[proofplan] We show that every $u \in H^s(\mathbb{R}^n)$ with $s > n/2$ has a bounded continuous representative. The strategy is: (1) the Fourier representative $(\hat{u})_{\mathrm{rep}} \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ is shown to be globally $L^1$ via a Cauchy--Schwarz argument where the condition $s > n/2$ ensures the weight $(1+|\xi|^2)^{-s}$ is integrable; (2) the inverse Fourier transform of an $L^1$ function is bounded and continuous; (3) we identify this inverse Fourier transform with $u$ in $\mathcal{S}'(\mathbb{R}^n)$ using Fubini's theorem. [/proofplan] [step:Obtain the Fourier representative and prove it belongs to $L^1(\mathbb{R}^n)$] Let $u \in H^s(\mathbb{R}^n)$. By the integral formula for the inhomogeneous Sobolev norm, the Fourier transform $\hat{u} \in \mathcal{S}'(\mathbb{R}^n)$ is a [regular distribution](/page/Regular%20Distribution): there exists a unique $(\hat{u})_{\mathrm{rep}} \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ such that $\hat{u} = T_{(\hat{u})_{\mathrm{rep}}}$, and \begin{align*} \|u\|_{H^s}^2 = \int_{\mathbb{R}^n} (1 + |\xi|^2)^{s}\, |(\hat{u})_{\mathrm{rep}}(\xi)|^2\, d\mathcal{L}^n(\xi). \end{align*} [claim:$L^1$ integrability of the Fourier representative] If $s > n/2$, then $(\hat{u})_{\mathrm{rep}} \in L^1(\mathbb{R}^n)$ with $\|(\hat{u})_{\mathrm{rep}}\|_{L^1} \le C_{n,s}\|u\|_{H^s}$. [/claim] [proof] Insert the weight $(1 + |\xi|^2)^{s/2}$ and its reciprocal, then apply the Cauchy--Schwarz inequality: \begin{align*} \int_{\mathbb{R}^n} |(\hat{u})_{\mathrm{rep}}(\xi)|\, d\mathcal{L}^n(\xi) &= \int_{\mathbb{R}^n} (1 + |\xi|^2)^{-s/2} \cdot (1 + |\xi|^2)^{s/2}|(\hat{u})_{\mathrm{rep}}(\xi)|\, d\mathcal{L}^n(\xi) \\ &\le \underbrace{\left(\int_{\mathbb{R}^n} (1 + |\xi|^2)^{-s}\, d\mathcal{L}^n(\xi)\right)^{1/2}}_{=:\, C_{n,s}} \cdot \underbrace{\left(\int_{\mathbb{R}^n} (1 + |\xi|^2)^{s}\, |(\hat{u})_{\mathrm{rep}}(\xi)|^2\, d\mathcal{L}^n(\xi)\right)^{1/2}}_{=\, \|u\|_{H^s}}. \end{align*} It remains to show $C_{n,s} < \infty$ when $s > n/2$. Since the integrand is radial, apply the [coarea formula](/theorems/24) with the radial function $\xi \mapsto |\xi|$: \begin{align*} \int_{\mathbb{R}^n} (1 + |\xi|^2)^{-s}\, d\mathcal{L}^n(\xi) = n\omega_n \int_0^\infty (1 + r^2)^{-s}\, r^{n-1}\, d\mathcal{L}^1(r), \end{align*} where $\omega_n$ is the volume of the unit ball and $\mathcal{H}^{n-1}(\partial B(0, r)) = n\omega_n r^{n-1}$. On $[0, 1]$, the integrand is bounded by $r^{n-1}$, so $\int_0^1 (1 + r^2)^{-s} r^{n-1}\, d\mathcal{L}^1(r) \le 1/n < \infty$. On $[1, \infty)$, using $(1 + r^2)^{-s} \le r^{-2s}$: \begin{align*} \int_1^\infty (1 + r^2)^{-s}\, r^{n-1}\, d\mathcal{L}^1(r) \le \int_1^\infty r^{n-1-2s}\, d\mathcal{L}^1(r), \end{align*} which converges if and only if $n - 1 - 2s < -1$, i.e., $s > n/2$. [/proof] [guided] The Cauchy--Schwarz step inserts the weight $(1+|\xi|^2)^{s/2}$ and its reciprocal: \begin{align*} \int_{\mathbb{R}^n}|(\hat{u})_{\mathrm{rep}}(\xi)|\,d\mathcal{L}^n(\xi) &= \int_{\mathbb{R}^n}(1+|\xi|^2)^{-s/2}\cdot(1+|\xi|^2)^{s/2}|(\hat{u})_{\mathrm{rep}}(\xi)|\,d\mathcal{L}^n(\xi) \\ &\leq \underbrace{\left(\int_{\mathbb{R}^n}(1+|\xi|^2)^{-s}\,d\mathcal{L}^n(\xi)\right)^{1/2}}_{C_{n,s}} \cdot \|u\|_{H^s}. \end{align*} The role of $s > n/2$ is to ensure $C_{n,s} < \infty$. Switching to polar coordinates via the [coarea formula](/theorems/24): \begin{align*} \int_{\mathbb{R}^n}(1+|\xi|^2)^{-s}\,d\mathcal{L}^n(\xi) = n\omega_n\int_0^\infty (1+r^2)^{-s}\,r^{n-1}\,d\mathcal{L}^1(r). \end{align*} On $[0, 1]$, the integrand is bounded by $r^{n-1}$, contributing a finite amount. On $[1, \infty)$, using $(1+r^2)^{-s} \leq r^{-2s}$: \begin{align*} \int_1^\infty (1+r^2)^{-s}\,r^{n-1}\,d\mathcal{L}^1(r) \leq \int_1^\infty r^{n-1-2s}\,d\mathcal{L}^1(r), \end{align*} which converges if and only if $n - 1 - 2s < -1$, i.e., $s > n/2$. Near the origin, $(1+|\xi|^2)^{-s}$ is bounded (it equals $1$ at $\xi = 0$), so there is no singularity. This is the opposite regime from the homogeneous embedding (theorem 225), which required $s < n/2$ to control behaviour near the origin (where $|\xi|^{-2s}$ blows up). Here the inhomogeneous weight $(1+|\xi|^2)^{-s}$ tames both regimes: bounded at the origin, decaying at infinity. The resulting bound $\|(\hat{u})_{\mathrm{rep}}\|_{L^1} \leq C_{n,s}\|u\|_{H^s}$ is what enables us to define a pointwise representative via the inverse Fourier transform. [/guided] [/step] [step:Construct the bounded continuous representative via inverse Fourier transform] Since $(\hat{u})_{\mathrm{rep}} \in L^1(\mathbb{R}^n)$, define \begin{align*} v: \mathbb{R}^n &\to \mathbb{C} \\ x &\mapsto (2\pi)^{-n/2}\int_{\mathbb{R}^n} e^{ix \cdot \xi}\, (\hat{u})_{\mathrm{rep}}(\xi)\, d\mathcal{L}^n(\xi). \end{align*} **Boundedness.** For every $x \in \mathbb{R}^n$, $|e^{ix \cdot \xi}| = 1$, so \begin{align*} |v(x)| \le (2\pi)^{-n/2}\|(\hat{u})_{\mathrm{rep}}\|_{L^1} \le (2\pi)^{-n/2}\, C_{n,s}\, \|u\|_{H^s}. \end{align*} **Continuity.** Let $x_k \to x$ in $\mathbb{R}^n$. The integrand $e^{ix_k \cdot \xi}(\hat{u})_{\mathrm{rep}}(\xi) \to e^{ix \cdot \xi}(\hat{u})_{\mathrm{rep}}(\xi)$ pointwise for every $\xi$, and $|e^{ix_k \cdot \xi}(\hat{u})_{\mathrm{rep}}(\xi)| = |(\hat{u})_{\mathrm{rep}}(\xi)| \in L^1(\mathbb{R}^n)$. The [Dominated Convergence Theorem](/theorems/4) gives $v(x_k) \to v(x)$. Therefore $v \in L^\infty(\mathbb{R}^n) \cap C^0(\mathbb{R}^n)$. [/step] [step:Identify $u = T_v$ in $\mathcal{S}'(\mathbb{R}^n)$] We verify that for every $\phi \in \mathcal{S}(\mathbb{R}^n)$, \begin{align*} u(\phi) = \int_{\mathbb{R}^n} v(x)\, \phi(x)\, d\mathcal{L}^n(x). \end{align*} By definition of the distributional Fourier transform, $u(\phi) = \hat{u}(\hat{\phi})$. Since $\hat{u} = T_{(\hat{u})_{\mathrm{rep}}}$: \begin{align*} u(\phi) = \int_{\mathbb{R}^n} (\hat{u})_{\mathrm{rep}}(\xi)\, \hat{\phi}(\xi)\, d\mathcal{L}^n(\xi). \end{align*} On the other hand, substituting the definition of $v$ and applying Fubini's theorem (justified since $(\hat{u})_{\mathrm{rep}} \in L^1$ and $\phi \in \mathcal{S}$): \begin{align*} \int_{\mathbb{R}^n} v(x)\, \phi(x)\, d\mathcal{L}^n(x) &= (2\pi)^{-n/2}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} e^{ix \cdot \xi}\, (\hat{u})_{\mathrm{rep}}(\xi)\, \phi(x)\, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(x) \\ &= \int_{\mathbb{R}^n} (\hat{u})_{\mathrm{rep}}(\xi) \underbrace{\left((2\pi)^{-n/2}\int_{\mathbb{R}^n} e^{ix \cdot \xi}\, \phi(x)\, d\mathcal{L}^n(x)\right)}_{=\, \hat{\phi}(\xi)}\, d\mathcal{L}^n(\xi). \end{align*} The two expressions agree, so $u = T_v$ in $\mathcal{S}'(\mathbb{R}^n)$. Setting $u_{\mathrm{rep}} := v$ completes the proof. [/step]