[proofplan] The embedding follows from the pointwise inequality $|\xi|^{2s} \le (1+|\xi|^2)^s$ for $s \ge 0$, which holds because adding $1$ to $|\xi|^2$ can only increase the $s$-th power. We integrate this inequality against $|\hat{u}_{T\text{-rep}}(\xi)|^2$ to pass directly from the inhomogeneous norm $\|u\|_{H^s}$ to the homogeneous norm $\|u\|_{\dot{H}^s}$. [/proofplan] [step:Establish the pointwise inequality $|\xi|^{2s} \le (1+|\xi|^2)^s$] For $s \ge 0$ and every $\xi \in \mathbb{R}^n$, the inequality $|\xi|^{2s} \le (1+|\xi|^2)^s$ holds because the function $r \mapsto (1+r)^s$ satisfies $(1+r)^s \ge r^s$ for all $r = |\xi|^2 \ge 0$ and $s \ge 0$. [/step] [step:Identify the $T$-representative of $\hat{u}$ and express the homogeneous norm] Let $u \in H^s(\mathbb{R}^n)$ and let $g_s \in L^2(\mathbb{R}^n)$ be the function representing $(1+|\xi|^2)^{s/2}\hat{u}$, so that $\|u\|_{H^s} = \|g_s\|_{L^2}$ by the [well-definedness theorem](/theorems/466). Membership $u \in H^s$ implies that $\hat{u}$ is a [regular distribution](/page/Regular%20Distribution) with $T$-representative $\hat{u}_{T\text{-rep}}(\xi) = g_s(\xi)(1+|\xi|^2)^{-s/2} \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$. Since $|\xi|^s|\hat{u}_{T\text{-rep}}(\xi)| = |\xi|^s(1+|\xi|^2)^{-s/2}|g_s(\xi)| \le |g_s(\xi)|$ for $\mathcal{L}^n$-a.e. $\xi$ (using $|\xi|^{2s} \le (1+|\xi|^2)^s$), the function $|\xi|^s\hat{u}_{T\text{-rep}}$ belongs to $L^2(\mathbb{R}^n)$. [/step] [step:Integrate the pointwise bound to obtain $\|u\|_{\dot{H}^s} \le \|u\|_{H^s}$] The homogeneous norm is \begin{align*} \|u\|_{\dot{H}^s}^2 &= \int_{\mathbb{R}^n} |\xi|^{2s}\,|\hat{u}_{T\text{-rep}}(\xi)|^2 \, d\mathcal{L}^n(\xi) \le \int_{\mathbb{R}^n} (1+|\xi|^2)^s\,|\hat{u}_{T\text{-rep}}(\xi)|^2 \, d\mathcal{L}^n(\xi) = \|g_s\|_{L^2}^2 = \|u\|_{H^s}^2, \end{align*} where every integrand involves pointwise values of the genuine function $\hat{u}_{T\text{-rep}}$. Taking square roots gives $\|u\|_{\dot{H}^s} \le \|u\|_{H^s}$. [/step]