Find all $\alpha$ such that $n^\alpha\chi_{[n,n+1]}$ converges weakly to 0 in $L^p$.

Question

Edit: $1 < p < \infty$

Let $f_n(x) = n^\alpha \chi_{[n,n+1]}.$ Then $$ \begin{align} \left|\int_{-\infty}^{\infty} n^\alpha \chi_{[n,n+1]}g(x)dx\right| &\le n^\alpha \lvert\lvert\chi_{[n,n+1]}g(x)\rvert\rvert_1\\ &\le n^\alpha \lvert\lvert\chi_{[n,n+1]}\rvert\rvert_p \lvert\lvert g(x)\rvert\rvert_q\\ &= n^\alpha \lvert\lvert g(x)\rvert\rvert_q \end{align} $$ which goes to 0 if $\alpha < 0$. Suppose $\alpha \ge 0$ and let $g(x) = \chi_{[n,n+1]}$. Then $g\in L^q$, and $$\left|\int_{-\infty}^{\infty} n^\alpha \chi_{[n,n+1]}g(x)dx\right| = n^\alpha \to \infty.$$

Did I miss anything here? Thanks.

Answer 1

Hints, assuming $1<p<\infty$ (the answer is slightly different for the endpoint cases):

First, if $f_n\to0$ weakly in $L^p$ then $||f_n||_p$ is bounded (by the Uniform Boundedness Principle). Second, if $g\in L^q$ then $\sum\int_n^{n+1}|g|^q<\infty$, hence $\int_n^{n+1}|g|^q\to0$.

Answer 2

The case $\alpha < 0$ is ok.

The next calculation is also ok, but not the conclusion (as pointed out by Arctic chair). However, if you replace your $g$ by $g_n$ and observe that $g_n$ has constant norm in $L^q$, you find that the norm of $f_n$ has to be unbounded in $L^p$ if $\alpha > 0$ (note that $n^\alpha \equiv 1$ for $\alpha = 0$).

For $\alpha = 0$ you need a different treatment.