I want to show if the following sequences are pointwise convergent and if uniformly convergent.
a) $f_n(x)$ =
- $(-1)^n\: if\:x\in [0,\frac1n]$
- $0 \: if\:x\in (\frac1n,1]$. Over [0,1],
b) $f_n(x)$=
- $0\: if\: x=0\: or\:x\in (\frac1n,1]$
- $(-1)^n n \:if\: x \in (0,\frac1n]$. Over [0,1],
Here's what I've done, I'm not sure if I'm doing it right:
a)If $x\in[0,1]$, then $f_n(x) = (-1)^n$ for all $n\leq1/x$. $f_n(x)=0\rightarrow0$ for all $n>1/x$. Since $(-1)^n$ doesn't converge then $f_n$ is not pontwise convergent.
b) If $x\in (0,1]$, then $f_n(x)=0 \rightarrow0$ for all $n>1/x$. and $f_n(x)=(-1)^nn$ for all $n\leq1/x$ which doesn't converge pointwise also f(0) = 0.
I'm not sure if the convergences are correct?
I think your answer in part a) is correct, but it's hard to follow what you're saying. I would rephrase the answer something like
a) If $0<x\le 1,$ then $f_n(x) = 0$ when $n > 1/x,$ so $f_n(x) \rightarrow 0.$ However, $f_n(0) = \pm 1$ for all $n,$ so $f_n$ does not converge at $x=0.$
Now in b) your answer is incorrect, and I can't help thinking it's mainly because you aren't expressing your ideas clearly, so you're getting confused.
It's the same idea; when $n >1/x, f_n(x) = 0,$ so $f_n(x) \rightarrow 0.$ However, this time we also have $f_n(0) \rightarrow 0$ The statement you make just seems to assert the the sequence doesn't converge without any support, but it's hard for me to be sure. I don't know if the beginning of your answer is just restating the question, or if there's supposed to be some reasoning there that I'm missing.