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Since $\phi$ is $C^\infty_c$ we know that $\widehat{\phi} \in \mathcal{S}(\mathbb{R})$, in particular $\widehat{\phi}(\xi) \lesssim (1+|\xi|)^{-100}$. On the other hand $\widehat{\phi_k}(\xi) = 2^k \widehat{\phi}(\xi 2^k)$ which by our previous estimate satisfies $|\widehat{\phi_k}(\xi)| \lesssim 2^k (1+|2^k \xi|)^{-100}$. Now, $\widehat{G_k} = G \ast \widehat{\phi_k}$ so by Cauchy-Schwartz Cauchy-Schwarz we can estimate the $L^\infty$ norm of $\widehat{G_k}$ by $\lesssim 2^{-90k} \|G\|_{L^2}$. So by the dominated convergence theorem $\|\widehat{G_k}\|_{L^1} \to 0$ as $k \to \infty$.
Since $\phi$ is $C^\infty_c$ we know that $\widehat{\phi} \in \mathcal{S}(\mathbb{R})$, in particular $\widehat{\phi}(\xi) \lesssim (1+|\xi|)^{-100}$. On the other hand $\widehat{\phi_k}(\xi) = 2^k \widehat{\phi}(\xi 2^k)$ which by our previous estimate satisfies $|\widehat{\phi_k}(\xi)| \lesssim 2^k (1+|2^k \xi|)^{-100}$. Now, $\widehat{G_k} = G \ast \widehat{\phi_k}$ so by Cauchy-Schwartz we can estimate the $L^\infty$ norm of $\widehat{G_k}$ by $\lesssim 2^{-90k} \|G\|_{L^1}$. |G\|_{L^2}$. So by the dominated convergence theorem$\|\widehat{G_k}\|_{L^1} \to 0$as$k \to \infty$. 1 Since$\phi$is$C^\infty_c$we know that$\widehat{\phi} \in \mathcal{S}(\mathbb{R})$, in particular$\widehat{\phi}(\xi) \lesssim (1+|\xi|)^{-100}$. On the other hand$\widehat{\phi_k}(\xi) = 2^k \widehat{\phi}(\xi 2^k)$which by our previous estimate satisfies$|\widehat{\phi_k}(\xi)| \lesssim 2^k (1+|2^k \xi|)^{-100}$. Now,$\widehat{G_k} = G \ast \widehat{\phi_k}$so we can estimate the$L^\infty$norm of$\widehat{G_k}$by$\lesssim 2^{-90k} \|G\|_{L^1}$. So by the dominated convergence theorem$\|\widehat{G_k}\|_{L^1} \to 0$as$k \to \infty\$.