The strategy is to express all three norms in terms of the [Fourier transform](/page/Fourier%20Transform), decompose the $H^r$ integrand as a product of two factors corresponding to $L^2$ and $H^s$, and apply Hölder's inequality. **Step 1: Fourier representation of the norms.** By the [Parseval Identity](/theorems/248), for any $\sigma \ge 0$: \begin{align*} \|f\|_{H^\sigma}^2 &= \|\langle\nabla\rangle^\sigma f\|_{L^2}^2 = \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} \langle\xi\rangle^{2\sigma}\,|\hat{f}(\xi)|^2\,d\mathcal{L}^n(\xi), \end{align*} where $\langle\xi\rangle := (1 + |\xi|^2)^{1/2}$ and $\hat{f}(\xi) = \int_{\mathbb{R}^n} f(x)\,e^{-i\xi \cdot x}\,d\mathcal{L}^n(x)$. In particular, $\|f\|_{L^2}^2 = \frac{1}{(2\pi)^n}\|\hat{f}\|_{L^2}^2$ corresponds to $\sigma = 0$. **Step 2: Pointwise factorisation.** Write $\theta := r/s \in [0,1]$. For each $\xi \in \mathbb{R}^n$: \begin{align*} \langle\xi\rangle^{2r}\,|\hat{f}(\xi)|^2 &= \bigl(|\hat{f}(\xi)|^2\bigr)^{1-\theta} \cdot \bigl(\langle\xi\rangle^{2s}\,|\hat{f}(\xi)|^2\bigr)^\theta, \end{align*} since the right-hand side equals $|\hat{f}|^{2(1-\theta)} \cdot \langle\xi\rangle^{2s\theta} \cdot |\hat{f}|^{2\theta} = \langle\xi\rangle^{2r}\,|\hat{f}|^2$. **Step 3: Hölder's inequality.** Apply the [Holder Inequality](/theorems/516) with exponents $p = 1/(1-\theta)$ and $q = 1/\theta$ (satisfying $1/p + 1/q = 1$) to the factorisation in Step 2: \begin{align*} \int_{\mathbb{R}^n} \langle\xi\rangle^{2r}\,|\hat{f}|^2\,d\mathcal{L}^n &\le \Bigl(\int_{\mathbb{R}^n} |\hat{f}|^2\,d\mathcal{L}^n\Bigr)^{1-\theta} \cdot \Bigl(\int_{\mathbb{R}^n} \langle\xi\rangle^{2s}\,|\hat{f}|^2\,d\mathcal{L}^n\Bigr)^\theta. \end{align*} Multiplying both sides by $\frac{1}{(2\pi)^n}$ and using $\frac{1}{(2\pi)^n} = \bigl(\frac{1}{(2\pi)^n}\bigr)^{1-\theta} \cdot \bigl(\frac{1}{(2\pi)^n}\bigr)^\theta$: \begin{align*} \|f\|_{H^r}^2 &\le \bigl(\|f\|_{L^2}^2\bigr)^{1-\theta} \cdot \bigl(\|f\|_{H^s}^2\bigr)^\theta = \|f\|_{L^2}^{2(1-r/s)}\,\|f\|_{H^s}^{2r/s}. \end{align*} Taking square roots gives $\|f\|_{H^r} \le \|f\|_{L^2}^{1-r/s}\,\|f\|_{H^s}^{r/s}$.