The case $s = 0$ gives $p = 2$ and the embedding $\dot{H}^0 \cong L^2 \hookrightarrow L^2$ is trivial. For the remainder, assume $s > 0$.

Since $s < n/2$, the [functional realization theorem](/theorems/224) provides a canonical representative $f \in \mathcal{S}'(\mathbb{R}^n)$ of the class $[f]$, and $\hat{f}_{T\text{-rep}}$ extends to a function in $L^1_{\mathrm{loc}}(\mathbb{R}^n)$. The hypothesis $[f] \in \dot{H}^s$ means $|\xi|^s \hat{f}_{T\text{-rep}} \in L^2(\mathbb{R}^n)$.

Write $\alpha := n/2 - s > 0$, so that $p = n/\alpha$.

**Step 1 (Frequency decomposition).** Fix $N > 0$. Define $f_{\le N}$ and $f_{>N}$ by \begin{align*} \hat{f}_{\le N} := \mathbb{1}_{B(0,N)}\,\hat{f}_{T\text{-rep}}, \qquad \hat{f}_{>N} := \mathbb{1}_{B(0,N)^c}\,\hat{f}_{T\text{-rep}}. \end{align*} Since $|\xi|^s \hat{f}_{T\text{-rep}} \in L^2$, the function $\hat{f}_{>N}$ lies in $L^2$ (as $|\xi|^{-s}$ is bounded on $\{|\xi| > N\}$). The function $\hat{f}_{\le N}$ is compactly supported and lies in $L^2$ (since $|\xi|^{-s}$ is in $L^2(B(0,N))$ when $s < n/2$), hence also in $L^1$. By Plancherel, $f_{>N} \in L^2(\mathbb{R}^n)$, and the Fourier inversion formula applies to $f_{\le N}$, yielding a bounded continuous function.

[claim:Low Frequency Bound] $\|f_{\le N}\|_{L^\infty} \le C_1\,N^\alpha\,\|[f]\|_{\dot{H}^s}$, where $C_1 > 0$ depends only on $n$ and $s$. [/claim]

[proof] By the Fourier inversion formula and Cauchy–Schwarz: \begin{align*} \|f_{\le N}\|_{L^\infty} &\le (2\pi)^{-n}\int_{|\xi| \le N} |\hat{f}_{T\text{-rep}}(\xi)|\,d\mathcal{L}^n(\xi) \\ &\le (2\pi)^{-n}\left(\int_{|\xi| \le N} |\xi|^{-2s}\,d\mathcal{L}^n(\xi)\right)^{1/2}\|[f]\|_{\dot{H}^s}. \end{align*} Switching to polar coordinates: \begin{align*} \int_{|\xi| \le N} |\xi|^{-2s}\,d\mathcal{L}^n(\xi) = \frac{n\omega_n}{n - 2s}\,N^{2\alpha}, \end{align*} where $\omega_n$ is the volume of the unit ball and the integral converges because $n - 2s > 0$. Absorbing all constants into $C_1$ gives the claim. [/proof]

**Step 2 (Layer cake with Fubini).** We show $\|f\|_{L^p}^p \le C\,\|[f]\|_{\dot{H}^s}^p$.

For each $\lambda > 0$, choose $N(\lambda)$ so that $C_1\,N(\lambda)^\alpha\,\|[f]\|_{\dot{H}^s} = \lambda/2$: \begin{align*} N(\lambda) := \left(\frac{\lambda}{2C_1\,\|[f]\|_{\dot{H}^s}}\right)^{1/\alpha}. \end{align*} Then $\|f_{\le N(\lambda)}\|_{L^\infty} \le \lambda/2$, so whenever $|f(x)| > \lambda$, the triangle inequality gives $|f_{>N(\lambda)}(x)| > \lambda/2$. By Chebyshev and Plancherel: \begin{align*} \mathcal{L}^n(\{|f| > \lambda\}) &\le \frac{4}{\lambda^2}\,\|f_{>N(\lambda)}\|_{L^2}^2 = \frac{4(2\pi)^{-n}}{\lambda^2}\int_{|\xi| > N(\lambda)} |\hat{f}_{T\text{-rep}}(\xi)|^2\,d\mathcal{L}^n(\xi). \end{align*} Substituting into the layer cake representation: \begin{align*} \|f\|_{L^p}^p &= p\int_0^\infty \lambda^{p-1}\,\mathcal{L}^n(\{|f| > \lambda\})\,d\mathcal{L}^1(\lambda) \\ &\le 4p(2\pi)^{-n}\int_0^\infty \lambda^{p-3}\int_{|\xi| > N(\lambda)} |\hat{f}_{T\text{-rep}}(\xi)|^2\,d\mathcal{L}^n(\xi)\,d\mathcal{L}^1(\lambda). \end{align*} The integrand is non-negative, so Tonelli's theorem justifies swapping the order of integration. The condition $|\xi| > N(\lambda)$ is equivalent to $\lambda < \Lambda_\xi$, where \begin{align*} \Lambda_\xi := 2C_1\,\|[f]\|_{\dot{H}^s}\,|\xi|^\alpha \end{align*} is obtained by inverting $N(\Lambda_\xi) = |\xi|$. After the swap: \begin{align*} \|f\|_{L^p}^p &\le 4p(2\pi)^{-n}\int_{\mathbb{R}^n} |\hat{f}_{T\text{-rep}}(\xi)|^2 \int_0^{\Lambda_\xi} \lambda^{p-3}\,d\mathcal{L}^1(\lambda)\,d\mathcal{L}^n(\xi). \end{align*} Since $p > 2$ (because $s > 0$), the exponent $p - 3 > -1$ and the inner integral converges: \begin{align*} \int_0^{\Lambda_\xi} \lambda^{p-3}\,d\mathcal{L}^1(\lambda) = \frac{\Lambda_\xi^{p-2}}{p - 2}. \end{align*} Substituting $\Lambda_\xi^{p-2} = (2C_1)^{p-2}\,\|[f]\|_{\dot{H}^s}^{p-2}\,|\xi|^{\alpha(p-2)}$, the key identity is \begin{align*} \alpha(p - 2) = \frac{n - 2s}{2}\cdot\frac{4s}{n - 2s} = 2s, \end{align*} using $p - 2 = 4s/(n-2s)$. Therefore \begin{align*} \|f\|_{L^p}^p &\le \frac{4p(2\pi)^{-n}(2C_1)^{p-2}}{p-2}\,\|[f]\|_{\dot{H}^s}^{p-2}\int_{\mathbb{R}^n} |\xi|^{2s}\,|\hat{f}_{T\text{-rep}}(\xi)|^2\,d\mathcal{L}^n(\xi) \\ &= C_{n,s}'\,\|[f]\|_{\dot{H}^s}^{p-2}\,\|[f]\|_{\dot{H}^s}^2 = C_{n,s}'\,\|[f]\|_{\dot{H}^s}^p. \end{align*} Taking $p$-th roots gives $\|f\|_{L^p} \le C_{n,s}\,\|[f]\|_{\dot{H}^s}$ with $C_{n,s} = (C_{n,s}')^{1/p}$.