# Show that $f_n$ doesn't converge to $0$ in $L^p$ where $f_n = n^{-1/p}\chi_{[0,1]}$

Let $\mathfrak{B}$ be the Borel $\sigma$-algebra on $\mathbb{R}$ and $m$ be the Lebesgue measure; let $1 \leq p < \infty$.

Show that $n^{-1/p} \chi_{[0,1]}$ does not converge to $0$ in $L^p$.

As far as I understand this, we need to show that:

$$\left(\int_{\mathbb{R}} |n^{-1/p}\chi_{[0,1]}|^p \,dm\right)^{1/p} = \left(\int_{[0,1]}n^{-1}\,dm\right)^{1/p}$$

doesn't converge to $0$ .

But in the limit as $n \rightarrow \infty$, $1/n \rightarrow 0$ and so wouldn't the right hand side tend to $0$? But this is precisely what I am trying not to show.

What am I doing wrong here? All help is much appreciated.

-

As far as I can see, you're doing nothing wrong; I have also checked the definition of $L^p$-convergence, and you use the correct one.
Are you sure it's not $f_n = n^{-1/p}\chi_{[0,n]}$ instead? That function converges to $0$ pointwise but not in $L^p$, as plugging it in readily shows.
D'oh! It was $\chi_{[0,n]}$. –  Jack Rousseau Oct 15 '12 at 10:18
We have $\lVert n^{-1/p}\chi_{[0,1]}\rVert_p=n^{-1/p}\lVert \chi_{[0,1]}\rVert_p$ which converges to $0$ as $-1/p<0$, but if we take $f_n=n^{-1/p}\chi_{[0,n]}$,we have $\lVert n^{-1/p}\chi_{[0,n]}\rVert_p=1$.