Showing that this sequence converges weakly to $0$

Question

I was reading on weak convergence in $L^p$ spaces and an examples was provided, namely, $f_n=n^{1/p}g(nx)$ for $g\in L^p(R)$ and $1\leq p<\infty$ which converges weakly to $0$ but to nothing strongly. I know that for weak convergence one most show that $$\lim_{n\to\infty}\int_{R}f_nh\,d\mu=\int fh\,d\mu,\quad h\in L^{p'}(R)$$ where $p$ and $p'$ are Holder's conjugate, but I can't seem to show this. If $$\lim_{n\to\infty}\int_{R}f_nh\,d\mu=\lim_{n\to\infty}n^{1/p}\int_{R}g(nx)h(x)\,dx$$ now for $r>0$, let $g(nx)=g_1+g_2=\chi_{(-r,r)}g(nx)+\left(1-\chi_{(-r,r)}\right)g(nx)$, then $$\lim_{n\to\infty}n^{1/p}\int_{R}g(nx)h(x)\,dx=\lim_{n\to\infty}n^{1/p}\left(\int_Rg_1h(x)\,dx+\int_Rg_2h(x) \,dx\right)$$ By Holder Ineq., $$\int_R g_1h(x)\,dx=\int_{-r}^rg(nx)h(x)\,dx\leq\|g\|_1\|h\|_\infty$$ and similarly may be done with $g_2$ leading to $$\lim_{n\to\infty}n^{1/p}\int_{R}g(nx)h(x)\,dx\leq\lim_{n\to\infty}n^{1/p}2\|g\|_1\|h\|_\infty$$ which does not go to $0$. I believe my upperbound does not work. Any ideas in proving this?

Answer 1

The target function is that $f=0$, we are to show that \begin{align*} \left|\int f_{n}(x)h(x)\right|\rightarrow 0. \end{align*} For simplicity, I will slightly modify the notation in the question. I write $f_{n}(x)=n^{1/p}f(nx)$ for $f\in L^{p}$ and show that $f_{n}\rightarrow 0$ weakly.

Assume first that $f$ has compact support, say, $\text{supp}(f)\subseteq\{|x|\leq M\}$ for some $M>0$. Note that \begin{align*} &\left|\int f_{n}(x)h(x)dx\right|\\ &=\left|\int n^{1/p}f(nx)h(x)dx\right|\\ &=\left|\int_{|x|\leq M/n}n^{1/p}f(nx)h(x)dx\right|\\ &\leq\left(\int_{|x|\leq M/n}n|f(nx)|^{p}dx\right)^{1/p}\left(\int_{|x|\leq M/n}|h(x)|^{p'}dx\right)^{1/p'}\\ &=\|f\|_{L^{p}}\left(\int_{|x|\leq M/n}|h(x)|^{p'}dx\right)^{1/p'}\\ &\rightarrow 0, \end{align*} since $h\in L^{p'}$.

In general, since continuous functions with compact support are dense in $L^{p}$, we can pick a $\varphi$ such that $\|f-\varphi\|_{L^{p}}$ is small.

As before, denote that $\varphi_{n}(x)=n^{1/p}\varphi(nx)$.

As a result, \begin{align*} &\left|\int f_{n}(x)h(x)dx\right|\\ &\leq\left|\int(f_{n}(x)-\varphi_{n}(x))h(x)dx\right|+\left|\int\varphi_{n}(x)h(x)dx\right|\\ &\leq\|f_{n}-\varphi_{n}\|_{L^{p}}\|h\|_{L^{p'}}+\left|\int\varphi_{n}(x)h(x)dx\right|. \end{align*} We also note that \begin{align*} \|f_{n}-\varphi_{n}\|_{L^{p}}&=\left(\int|f_{n}(x)-\varphi_{n}(x)|^{p}dx\right)^{1/p}\\ &=\left(n\int|f(nx)-\varphi(nx)|^{p}dx\right)^{1/p}\\ &=\|f-\varphi\|_{L^{p}}, \end{align*} which is also small.