Uniform convergence

**tholan** · November 1st 2010, 03:48 AM

I am asked to show that $f_n (x) = x^n$ is uniformly convergent in $(0, 1)$ but not in $[0, 1]$ .

I proceed by computing the pointwise convergence to $f(x) = 0$ for $x \in [0, 1)$ and $f(x) = 1$ for $x = 1$ .

Then I compute for $x \in [0, 1)$ :

$\vert f_n(x) - f(x) \vert < \epsilon \Leftrightarrow \vert x^n \vert < \epsilon \Leftrightarrow x < (\epsilon)^{1/n}$

This is not fullfilled for all $x \in (0, 1)$ (since I can always find an $x : (\epsilon)^{1/n} < x < 1$ ) and therefore I have shown that $f_n (x) = x^n$ is not uniformly convergent in $(0, 1)$ .

I am confused since I was asked to show that it is unformly convergent in $(0, 1)$ . What is wrong?

**Opalg** · November 1st 2010, 06:23 AM

Originally Posted by tholan

I am asked to show that $f_n (x) = x^n$ is uniformly convergent in $(0, 1)$ but not in $[0, 1]$ .

I proceed by computing the pointwise convergence to $f(x) = 0$ for $x \in [0, 1)$ and $f(x) = 1$ for $x = 1$ .

Then I compute for $x \in [0, 1)$ :

$\vert f_n(x) - f(x) \vert < \epsilon \Leftrightarrow \vert x^n \vert < \epsilon \Leftrightarrow x < (\epsilon)^{1/n}$

This is not fullfilled for all $x \in (0, 1)$ (since I can always find an $x : (\epsilon)^{1/n} < x < 1$ ) and therefore I have shown that $f_n (x) = x^n$ is not uniformly convergent in $(0, 1)$ .

I am confused since I was asked to show that it is unformly convergent in $(0, 1)$ . What is wrong?

The question is wrong. The sequence of functions $f_n (x) = x^n$ is not uniformly convergent in $(0, 1)$ , as the above argument correctly shows.

**tholan** · November 1st 2010, 10:31 AM

Thank you.

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