The inequality interpolates between two known bounds on $f$: the hypothesis $f \in L^2(\mathbb{R}^n)$ and the [Sobolev embedding](/theorems/225) $\|f\|_{L^{p^*}} \le C_{n,s}\,\|[f]\|_{\dot{H}^s}$ at the critical exponent $p^* := 2n/(n-2s)$. The strategy is to verify that the exponent relation $1/q = 1/2 - \theta s/n$ is equivalent to the convex combination $1/q = (1-\theta)/2 + \theta/p^*$, and then apply the [Hölder inequality](/theorems/516) to obtain the $L^q$ bound from the $L^2$ and $L^{p^*}$ bounds. **Step 1 (Endpoint cases).** When $q = 2$, the relation $1/q = 1/2 - \theta s/n$ gives $\theta = 0$, and the inequality reduces to $\|f\|_{L^2} \le C\,\|f\|_{L^2}$, which holds with $C = 1$. When $q = p^*$, the relation gives $1/p^* = 1/2 - \theta s/n = 1/2 - s/n = (n-2s)/(2n) = 1/p^*$, so $\theta = 1$. The inequality reduces to $\|f\|_{L^{p^*}} \le C\,\|[f]\|_{\dot{H}^s}$, which is the [Sobolev embedding for homogeneous spaces](/theorems/225). This theorem applies because $[f] \in \dot{H}^s(\mathbb{R}^n)$ and $0 \le s < n/2$. **Step 2 (Interior case: $2 < q < p^*$).** Define $\theta := n(q-2)/(2sq) \in (0,1)$. The exponent $\theta$ lies in $(0,1)$ because $q > 2$ gives $\theta > 0$ and $q < p^* = 2n/(n-2s)$ gives $\theta < 1$ (since $q < 2n/(n-2s)$ rearranges to $n(q-2) < 2sq$). [claim:Exponent Identity] The exponent $\theta = n(q-2)/(2sq)$ satisfies both the relation $1/q = 1/2 - \theta s/n$ stated in the theorem and the convex combination $1/q = (1-\theta)/2 + \theta/p^*$. [/claim] [proof] For the first relation: \begin{align*} \frac{1}{2} - \frac{\theta s}{n} = \frac{1}{2} - \frac{s}{n}\cdot\frac{n(q-2)}{2sq} = \frac{1}{2} - \frac{q-2}{2q} = \frac{q - (q-2)}{2q} = \frac{1}{q}. \end{align*} For the second, using $1/p^* = 1/2 - s/n$: \begin{align*} \frac{1-\theta}{2} + \frac{\theta}{p^*} &= \frac{1-\theta}{2} + \theta\!\left(\frac{1}{2} - \frac{s}{n}\right) = \frac{1}{2} - \frac{\theta s}{n} = \frac{1}{q}, \end{align*} where the last equality is the relation just verified. [/proof] [claim:Interpolation Bound] Let $1 \le r_0 < r_1 \le \infty$, $\eta \in (0,1)$, and $1/r = (1-\eta)/r_0 + \eta/r_1$. If $g \in L^{r_0}(\mathbb{R}^n) \cap L^{r_1}(\mathbb{R}^n)$, then $g \in L^r(\mathbb{R}^n)$ and \begin{align*} \|g\|_{L^r} \le \|g\|_{L^{r_0}}^{1-\eta}\,\|g\|_{L^{r_1}}^{\eta}. \end{align*} [/claim] [proof] Write $|g|^r = |g|^{r(1-\eta)} \cdot |g|^{r\eta}$. The relation $1/r = (1-\eta)/r_0 + \eta/r_1$ rearranges to \begin{align*} \frac{r(1-\eta)}{r_0} + \frac{r\eta}{r_1} = 1, \end{align*} so the exponents $r_0/(r(1-\eta))$ and $r_1/(r\eta)$ are conjugate. The [Hölder inequality](/theorems/516) applied with these exponents gives \begin{align*} \int_{\mathbb{R}^n} |g|^r\,d\mathcal{L}^n &\le \left(\int_{\mathbb{R}^n} |g|^{r_0}\,d\mathcal{L}^n\right)^{r(1-\eta)/r_0} \left(\int_{\mathbb{R}^n} |g|^{r_1}\,d\mathcal{L}^n\right)^{r\eta/r_1}, \end{align*} which is $\|g\|_{L^r}^r \le \|g\|_{L^{r_0}}^{r(1-\eta)}\,\|g\|_{L^{r_1}}^{r\eta}$. Taking $r$-th roots yields the claim. [/proof] **Step 3 (Conclusion).** The hypothesis $f \in L^2(\mathbb{R}^n)$ and the [Sobolev embedding](/theorems/225) $f \in L^{p^*}(\mathbb{R}^n)$ (established in Step 1) provide the two [integrability](/page/Integral) conditions required by the Interpolation Bound. The Exponent Identity gives $1/q = (1-\theta)/2 + \theta/p^*$, so applying the Interpolation Bound with $r_0 = 2$, $r_1 = p^*$, $\eta = \theta$: \begin{align*} \|f\|_{L^q} \le \|f\|_{L^2}^{1-\theta}\,\|f\|_{L^{p^*}}^{\theta} \le C_{n,s}^{\theta}\,\|[f]\|_{\dot{H}^s}^{\theta}\,\|f\|_{L^2}^{1-\theta}. \end{align*} Setting $C := C_{n,s}^{\theta}$ (which depends only on $n$, $s$, and $q$ through $\theta$) completes the proof.