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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-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?]
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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?]
show/hide this revision's text 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.
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