For every $n\in\mathbb{N^+}$, let $f_n:(0,+\infty)\to\mathbb{R}$ be as defined:
$$f_n(x)=\frac{x-n}{x^2}\cdot\chi_{(n,+\infty)}(x).$$
Study the convergence of the sequence $\{f_n\}_{n\in\mathbb{N^+}}$ in $L^p((0,+\infty))$.
The pointwise limit I think is $0$ because the indicator function on a smaller and smaller interval going to infinity.
Moreover, since $f_n(x)<0$ when $x<n$ so in this interval the indicator function is $0$, so the $L^1$ norm that is
$$\int_n^{+\infty}\frac{1}{x}-\frac{n}{x^2}dx=(\frac{n}{x}+\mathrm{log}(x)|_n^{+\infty}=+\infty$$
Since the $L^1$ norm is not finite, can we conclude that the sequence does not converge weakly (and so strongly) in $L^1$?
Moreover, can we say that the $L^p$ norms are also infinite?
Best Answer
First note that $f_n(x)\leq x^{-1}\chi_{(n,\infty)}(x)$, so we don't have to take powers of the sum in the numerator.
If $x\leq n$, then $f_n(x)=0$ because of the indicator function. If $x>n$, then $x-n>0$ and thus $f_n(x)>0$. In any case, $f_n(x)\geq 0$. Hence $f_n(x)^p\leq x^{-p}\chi_{(n,\infty)}(x)$.
Now you can either compute $\int_n^\infty x^{-p}\,dx$ and see that it converges to zero, or, if you are as lazy as me and already know that $x^{-p}$ is integrable at $\infty$, you can just evoke the dominated convergence theorem (clearly $x^{-p}\chi_{(n,\infty)}(x)\to 0$ pointwise and $x^{-p}\chi_{(n,\infty)}(x)\leq x^{-p}\chi_{(1,\infty)}(x)$).