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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.

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$L^p$ spaces for $p \geq 1$ are complete so if it were cauchy here, it would converge. However, for $p>1$, the norm goes to infinity (contradiction). This goes to the 0 function in measure so by uniqueness of limits, for $p=1$, the norm would have to go to zero, which does not happen. It appears it only converges for $p<1$ (if it does at all) – Blitzer Dec 23 '12 at 1:11
up vote 3 down vote accepted

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, $$\int \vert f_m(x) - f_n(x) \vert^p d \mu = (m-n)^p \times \dfrac1m + n^p \left(\dfrac1n - \dfrac1m\right)$$

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I can see the point you are making here. But I am sorry, I have difficulty to follow the calculation you made. Can you tell me a source to understand the manipulation of characteristic function under the exponent. – Deepak Dec 23 '12 at 1:30
@Deepak Which one are you finding it difficult? – user17762 Dec 23 '12 at 1:34
@Deepak I have added some more details. Hope it is clear now. – user17762 Dec 23 '12 at 1:43
That is absolutely super cool. Every single thing does make sense now. Thanks a lot. – Deepak Dec 23 '12 at 1:45

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 :)

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