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Let $s$ be a non-negative integer and $$H^{s}(\mathbb{R}^{n}) = \{f \in L^{2}(\mathbb{R}^{n}) : \frac{\partial^{\alpha}}{\partial x^{\alpha}}f \in L^{2}(\mathbb{R}^{n})\text{ for all $\alpha$ with $|\alpha| = \alpha_{1} + \cdots + \alpha_{n} \leq s$}\}$$ where the derivatives are to be interpreted in the sense of distributions. If $s > \frac{n}{2}$, to show that each $u \in H^{s}(\mathbb{R}^{n})$ is bounded and continuous, why does it suffice to show that $\widehat{u}(\xi) \in L^{1}(\mathbb{R}^{n})$?

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By the Fourier inversion formula, $$u(x) = \int_{\mathbb{R}^n} \hat{u}(\xi) e^{i x \cdot \xi} ~d\xi,$$ from which it certainly follows that $u(x)$ is both bounded and continuous if $\hat{u} \in L^1(\mathbb{R}^n)$ (note that $\hat{u}$ is automatically continuous). By the Cauchy-Schwarz inequality, $$\int_{\mathbb{R}^n} |\hat{u}(\xi)| ~d\xi \leq \left( \int_{\mathbb{R}^n} (1 + |\xi|)^s |\hat{u}(\xi)|^2 ~d\xi\right)^{1/2} \left( \int_{\mathbb{R}^n} (1 + |\xi|)^{-s} ~d\xi \right)^{1/2} < \infty$$ when $s > n/2$, so the result follows.

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In fact $u$ is not only bounded but vanishes at infinity, thanks to the Riemann-Lebesgue lemma. –  Nate Eldredge May 1 '12 at 3:15

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