- #1

twoflower

- 368

- 0

Hi, let's suppose

[tex]

f_{n}(x) = \frac{x^{n}}{1+x^{n}},

[/tex]

[tex]

a) x \in [ 0, 1 - \epsilon]

[/tex]

[tex]

b) x \in [ 1 - \epsilon, 1 + \epsilon]

[/tex]

[tex]

c) x \in [ 1 + \epsilon, \infty]

[/tex]

Where [itex]\epsilon \in \left( 0, 1 \right)[/itex]. Analyse pointwise, uniform and locally uniform convergence.

Well, we have

[tex]

\lim_{n \rightarrow \infty} f_{n} = \left\{ \begin{array}{rcl}

0 & x \in [0, 1)

\\ \frac{1}{2} & x = 1 \\

1 & x > 1

\end{array}\right.

[/tex]

I'll take the first case ([itex]x \in [ 0, 1 - \epsilon][/itex]). I want to find the supremum of

[tex]

\left|\frac{x^{n}}{1+x^{n}}\right|, x \in [0, 1-\epsilon]

[/tex]

To see if there are some extrems on this interval, I find the derivative:

[tex]

f^{'} = \frac{nx^{n-1}}{\left(1+x^{n}\right)^2} = 0 \Leftrightarrow x = 0 \notin [0, 1 - \epsilon]

[/tex]

So no extreme on this interval. Because

[tex]

f^{'} > 0

[/tex]

on this interval, [itex]f_{n}[/itex] is increasing. Thus the supremum is in the utmoust point of the interval, ie. [itex]\frac{1}{2}[/itex].

Thus I would say

[itex]f_{n}[/itex] doesn't converge uniformly on this interval.

Problem no. 1 - It's wrong.

Problem no. 2 - How would I analyse locally uniform convergence on this interval?

Thank you for any ideas.

[tex]

f_{n}(x) = \frac{x^{n}}{1+x^{n}},

[/tex]

[tex]

a) x \in [ 0, 1 - \epsilon]

[/tex]

[tex]

b) x \in [ 1 - \epsilon, 1 + \epsilon]

[/tex]

[tex]

c) x \in [ 1 + \epsilon, \infty]

[/tex]

Where [itex]\epsilon \in \left( 0, 1 \right)[/itex]. Analyse pointwise, uniform and locally uniform convergence.

Well, we have

[tex]

\lim_{n \rightarrow \infty} f_{n} = \left\{ \begin{array}{rcl}

0 & x \in [0, 1)

\\ \frac{1}{2} & x = 1 \\

1 & x > 1

\end{array}\right.

[/tex]

I'll take the first case ([itex]x \in [ 0, 1 - \epsilon][/itex]). I want to find the supremum of

[tex]

\left|\frac{x^{n}}{1+x^{n}}\right|, x \in [0, 1-\epsilon]

[/tex]

To see if there are some extrems on this interval, I find the derivative:

[tex]

f^{'} = \frac{nx^{n-1}}{\left(1+x^{n}\right)^2} = 0 \Leftrightarrow x = 0 \notin [0, 1 - \epsilon]

[/tex]

So no extreme on this interval. Because

[tex]

f^{'} > 0

[/tex]

on this interval, [itex]f_{n}[/itex] is increasing. Thus the supremum is in the utmoust point of the interval, ie. [itex]\frac{1}{2}[/itex].

Thus I would say

[itex]f_{n}[/itex] doesn't converge uniformly on this interval.

Problem no. 1 - It's wrong.

Problem no. 2 - How would I analyse locally uniform convergence on this interval?

Thank you for any ideas.

Last edited: