1
$\begingroup$

I'm stuck on a homework problem

For each natural number $n$ and each number $x \in [0,1]$ let $f_n(x) = \frac{x}{nx+1}$ Find the function $f:[0,1] \to \mathbb{R}$ to which the sequence $\{f_n : [0,1] \to \mathbb{R}\}$ converges pointwise. Prove the convergence is uniform.

Any ideas? Thanks!

  • 1
    Where did you get stuck? Do you know what pointwise convergence means? Did you try to figure out the limit of $f_n(1)$, say? Do you know what uniform convergence is?2011-12-02

1 Answers 1

1

$f_n$ clearly converges to $g(x)=0$ pointwise.

Indeed, $f_n(0)=0$ for all $n$. While, for $0, we have ${1\over x}\ge1 $; whence:

$ \tag{1}|f_n(x)|=| {x\over nx+1}| ={1\over n+{1\over x}}\le{1\over n+1}\ \buildrel {n\rightarrow\infty}\over{\longrightarrow} \ 0. $

The above actually shows that the convergence is uniform: we can make $f_n(x)$ small for all $x$ by taking $n$ sufficiently large.

To be formal:

Let $\epsilon>0$. Choose $N$ so that ${1\over N+1}<\epsilon$. Then if $n\ge N$, we have, using (1): $|f_n(x)-0|=|f_n(x)|\le{1\over n+1} \le{1\over N+1}<\epsilon$ for all $0< x\le1$. Also, $|f_n(0)|=0<\epsilon$. Thus, the convergence is uniform.

  • 2
    Hmm, isn't there a policy for questions about homework here http://meta.math.stackexchange.com/questions/106/what-is-the-proper-way-to-handle-homework-questions ?2011-12-02