1. ## Pointwise convergence.

I'm ask to prove whether or not the sequence of functions $\displaystyle {f_n}$ defined on [-1,1] is pointwise convergent, where $\displaystyle f_n = (1-|x|)^n$ for each natural number n. I was hoping someone could look over my proof and see if I made any mistakes. The cases where x=0 and |x|=1 are obvious, so consider those as already proven.

Proof: Let $\displaystyle \epsilon >0$ be given and let $\displaystyle x \in (-1, 1)$ where $\displaystyle x \ne 0$. Then, $\displaystyle 0<1-|x|<1$. Let N be a positive integer such that $\displaystyle N> \frac{|ln(\epsilon)|}{|ln(1-|x|)|}$. Then, $\displaystyle \epsilon >(1-|x|)^N$ (I'm not 100% sure this is right, so please correct me if I'm wrong). So, for any positive integer $\displaystyle n>N$, we have $\displaystyle |(1-|x|)^n-0|=(1-|x|)^n<(1-|x|)^N<\epsilon$. Therefore, $\displaystyle {f_n}$ converges pointwise to the zero function on (-1,1)-{0}.

2. Originally Posted by spoon737
I'm ask to prove whether or not the sequence of functions $\displaystyle {f_n}$ defined on [-1,1] is pointwise convergent, where $\displaystyle f_n = (1-|x|)^n$ for each natural number n. I was hoping someone could look over my proof and see if I made any mistakes. The cases where x=0 and |x|=1 are obvious, so consider those as already proven.
If $\displaystyle x\not = 0$ then $\displaystyle 0<1-|x| < 1$ and so $\displaystyle (1-|x|)^n \to 0$. Otherwise, if $\displaystyle x=0$ then $\displaystyle (1-|x|)^n = 1^n \to 1$. Thus, $\displaystyle f_n\to f$ where $\displaystyle f(x) = \left\{ \begin{array}{c} 1\mbox{ if }x=0\\0\mbox{ if }-1\leq x\leq 1 \mbox{ and }x\not = 0 \end{array} \right.$.

3. It was easy enough for me to see that $\displaystyle (1-|x|)^n$ converges to zero where $\displaystyle 0<|x|\le 1$, but my professor wants us to provide more details. I guess my main concern is if I made the right choice for N. Will that work, or did I make a mistake?