[proofplan] The extension theorem is the Banach-space adjoint of the [Stein--Tomas Restriction Theorem](/theorems/3202). We define the restriction operator $T : L^p(\mathbb{R}^n) \to L^2(S^{n-1}, d\sigma)$ at the endpoint $p_0 = 2(n+1)/(n+3)$, identify its Hilbert--Banach adjoint $T^* : L^2(S^{n-1}, d\sigma) \to L^{p_0'}(\mathbb{R}^n)$, and verify that $T^*$ is precisely the extension operator $\mathcal{E}$ when restricted to a dense subspace of $L^2(S^{n-1}, d\sigma)$. The general norm identity $\|T\|_{X \to H} = \|T^*\|_{H \to X^*}$ for bounded linear maps from a Banach space $X$ to a Hilbert space $H$ then converts the restriction estimate into the extension estimate at $p_0$. The full range $p' \ge 2(n+1)/(n-1)$ follows by interpolation between this endpoint and the elementary $L^2(d\sigma) \to L^\infty(\mathbb{R}^n)$ bound. [/proofplan] [step:Set up the dual operators and dense subspaces] We use the symmetric normalisation throughout: \begin{align*} \hat f : \mathbb{R}^n &\to \mathbb{C} \\ \xi &\mapsto \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n} f(x)\, e^{-i\xi \cdot x}\, d\mathcal{L}^n(x), \qquad f \in L^1(\mathbb{R}^n), \end{align*} extended to $\mathcal{S}'(\mathbb{R}^n)$ by duality. By the [Stein--Tomas Restriction Theorem](/theorems/3202), the restriction map \begin{align*} T_0 : \mathcal{S}(\mathbb{R}^n) &\to L^2(S^{n-1}, d\sigma) \\ f &\mapsto \hat f|_{S^{n-1}} \end{align*} extends to a bounded linear operator $T : L^p(\mathbb{R}^n) \to L^2(S^{n-1}, d\sigma)$ for every $1 \le p \le p_0 := 2(n+1)/(n+3)$, with \begin{align*} \|T f\|_{L^2(S^{n-1}, d\sigma)} \le C_n\, \|f\|_{L^p(\mathbb{R}^n)}, \qquad C_n = C_n(p) > 0. \end{align*} The space $L^p(\mathbb{R}^n)$ is reflexive for $1 < p < \infty$ with dual $L^{p'}(\mathbb{R}^n)$ under the standard pairing \begin{align*} \langle f, h \rangle := \int_{\mathbb{R}^n} f(x)\, \overline{h(x)}\, d\mathcal{L}^n(x), \qquad f \in L^p, \quad h \in L^{p'}. \end{align*} The space $L^2(S^{n-1}, d\sigma)$ is a Hilbert space under the inner product \begin{align*} (\phi, \psi)_{L^2(d\sigma)} := \int_{S^{n-1}} \phi(\omega)\, \overline{\psi(\omega)}\, d\sigma(\omega). \end{align*} Since $T : L^p \to L^2(d\sigma)$ is bounded between a reflexive Banach space and a Hilbert space, the Hilbert-Banach adjoint \begin{align*} T^* : L^2(S^{n-1}, d\sigma) &\to L^{p'}(\mathbb{R}^n) \end{align*} exists, is bounded, and satisfies $\|T^*\|_{L^2(d\sigma) \to L^{p'}} = \|T\|_{L^p \to L^2(d\sigma)} \le C_n$. [/step] [step:Identify $T^*$ as the extension operator $\mathcal{E}$ on a dense subspace] We claim $T^* g = \widehat{g\, d\sigma}$ for $g \in C^\infty(S^{n-1})$, identified with an element of $L^{p'}(\mathbb{R}^n)$. For $g \in C^\infty(S^{n-1})$, the measure $g\, d\sigma$ is a finite, compactly supported (and in particular tempered) measure on $\mathbb{R}^n$. Its Fourier transform \begin{align*} \widehat{g\, d\sigma} : \mathbb{R}^n &\to \mathbb{C} \\ x &\mapsto \int_{S^{n-1}} g(\omega)\, e^{i x \cdot \omega}\, d\sigma(\omega) \end{align*} is a smooth bounded function on $\mathbb{R}^n$ (smooth because differentiation under the integral is justified by uniform integrability on the compact set $S^{n-1}$, bounded by $|g|_\infty\, \sigma(S^{n-1}) < \infty$). For $f \in \mathcal{S}(\mathbb{R}^n)$, Fubini's theorem applies (the integrand $g(\omega)\, \overline{f(x)}\, e^{i x \cdot \omega}$ is integrable on $\mathbb{R}^n \times S^{n-1}$ because $f \in L^1(\mathbb{R}^n)$ and $g \in L^\infty(S^{n-1}, d\sigma)$ with $\sigma(S^{n-1}) < \infty$), so \begin{align*} (T_0 f, g)_{L^2(d\sigma)} &= \int_{S^{n-1}} \hat f(\omega)\, \overline{g(\omega)}\, d\sigma(\omega) \\ &= \int_{S^{n-1}} \!\left( \int_{\mathbb{R}^n} f(x)\, e^{-i\omega \cdot x}\, d\mathcal{L}^n(x) \right) \overline{g(\omega)}\, d\sigma(\omega) \\ &= \int_{\mathbb{R}^n} f(x)\, \overline{ \int_{S^{n-1}} g(\omega)\, e^{i\omega \cdot x}\, d\sigma(\omega)}\, d\mathcal{L}^n(x) \\ &= \int_{\mathbb{R}^n} f(x)\, \overline{\widehat{g\, d\sigma}(x)}\, d\mathcal{L}^n(x) = \langle f, \widehat{g\, d\sigma} \rangle. \end{align*} This identity says exactly $(T_0 f, g)_{L^2(d\sigma)} = \langle f, \widehat{g\, d\sigma} \rangle$ for all $f \in \mathcal{S}(\mathbb{R}^n)$ and all $g \in C^\infty(S^{n-1})$. By the definition of the Hilbert--Banach adjoint, this means $T^* g = \widehat{g\, d\sigma}$ as elements of $(L^p)^* = L^{p'}$, provided we know a priori that $\widehat{g\, d\sigma} \in L^{p'}$. The boundedness $T^* : L^2(d\sigma) \to L^{p'}$ guarantees that $T^* g \in L^{p'}(\mathbb{R}^n)$ for every $g \in L^2(d\sigma)$, in particular for $g \in C^\infty(S^{n-1}) \subset L^2(d\sigma)$. The Riesz representation of $T^* g$ as a bounded linear functional on $L^p(\mathbb{R}^n)$ via $(T_0 f, g)_{L^2(d\sigma)}$ matches the integral pairing $\langle f, \widehat{g\, d\sigma}\rangle$, and a Riesz representative is unique a.e., so $T^* g = \widehat{g\, d\sigma}$ a.e. on $\mathbb{R}^n$ for every $g \in C^\infty(S^{n-1})$. [/step] [step:Extend the identification $T^* g = \widehat{g\, d\sigma}$ to all $g \in L^2(d\sigma)$] The map \begin{align*} \mathcal{E} : L^2(S^{n-1}, d\sigma) &\to \mathcal{S}'(\mathbb{R}^n) \\ g &\mapsto \widehat{g\, d\sigma} \end{align*} is well-defined and continuous as a map into $\mathcal{S}'$ because $g \in L^2(S^{n-1}, d\sigma) \subset L^1(S^{n-1}, d\sigma)$ (since $\sigma(S^{n-1}) < \infty$, so $L^2 \hookrightarrow L^1$ on this finite measure space by Cauchy--Schwarz), so $g\, d\sigma$ is a finite tempered measure on $\mathbb{R}^n$ and its Fourier transform is a bounded continuous function on $\mathbb{R}^n$. Both maps $\mathcal{E}$ and $T^*$ are continuous from $L^2(S^{n-1}, d\sigma)$ into $\mathcal{S}'(\mathbb{R}^n)$ (the former with values in $C_b(\mathbb{R}^n) \hookrightarrow \mathcal{S}'$, the latter with values in $L^{p'}(\mathbb{R}^n) \hookrightarrow \mathcal{S}'$ by the standard embedding $h \mapsto T_h$). They agree on the dense subset $C^\infty(S^{n-1}) \subset L^2(S^{n-1}, d\sigma)$ by the previous step. Density and continuity of both into $\mathcal{S}'(\mathbb{R}^n)$ imply $T^* g = \mathcal{E} g$ in $\mathcal{S}'(\mathbb{R}^n)$ for every $g \in L^2(S^{n-1}, d\sigma)$. Since $T^* g \in L^{p'}(\mathbb{R}^n)$ by the boundedness of $T^*$, we obtain $\widehat{g\, d\sigma} = T^* g \in L^{p'}(\mathbb{R}^n)$ for every $g \in L^2(S^{n-1}, d\sigma)$, and \begin{align*} \|\widehat{g\, d\sigma}\|_{L^{p'}(\mathbb{R}^n)} = \|T^* g\|_{L^{p'}} \le \|T^*\|_{L^2(d\sigma) \to L^{p'}}\, \|g\|_{L^2(d\sigma)} \le C_n\, \|g\|_{L^2(d\sigma)}. \end{align*} [/step] [step:Establish the endpoint estimate at $p_0' = 2(n+1)/(n-1)$] Choose $p = p_0 = 2(n+1)/(n+3)$ in the previous step, so $p_0' = 2(n+1)/(n-1)$. The conclusion of the previous step at this exponent reads \begin{align*} \|\widehat{g\, d\sigma}\|_{L^{p_0'}(\mathbb{R}^n)} \le C(n)\, \|g\|_{L^2(S^{n-1}, d\sigma)} \qquad \text{for every } g \in L^2(S^{n-1}, d\sigma). \end{align*} This is the endpoint of the desired range $p' \ge p_0' = 2(n+1)/(n-1)$. [/step] [step:Extend to all $p' \ge 2(n+1)/(n-1)$ by interpolation with the elementary $L^\infty$ bound] For $g \in L^2(S^{n-1}, d\sigma) \subset L^1(S^{n-1}, d\sigma)$, the boundedness of the trigonometric kernel gives directly \begin{align*} \|\widehat{g\, d\sigma}\|_{L^\infty(\mathbb{R}^n)} = \sup_{x \in \mathbb{R}^n} \left| \int_{S^{n-1}} g(\omega)\, e^{i x \cdot \omega}\, d\sigma(\omega) \right| \le \|g\|_{L^1(S^{n-1}, d\sigma)} \le \sigma_n^{1/2}\, \|g\|_{L^2(S^{n-1}, d\sigma)}, \end{align*} where the last step uses Cauchy--Schwarz and $\sigma_n := \sigma(S^{n-1})$. Hence \begin{align*} \mathcal{E} : L^2(S^{n-1}, d\sigma) \to L^\infty(\mathbb{R}^n) \quad \text{with} \quad \|\mathcal{E}\|_{L^2(d\sigma) \to L^\infty} \le \sigma_n^{1/2}. \end{align*} Apply the [Riesz--Thorin Interpolation Theorem](/theorems/???) to the linear operator $\mathcal{E} : L^2(S^{n-1}, d\sigma) \to L^q(\mathbb{R}^n)$, with input space fixed at $L^2(S^{n-1}, d\sigma)$ and output exponents interpolated between $q_0 = p_0' = 2(n+1)/(n-1)$ and $q_1 = \infty$. Riesz--Thorin (in the form valid when only the target exponent varies, applicable because the operator is the same on a fixed input space) gives, for every $\theta \in [0, 1]$ and $q$ defined by $1/q = (1-\theta)/q_0 + \theta/\infty = (1-\theta)/q_0$, \begin{align*} \|\mathcal{E}\|_{L^2(d\sigma) \to L^q(\mathbb{R}^n)} \le \|\mathcal{E}\|_{L^2(d\sigma) \to L^{q_0}}^{1-\theta}\, \|\mathcal{E}\|_{L^2(d\sigma) \to L^\infty}^\theta \le C(n)^{1-\theta}\, \sigma_n^{\theta/2}. \end{align*} As $\theta$ ranges over $[0, 1]$, $q$ ranges over $[q_0, \infty]$. Hence \begin{align*} \|\widehat{g\, d\sigma}\|_{L^{p'}(\mathbb{R}^n)} \le C(n)\, \|g\|_{L^2(S^{n-1}, d\sigma)} \qquad \text{for every } p' \ge \tfrac{2(n+1)}{n-1} \text{ and every } g \in L^2(S^{n-1}, d\sigma), \end{align*} where the constant $C(n)$ may now depend on $p'$ but is uniformly bounded as $p'$ ranges over any compact subset of $[p_0', \infty]$, and depends only on $n$. [/step] [step:Conclude] Combining the endpoint estimate from the fifth step with the interpolation argument of the sixth step, the bound \begin{align*} \|\widehat{g\, d\sigma}\|_{L^{p'}(\mathbb{R}^n)} \le C\, \|g\|_{L^2(S^{n-1}, d\sigma)} \end{align*} holds for every $g \in L^2(S^{n-1}, d\sigma)$ and every $p' \ge 2(n+1)/(n-1)$, with $C = C(n, p') > 0$ depending only on $n$ and $p'$. This is the asserted extension estimate. [/step]