[proofplan]
We prove the Sobolev embedding $\dot{H}^s(\mathbb{R}^n) \hookrightarrow L^p(\mathbb{R}^n)$ for $p = 2n/(n-2s)$ using a frequency decomposition and the layer cake representation, avoiding Littlewood--Paley theory. The low-frequency part $f_{\le N}$ is bounded in $L^\infty$ by Cauchy--Schwarz (exploiting $s < n/2$), while the high-frequency part $f_{>N}$ is controlled in $L^2$ by Plancherel. Choosing $N$ as a function of $\lambda$, we apply Chebyshev's inequality, swap integration order via Tonelli, and use the critical identity $\alpha(p-2) = 2s$ (where $\alpha = n/2 - s$) to close the $L^p$ estimate.
[/proofplan]
[step:Handle the case $s = 0$ and set up notation]
The case $s = 0$ gives $p = 2$ and the embedding $\dot{H}^0 \cong L^2 \hookrightarrow L^2$ is the identity. 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]
[step:Decompose into low and high frequencies and bound $\|f_{\le N}\|_{L^\infty}$]
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 $|\xi|^{-s} \in L^2(B(0,N))$ 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 $L^\infty$ 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 the Cauchy--Schwarz inequality:
\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 with $\omega_n$ the volume of the unit ball:
\begin{align*}
\int_{|\xi| \le N} |\xi|^{-2s}\,d\mathcal{L}^n(\xi) = \frac{n\omega_n}{n - 2s}\,N^{n-2s} = \frac{n\omega_n}{n - 2s}\,N^{2\alpha},
\end{align*}
where $n - 2s = 2\alpha > 0$ ensures convergence. Absorbing all constants into $C_1$ gives the claim.
[/proof]
[/step]
[step:Apply the layer cake representation with Tonelli to estimate $\|f\|_{L^p}^p$]
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's inequality 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 $\|f\|_{L^p}^p = p\int_0^\infty \lambda^{p-1}\,\mathcal{L}^n(\{|f| > \lambda\})\,d\mathcal{L}^1(\lambda)$:
\begin{align*}
\|f\|_{L^p}^p &\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*}
[guided]
The Tonelli swap is the heart of the argument.
Before the swap, we integrate first in $\xi$ for each fixed $\lambda$, computing the $L^2$ tail of $\hat{f}_{T\text{-rep}}$ beyond frequency $N(\lambda)$.
After the swap, we integrate first in $\lambda$ for each fixed $\xi$.
The condition $|\xi| > N(\lambda)$ defines a region in the $(\xi, \lambda)$ half-plane.
To invert, recall $N(\lambda) = (\lambda/(2C_1\|[f]\|_{\dot{H}^s}))^{1/\alpha}$, so $|\xi| > N(\lambda)$ is equivalent to $\lambda < 2C_1\|[f]\|_{\dot{H}^s}|\xi|^\alpha =: \Lambda_\xi$.
After the swap:
\begin{align*}
\|f\|_{L^p}^p \leq 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*}
The inner $\lambda$-integral is elementary since $p > 2$ ensures $p - 3 > -1$:
\begin{align*}
\int_0^{\Lambda_\xi}\lambda^{p-3}\,d\mathcal{L}^1(\lambda) = \frac{\Lambda_\xi^{p-2}}{p-2}.
\end{align*}
Since $\Lambda_\xi^{p-2} \propto |\xi|^{\alpha(p-2)}$, the outer integral becomes $\int |\xi|^{\alpha(p-2)}|\hat{f}_{T\text{-rep}}(\xi)|^2\,d\mathcal{L}^n(\xi)$.
For this to reproduce the $\dot{H}^s$ norm $\int|\xi|^{2s}|\hat{f}_{T\text{-rep}}|^2\,d\mathcal{L}^n$, we need $\alpha(p-2) = 2s$.
This is precisely the critical identity consumed in the next step.
[/guided]
[/step]
[step:Close the estimate using the critical identity $\alpha(p-2) = 2s$]
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 $p - 2 = \frac{2n}{n-2s} - 2 = \frac{4s}{n-2s}$. 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 $C_{n,s} = (C_{n,s}')^{1/p}$.
[/step]
