For which $L^p[0,1]$, $1\le p < \infty$ does the function $n^\alpha*\chi_{[0,1/n]}$ converge weakly to 0?

Question

My hypothesis is that this function converges to 0 weakly iff $ p < 1/\alpha$, but I am not sure how to prove this. We are working in the space $[0,1]$ with the Borel sets and Lebesgue measure. I can fairly easily show that it does weakly converge when $p < 1/\alpha$ but am not sure how to show the reverse direction. Help?

Answer 1

Let $f_n:=n^\alpha \chi_{[0,1/n]}$. We have $$\lVert f_n\rVert_p=n^{\alpha-\frac 1p}$$ hence if $\alpha <\frac 1p$, there is convergence in norm to $0$ hence weak convergence, and if $\alpha>1/p$, the sequence $\lVert f_n\rVert_p$ is not bounded in $\mathbb L^p$, hence it cannot be weakly convergent.

It remains to deal with the case $\alpha=1/p$. Recall that if $1\leqslant p<\infty$, a linear functional can be represented by a function in $\mathbb L^q$ where $\frac 1p+\frac 1q=1$. By an approximation argument, we deal with the case $g=\chi_B$ , where $B\subset [0,1]$ and we get weak convergence to $0$ when $p\gt 1$.

When $p=1=\alpha$, things are different, since $\int f_n\cdot 1=1$.