Convergence in $L^{p}$ spaces Set $$f_n= n1_{[0,1/n]}$$
For $0<p\le\infty $ , one has that  $\{f_n\}_n$ is in $L^p(\mathbb R)$. For which values of $p$ is  $\{f_n\}_n$ a Cauchy sequence in $L^p$? Justify your answer.
This was a Comp question I was not able to answer. I don't mind getting every details of the proof.
What I know for sure is for $p=1$,  $\{f_n\}_n$ is Cauchy in $L^p$ because when you get the integral of the function that is going to equal 1 no matter the value of $n$. So the sequence is not convergent in $L^1$, and hence is not Cauchy. I do not know how can I be more rigorous on this problem.  Any help much appreciated. 
 A: Note that, we have $$\Vert f_{2n} -f_n\Vert_p^p = n^p \left(\dfrac1n - \dfrac1{2n}\right) + (2n-n)^p \dfrac1{2n} = \dfrac{n^{p-1}}2 + \dfrac{n^{p-1}}2 \geq 1 \,\,\,\,\,\,\, \forall p \geq 1$$
For $p<1$, and $m>n$ we have
$$\Vert f_m - f_n\Vert_p^p = n^p \left(\dfrac1n - \dfrac1m\right) + (m-n)^p \dfrac1m < n^p \dfrac1n + \dfrac{m^p}m = n^{p-1} + m^{p-1} < 2 n^{p-1} \to 0$$
EDIT
Note that
\begin{align}
f_m(x) & =\begin{cases} m & x \in [0,1/m]\\ 0 & \text{else}\end{cases}\\
f_n(x) & =\begin{cases} n & x \in [0,1/n]\\ 0 & \text{else}\end{cases}
\end{align}
Since $m>n$, we have
$$f_m(x) - f_n(x) = \begin{cases} (m-n) & x \in [0,1/m]\\ -n & x \in [1/m,1/n]\\ 0 & \text{else}\end{cases}$$
Hence,
$$\vert f_m(x) - f_n(x) \vert^p = \begin{cases} (m-n)^p & x \in [0,1/m]\\ n^p & x \in [1/m,1/n]\\ 0 & \text{else}\end{cases}$$
Hence, $$\int \vert f_m(x) - f_n(x) \vert^p d \mu = (m-n)^p \times \dfrac1m + n^p \left(\dfrac1n - \dfrac1m\right)$$
A: You need to compute $\int_0^1 |f_n(x) - f_m(x)|^p dx$ by hand. For $p \ge 1$, this is not a Cauchy sequence, since $\Vert f_n \Vert_{L^p}^p = n^{p-1}$. For $p < 1$ there is hope :)
