Suppose we have a sequence of $L^1(\mathbb{R})$ functions $p_\epsilon$ with $\|p_\epsilon\|_{L^1} \leq 1$ for all $n$. Suppose we know that $p_\epsilon \to 0$ in distributions. Is it obvious that $\epsilon p_\epsilon(\epsilon x) \to 0$ in distributions as $\epsilon \to 0$ as well? It seems that oscillation and concentration are both not possible for $\epsilon p_\epsilon(\epsilon x)$.
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This is wrong. Take, for example, $p_\epsilon(x)=\frac1\epsilon \varphi(x/\epsilon)$ and $\varphi(x)=\varphi_0(x-1)-\varphi_0(x+1)$, where $\varphi_0$ is a smooth cap supported on $(-1,1)$. Then $$ \int \epsilon p_\epsilon(\epsilon x)\varphi(x)dx=\int \varphi(x)^2dx, $$ which does not tend to zero. On the other hand, all $p_\epsilon$ have constant $L_1$-norm, and $p_\epsilon/\epsilon$ converges to the derivative of the delta-function (which means that $p_\epsilon$ itself converges to zero).
