I need some help with this exercise: Let $n \in N$ and $f_n:[0,1]\Rightarrow R$ with $f_n(x):=\frac{n^\alpha \cdot \ x}{1+n^2x^2}$, and $\alpha\ge0$ a) For which $\alpha$ is $f_n$ pointwise convergent on $[0,1]$? And what is $f$ in this case? Is $f$ L-integrable on $[0,1]$?
I tried the convergents for $\alpha=0,\alpha=1,\alpha=2$ and so i think it convergent for $\alpha\le1$ because for $\alpha=2$ I have $1/x$ which i dont definied for $x=0$
Maybe you can help me, because if its convergent I can maybe use Lebesgue theorem of dominated convergence or monton convergence theorem
Thanks :)
It converges pointwise if $0\leqslant\alpha\leqslant2$. If $\alpha\in[0,2)$, it converges pointwise to the null function and, if $\alpha=2$, it converges pointwise to$$\begin{array}{ccc}[0,1]&\longrightarrow&\mathbb R\\x&\mapsto&\begin{cases}0&\text{ if }x=0\\\frac1x&\text{ otherwise.}\end{cases}\end{array}$$If $\alpha>3$, it doesn't converge pointwise.
In particular, your sequence converges pointwise to a Lebesgue-integrable function if and only if $\alpha\in[0,2)$.