Convergence of fn: R->R to f: R->R

[tomboi03]

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

Let f : R $$\rightarrow$$ R be the zero function.
a. Show that f_n(x) $$\rightarrow$$ f(x) for each x $$\in$$ R
b. Show that f_n 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 f_n(x) $$\rightarrow$$ f(x) for each x $$\in$$ R
b. Theorem 21.6 states,
" let f_n: X$$\rightarrow$$ Y be a sequence of continuous functions from the topological space X to the metric space Y. If (f_n) converges uniformly to f, then f is continuous. "

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

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

Can someone help me?
Thank You

 

[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).

 

[HallsofIvy]

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

Let f : R $$\rightarrow$$ R be the zero function.
a. Show that f_n(x) $$\rightarrow$$ f(x) for each x $$\in$$ R
b. Show that f_n 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...
$$\frac{1}{[x-1]^2+1}$$
$$\frac{1}{x^2-2x+2}$$

I'm not clear what you mean by "equivalent" here. I would recommend calculating a few more terms :
$$f_2(x)= \frac{1}{8(x- 1/2)^2+ 1}= \frac{1}{8x^2- 8x+ 3}$$
$$f_3(x)= \frac{1}{27(x- 1/3)^2+ 1}= \frac{1}{27x^2- 18x+ 4}$$

but I'm not sure how to show f_n(x) $$\rightarrow$$ f(x) for each x $$\in$$ R

Surely, it is not all that difficult to calculate the general
$$f_n(x)= \frac{1}{n^3(x- 1/n)^2+ 1}= \frac{1}{n^3(x^2- (2/n)x+ 1/n^2+ 1}$$

b. Theorem 21.6 states,
" let f_n: X$$\rightarrow$$ Y be a sequence of continuous functions from the topological space X to the metric space Y. If (f_n) converges uniformly to f, then f is continuous. "

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

Yes, and you are asked to prove that is not true.

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

That is NOT what you are asked to prove. You are told that this sequence of functions is convergent. You are asked to show that the convergence is not "uniform". That is, you must show that the N in "Given any $\epsilon$> 0, there exist N such that if n> N, the $|f_n(x)- f(x)|< \epsilon$ must depend on x.

an someone help me?
Thank You