The strategy is to use the [integral formula](/theorems/929) to obtain a representative $(\hat{u})_{\mathrm{rep}}$ of the distributional Fourier transform $\hat{u}$, show that $(\hat{u})_{\mathrm{rep}} \in L^1(\mathbb{R}^n)$ via Cauchy–Schwarz (this is where $s > n/2$ enters), and then define $u_{\mathrm{rep}}$ as the inverse Fourier transform of the function $(\hat{u})_{\mathrm{rep}}$. Since the inverse Fourier transform of an $L^1$ function is bounded and continuous, this produces the desired representative.

**Step 1: The Fourier representative.** Let $u \in H^s(\mathbb{R}^n)$. By the [integral formula for the inhomogeneous Sobolev norm](/theorems/929), the Fourier transform $\hat{u} \in \mathcal{S}'(\mathbb{R}^n)$ is a regular distribution: there exists a unique $(\hat{u})_{\mathrm{rep}} \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ such that \begin{align*} \hat{u} = T_{(\hat{u})_{\mathrm{rep}}}, \end{align*} and the norm satisfies \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*}

[remark:Notation For The Representative] The notation $(\hat{u})_{\mathrm{rep}}$ denotes the $L^1_{\mathrm{loc}}$ function representing the tempered distribution $\hat{u}$. This is **not** the same as $\widehat{u_{\mathrm{rep}}}$ (the Fourier transform of some function $u_{\mathrm{rep}}$): at this stage, no function $u_{\mathrm{rep}}$ has been constructed, and the commutativity of hat and rep has not been established. The function $u_{\mathrm{rep}}$ will be defined in Step 2 as the inverse Fourier transform of $(\hat{u})_{\mathrm{rep}}$. [/remark]

[claim: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*} The second factor is finite by hypothesis. It remains to show $C_{n,s} < \infty$ when $s > n/2$.

Since the integrand $\xi \mapsto (1 + |\xi|^2)^{-s}$ is radial, apply the [coarea formula](/theorems/24) with $u(\xi) = |\xi|$ (which satisfies $|\nabla u| = 1$ on $\mathbb{R}^n \setminus \{0\}$, and $\{0\}$ has $\mathcal{L}^n$-measure zero): \begin{align*} \int_{\mathbb{R}^n} (1 + |\xi|^2)^{-s}\, d\mathcal{L}^n(\xi) &= \int_0^\infty \int_{\partial B(0, r)} (1 + |\xi|^2)^{-s}\, d\mathcal{H}^{n-1}(\xi)\, d\mathcal{L}^1(r). \end{align*} On the sphere $\partial B(0, r)$, every point $\xi$ satisfies $|\xi| = r$, so the integrand $(1 + |\xi|^2)^{-s}$ takes the constant value $(1 + r^2)^{-s}$. The inner integral therefore evaluates as \begin{align*} \int_{\partial B(0, r)} (1 + |\xi|^2)^{-s}\, d\mathcal{H}^{n-1}(\xi) = (1 + r^2)^{-s} \cdot \mathcal{H}^{n-1}(\partial B(0, r)). \end{align*} By the [surface area formula](/theorems/871), $\mathcal{H}^{n-1}(\partial B(0, r)) = n\omega_n r^{n-1}$. Substituting into the outer integral: \begin{align*} \int_{\mathbb{R}^n} (1 + |\xi|^2)^{-s}\, d\mathcal{L}^n(\xi) &= \int_0^\infty (1 + r^2)^{-s} \cdot n\omega_n r^{n-1}\, d\mathcal{L}^1(r) = n\omega_n \int_0^\infty (1 + r^2)^{-s}\, r^{n-1}\, d\mathcal{L}^1(r). \end{align*} Split the integral at $r = 1$. On $[0, 1]$, the integrand is bounded by $1$, so $\int_0^1 (1 + r^2)^{-s} r^{n-1}\, d\mathcal{L}^1(r) \le \int_0^1 r^{n-1}\, d\mathcal{L}^1(r) = 1/n < \infty$. On $[1, \infty)$, using $(1 + r^2)^{-s} \le r^{-2s}$ (since $1 + r^2 \ge r^2$): \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) = \frac{r^{n-2s}}{n - 2s}\bigg|_1^\infty, \end{align*} which converges if and only if $n - 2s < 0$, i.e. $s > n/2$. Hence $C_{n,s} < \infty$ precisely when $s > n/2$. [/proof]

**Step 2: Constructing a candidate representative.** Since $(\hat{u})_{\mathrm{rep}} \in L^1(\mathbb{R}^n)$, its inverse Fourier transform is well-defined as a pointwise integral. 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*}

[claim:Boundedness And Continuity Of The Candidate] The function $v$ belongs to $L^\infty(\mathbb{R}^n) \cap C^0(\mathbb{R}^n)$, with \begin{align*} \|v\|_{L^\infty} \le (2\pi)^{-n/2}\, C_{n,s}\, \|u\|_{H^s}. \end{align*} [/claim]

[proof] **Boundedness:** For every $x \in \mathbb{R}^n$, $|e^{ix \cdot \xi}| = 1$, so \begin{align*} |v(x)| \le (2\pi)^{-n/2}\int_{\mathbb{R}^n} |(\hat{u})_{\mathrm{rep}}(\xi)|\, d\mathcal{L}^n(\xi) = (2\pi)^{-n/2}\|(\hat{u})_{\mathrm{rep}}\|_{L^1}. \end{align*} Taking the supremum over $x$ and applying the $L^1$ bound from the claim above gives the stated estimate.

**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 gives $v(x_k) \to v(x)$. [/proof]

**Step 3: Identification.** We verify that $u = T_v$ in $\mathcal{S}'(\mathbb{R}^n)$, i.e. 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 $\hat{u} = T_{(\hat{u})_{\mathrm{rep}}}$ (from Step 1), this gives \begin{align*} u(\phi) = T_{(\hat{u})_{\mathrm{rep}}}(\hat{\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) \\ &= \int_{\mathbb{R}^n} (\hat{u})_{\mathrm{rep}}(\xi)\, \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.