\geq k,\quad|f_n(x)|\leq e^{-nk}\to_{n\to\infty}0$$ so the sequence is uniformly convergent on $[k,+\infty)$.