1. ## function sequence convergence

State where

$f_n(x)=x^{2n}-x^{n}, x \in \left[ -1, 1\right]$

is pointwisely convergent and find a limit function.

then assess if $f_n(x)=x^{2n}-x^{n}, x \in \left[ -1, 1\right]$ is uniformly convergent for $x \in \left[ 0, \frac{1}{2} \right]$

2. ## Re: function sequence convergence

Hint for the pointwise limit: deal with the cases $x=1$, $x=-1$, $x\in (-1,1)$.

3. ## Re: function sequence convergence

Originally Posted by girdav
Hint for the pointwise limit: deal with the cases $x=1$, $x=-1$, $x\in (-1,1)$.
so in case of $x=-1$ there is no limit. when $x \in (-1,1]$ limit is equal to $0$

yes?

4. ## Re: function sequence convergence

That's correct.

5. ## Re: function sequence convergence

So it isnt uniformly convergent when $x \in [-1,1]$, but it may be when $x \in [0,1/2]$ , yes?

6. ## Re: function sequence convergence

Yes, and that's what you have to determine.

7. ## Re: function sequence convergence

$sup|x^{2n} - x^{n}| \rightarrow 0$ , when $x \in [0, 1/2]$ and $n \rightarrow \infty$

So in that interval the function sequence is uniformly convergent. Is that ok and enough writing for an exam?

8. ## Re: function sequence convergence

Maybe you could justify that $\sup_{0\leq x\leq 1/2}|x^{2n}-x^n|\to 0$, for example using an upper bound.

9. ## Re: function sequence convergence

Originally Posted by girdav
Maybe you could justify that $\sup_{0\leq x\leq 1/2}|x^{2n}-x^n|\to 0$, for example using an upper bound.
could you explain?

10. ## Re: function sequence convergence

I just meant that $\sup_{0\leq x\leq 1/2}|x^{2n}-x^n|\leq \frac 1{2^n}$ in order to justify uniform convergence (just saying it is not enough).

11. ## Re: function sequence convergence

Maybe Im stupid but I just cant see why it is so...

12. ## Re: function sequence convergence

For $0\leq x\leq \frac 12$ and $n$ integer $|x^{2n}-x^n|=|x|^n|x^n-1|\leq \frac 1{2^n}(1-x^n)\leq \frac 1{2^n}$.

13. ## Re: function sequence convergence

aaa of course! thank you very much