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The Fourier transform $\hat{f}$ of $f$ is in $L^2$ for the weight $(1+|x|^2)^s$. Since $s>n/2$, the constant function $1$ is in that weighted $L^2$ space, so by Cauchy-Schwarz-Bunyakowsky, $\int_{\mathbb R^n} |\hat{f}| = \int |\hat{f}|\cdot 1\le |\hat{f}|s \hat{f}|_s \cdot |1|{-s}$, 1|_t$, where$t=-s$[to avoid TeX bug?] and the subscript denotes the weight. Thus,$\hat{f}$is in$L^1$. Thus, by Fourier inversion and Riemann-Lebesgue,$f=\hat{\hat{f}}$is uniformly continuous and goes to$0$at infinity. [Dangit, TeX is broken?] 3 deleted 4 characters in body; added 28 characters in body This is very standard, but perhaps buried in fancier things: The Fourier transform$\hat{f}$of$f$is in$L^2$for the weight$(1+|x|^2)^s$. Since$s>n/2$, the constant function$1$is in that weighted$L^2$space, so by Cauchy Schwarz BunyakowskyCauchy-Schwarz-Bunyakowsky, $$\int_{\mathbb \int_{\mathbb R^n} |\hat{f}| = \int |\hat{f}|\cdot 1\le |\hat{f}|s |1|{-s}$$, , where the subscript denotes the weight. Thus,$\hat{f}$is in$L^1$. Thus, by Fourier inversion and Riemann-Lebesgue,$f=\hat{\hat{f}}$is uniformly continuous and goes to$0$at infinity. [Dangit, TeX is broken?] 2 added 3 characters in body; deleted 7 characters in body This is very standard, but perhaps buried in fancier things: The Fourier transform$\hat{f}$of$f$is in$L^2$for the weight$(1+|x|^2)^s$. Since$s>n/2$, the constant function$1$is in that weighted$L^2$space, so by Cauchy Schwartz Schwarz Bunyakowsky,$\int_{\mathbb $\int_{\mathbb R^n} |\hat{f}| \;=\; = \int |\hat{f}|\cdot 1 \le 1\le ||\hat{f}||_s\cdot \hat{f}|s ||1||_s$ 1|{-s} , where the subscript denotes the weight. Thus, $\hat{f}$ is in $L^1$. Thus, by Fourier inversion and Riemann-Lebesgue, $f=\hat{\hat{f}}$ is uniformly continuous and goes to $0$ at infinity.