# Pointwise and uniform convergence

• Nov 15th 2009, 12:13 PM
thaopanda
Pointwise and uniform convergence
a) For each n $\in$ N consider $F_{n}$ : $[0, \infty) \rightarrow$ R be given by $F_{n}(x) := (x^n)e^{-nx}$. Show that { $F_{n}$} $_{n \in N}$ converges in a pointwise fashion to the zero function on $[0, \infty)$.

b) Decide whether the convergence in part a is uniform on $[0, \infty)$.

• Nov 15th 2009, 12:26 PM
redsoxfan325
Quote:

Originally Posted by thaopanda
a) For each n $\in$ N consider $F_{n}$ : $[0, \infty) \rightarrow$ R be given by $F_{n}(x) := (x^n)e^{-nx}$. Show that { $F_{n}$} $_{n \in N}$ converges in a pointwise fashion to the zero function on $[0, \infty)$.

b) Decide whether the convergence in part a is uniform on $[0, \infty)$.

a) Well, for all $x$, $\lim_{n\to\infty}\frac{x^n}{e^{nx}}=0$. One way to see this is that if you use L'Hopital's Rule $n$ times, you'll get $\lim_{n\to\infty}\frac{n!}{n!e^{nx}}=0$
b) The derivative of $x^ne^{-nx}$ is $nx^{n-1}e^{-nx}(1-x)$, so the max on $[0,\infty)$ occurs at $x=1$ (for all $n$), and it follows that $x^ne^{-nx}\leq e^{-n}$, which can clearly be made less than any $\epsilon$. (Choosing $n>-\ln\epsilon$ does it.)