Pointwise & Unifrom Convergence

[kingwinner]

Homework Statement

Find all intervals on which the sequence of functions
f_n(x) = x²ⁿ / [n + x²ⁿ], n≥1, converges uniformly.

Homework Equations

The Attempt at a Solution

I think we'll need to compute the pointwise limit first.
For x=0 or x=1, the pointwise limit is 0.
But what is its pointwise limit for x>0?

Any help is appreciated!

 

[Gib Z]

Thing thing to do would be to consider the cases (0,1), x=1 and x>1 separately.

 

[kingwinner]

OK, I think for all of the cases x=0,x=1,0<x<1, the pointwise limit would be 0.

But how can we compute the pointwise limit for x>1??

 

[Gib Z]

Write \frac{x^{2n}}{x^{2n}+n} =\frac{x^{2n}+n - n}{x^{2n}+n} = 1 - \frac{n}{n+ x^{2n}}

On the right hand side, the second term is certainly always less than \frac{n}{x^{2n}}. You have a linear term divided by an exponential term, so what is the limit of that term?

 

[kingwinner]

n/x²ⁿ ->0 for x>1

x²ⁿ / [n + x²ⁿ] ≥ 1 - n/x²ⁿ

1 - n/x²ⁿ->1, but still how can we compute the limit of x²ⁿ / [n + x²ⁿ]?

 

[Gib Z]

I'm not sure you know where all the inequalities are pointing at the moment. Can you see how

0 &lt; \frac{n}{n+x^{2n} } &lt; \frac{n}{x^{2n}}

You know what the term on the far right approaches, so what must the middle term approach? And so then what is the limit of the function if |x|> 1 ?

 

[kingwinner]

n/(n+x²ⁿ < n/x²ⁿ
=> -n/(n+x²ⁿ > -n/x²ⁿ
=> 1 -n/(n+x²ⁿ > 1 -n/x²ⁿ

Thus, x²ⁿ / [n + x²ⁿ] > 1 - n/x²ⁿ

1 - n/x²ⁿ -> 1

But still this doesn't say what x²ⁿ / [n + x²ⁿ] would converge to...

 

[Gib Z]

Okay, just read post 6 again and tell me what the term on the far right converges to. And so, what do you conclude that the middle term goes to?

 

[kingwinner]

That middle term goes to 0, so fn(x) -> 1 for |x|>1 and fn(x)->0 otherwise. Now how can we determine uniform convergence. I know that for any interval [a,b] containing 1 or -1 since it's discontinuous there, we won't have uniform convergence, but how can we prove uniform convergence on [a,b] where a>1?

 

[Gib Z]

Come on, try something! and Show me what you have done, and where you get stuck.

Of course you have to apply a definition of uniform convergence and see if the function satisfies it on some intervals.

 

[kingwinner]

fn(x) = x²ⁿ / [n + x²ⁿ]

For the pointwise limit, I got fn(x) -> 1 for |x|>1 and fn(x)->0 otherwise.

So I know that for any interval [a,b] containing 1 or -1, we won't have uniform convergence, but how can we prove uniform convergence on [a,b] where a>1??
This is what I got, but I'm stuck...
fn(x) <= b²ⁿ / [n + a²ⁿ] < b²ⁿ / [n + 1]
How can we prove that the RHS ->0 ?

Please help!
Thanks!