Approximate Identity

Question

Let ${({\varphi}_{n}})_{n=1}^{\infty}$ an Approximate Identity in Schwartz Space. Let $\alpha \in \mathbb{Z}^+$. Is it true or not the following statement? \begin{equation} \lim _{ n\longrightarrow \infty }{ \int _{ \mathbb{R} }^{ }{ { \left| y \right| }^{ \alpha }\left| { \varphi }_{ n }\left( y \right) \right| } } dy=0. \end{equation}

Answer 1

It is false. Let $\psi:\mathbb R \to [0,\infty)$ be a smooth function supported on $[-1,1]$ such that $\int_{\mathbb R} \psi = 1$, and suppose $s > 0$. Define $$ \varphi_n(x) = n (1-n^{-s}) \psi(nx) + n^{-s} \psi(x-n-1) .$$ It is easily seen to be an approximate identity. But $$ \int_{\mathbb R} |y|^\alpha |\varphi_n(y)| \, dy \ge \int_{n}^{n+2} y^\alpha n^{-s} \psi(y-n-1) \, dy \ge n^{-s} n^\alpha ,$$ and this diverges if $\alpha > s$.