Let $f \in C^\infty_c(\mathbb{R})$ and for $1 < p < \infty$ define $$g_n(x) := n^{-1/p}f(x/n)$$ for all $x \in \mathbb{R}$. Show that $g_n \rightharpoonup 0$ in $L^p(\mathbb{R})$.
A sequence $(x_n)_{n \in \mathbb{N}}$ converges weakly to $x$ in a normed space $X$ if and only if $f(x_n) \to f(x)$ for all $f \in X^*$. By the identification $(L^p(\mathbb{R}))^* \cong L^q(\mathbb{R})$ where $q$ is the dual exponent to $p$, we can equivalently show that $$\int_\mathbb{R} h(x) g_n(x) dx \to 0$$ for all $h \in L^q(\mathbb{R})$. I started with thinking about the support of $g_n$. We have that $\text{supp}(h_n) \subseteq [-Mn,Mn]$ if $\text{supp}(f) \subseteq [-M,M]$ for some $M > 0$. This does not help much. Hence I tried to apply Hölder's inequality: $$\int_\mathbb{R} |h(x)|n^{-1/p}|f(x/n)|dx = \int_\mathbb{R}|h(ny)|n^{1-1/p}|f(y)|dy \leq ||h(n\cdot)n^{1-1/p}||_q ||f||_p = ||h||_q||f||_p$$ which does also not help much. Has anybody an idea how to show the above statement?
Since $f$ is continuous with compact support, there exists $R$ and $M$ such that for all $u\in\mathbb R$, $\left\lvert f\left(u\right)\right\rvert\leqslant M\mathbf 1_{\left[-R,R\right]}\left(u\right)$ hence $\left\lvert g_n(x)\right\rvert\leqslant Mn^{-1/p}\mathbf 1_{\left[-Rn,Rn\right]}\left(x\right)$.
Defining $G_n\colon x\mapsto n^{-1/p}\mathbf 1_{\left[-Rn,Rn\right]}\left(x\right)$, we have to show that $G_n\to 0$ weakly in $\mathbb L^p$. Observe that the $\mathbb L^p$-norm of $G_n$ can be bounded independently of $n$ hence it suffices to show that $$\lim_{n\to +\infty}\int_{\mathbb R}G_n(x)h(x)\mathrm dx=0$$ for all $h\in D$, where $D$ is a dense subset of $\mathbb L^q$. It is not hard to do it when $D$ is the set of smooth functions with compact support, or when $D$ is the set of linear combinations of indicator functions of sets of finite Lebesgue measure.