$f_n(x) = n^{1/p}\chi(nx)$ in $L^p(\mathbb{R})$ weakly converges to 0 for $1 \leq p <\infty$

Question

I am trying to show that $f_n(x) = n^{1/p}\chi(nx)$ in $L^p(\mathbb{R})$ weakly converges to 0 when $1<p<\infty$. Here I have let $\chi$ denote the characteristic function on $[0,1]$. So far I have noticed that for $g\in L^q(\mathbb{R})$, for $q$ the conjugate pair of $p$, $$\int_\mathbb{R} f_n(x) g(x) d\mu(x) = \int_{[0,1/n]}n^{1/p}g(x) d\mu(x) = n^{1/p}\int_{[0,1/n]}g(x) d\mu(x)$$ From here, it seems like we should use some kind of change of variables like $x = u/n$. However, this does not seem to yield the final result. Also, why does the weak convergence fail when $p =1$? Thank you in advance!

Answer 1

$L^{p}(\mathbb R)$ always refers to $L^{p}$ space w.r.t. Lebesgue measure $\mu$ on $\mathbb R$.

By Holder's inequality, for $p>1$ and $g\in L_q$ where $\frac1q+\frac1p=1$, we have $$ n^{1/p}\int_0^{1/n}g(x) d\mu(x)$$ $$ \leq n^{1/p}\Big(\int_0^{1/n} |g(x)|^{q}d\mu(x)\Big)^{1/q} \Big(\int_0^{1/n} 1^{p}\Big)^{1/p} $$ $$=\Big(\int_0^{1/n} |g(x)|^{q}d\mu(x)\Big)^{1/q} \to 0.$$

When $p=1$ just take $g \equiv 1$ to see that your integral does not tend to $0$.