1. ## Uniform convergence

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

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

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

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

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

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

2. Originally Posted by tholan
I am asked to show that $\displaystyle f_n (x) = x^n$ is uniformly convergent in $\displaystyle (0, 1)$ but not in $\displaystyle [0, 1]$.

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

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

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

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

I am confused since I was asked to show that it is unformly convergent in $\displaystyle (0, 1)$. What is wrong?
The question is wrong. The sequence of functions $\displaystyle f_n (x) = x^n$ is not uniformly convergent in $\displaystyle (0, 1)$, as the above argument correctly shows.

3. Thank you.