# Convergence on R->R

1. Mar 5, 2009

### tomboi03

Let fn : R $$\rightarrow$$ R be the function
fn= $$\frac{1}{n^3 [x-(1/n)]^2+1}$$

Let f : R $$\rightarrow$$ R be the zero function.
a. Show that fn(x) $$\rightarrow$$ f(x) for each x $$\in$$ R
b. Show that fn does not converge uniformly to f. (This shows that the converse of Theorem 21.6 does not hold; the limit function f may be continuous even though the convergence is not uniform.)

a. i'm not sure...
is f(x) equivalent to f1(x)?
If it is.... then... the function would be...
$$\frac{1}{[x-1]^2+1}$$
$$\frac{1}{x^2-2x+2}$$
but i'm not sure how to show fn(x) $$\rightarrow$$ f(x) for each x $$\in$$ R
b. Theorem 21.6 states,
" let fn: X$$\rightarrow$$ Y be a sequence of continuous functions from the topological space X to the metric space Y. If (fn) converges uniformly to f, then f is continuous. "

The converse of this is...
"If f is continuous, then (fn) converges uniformly to f. "

i don't know how to prove a function is not convergent.

Can someone help me?
Thank You

2. Mar 5, 2009

### Havoide

ad a) I would say, just multiply out the brackets in the denominator and look at what happens if you take $$\lim_{n\rightarrow \infty} f_{n}(x)$$ for fixed $$x$$.
ad b) Uniform convergence means, that $$\sup_{x\in \mathbb{R}} |f_{n}(x)-f(x)| \rightarrow 0$$. Now try to choose your x such that you see, that this does not happen (you should see what to do by looking at the demoninator).

3. Mar 5, 2009