Csdrhrt

For every $ninmathbb{N^+}$, let $f_n:(0,+infty)tomathbb{R}$ be as defined:
$$f_n(x)=frac{x-n}{x^2}cdotchi_{(n,+infty)}(x).$$

Study the convergence of the sequence ${f_n}_{ninmathbb{N^+}}$ 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?

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 $xleq n$, 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 $f_n(x)^pleq x^{-p}chi_{(n,infty)}(x)$.

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 $x^{-p}chi_{(n,infty)}(x)leq x^{-p}chi_{(1,infty)}(x)$).