[proofplan] The proof reduces the equivalence of norms to a finite collection of Fourier multiplier estimates. The key observation is that the multipliers $\xi^\alpha\,(1+|\xi|^2)^{-k/2}$ (for $|\alpha| \le k$) and $(1+|\xi|^2)^{k/2}\,\xi^\alpha\,/(\sum_{|\beta| \le k}|\xi^\beta|^2)$ (for $|\alpha| \le k$, with appropriate normalisation) all satisfy the Mihlin condition. This converts each direction of the inequality into a sum of bounded multiplier operators on $L^p$. We use the [Mihlin Multiplier Theorem](/theorems/???) and the elementary identity $(1+|\xi|^2)^{k/2} = \sum_{|\alpha| \le k} \binom{k}{\alpha}^{1/2}\,\xi^\alpha \cdot (\text{cofactor})$ to express both norms in terms of $L^p$-multipliers of $J^{-k}f$ and $D^\alpha f$ respectively. The induction on $k$ is replaced by a single uniform bound across all multi-indices $|\alpha| \le k$. [/proofplan]

[step:Reduce to two finite collections of Fourier multiplier inequalities] The Sobolev norm $\|f\|_{W^{k,p}(\mathbb{R}^n)}$ is defined by \begin{align*} \|f\|_{W^{k,p}(\mathbb{R}^n)}^p = \sum_{|\alpha| \le k}\|D^\alpha f\|_{L^p(\mathbb{R}^n)}^p, \end{align*} where $D^\alpha = \partial_{x_1}^{\alpha_1}\cdots\partial_{x_n}^{\alpha_n}$ for $\alpha = (\alpha_1, \ldots, \alpha_n)$ with $|\alpha| := \alpha_1 + \cdots + \alpha_n$. The Bessel potential norm is $\|f\|_{H^{k,p}(\mathbb{R}^n)} = \|J^{-k}f\|_{L^p(\mathbb{R}^n)}$ where $J^{-k}$ has Fourier symbol $\langle\xi\rangle^k := (1 + |\xi|^2)^{k/2}$. To prove the equivalence, it suffices to establish the two inequalities \begin{align*} \text{(A)}\qquad \|D^\alpha f\|_{L^p(\mathbb{R}^n)} &\le C_{n,k,p}\,\|J^{-k}f\|_{L^p(\mathbb{R}^n)} \qquad \text{for every } |\alpha| \le k, \\ \text{(B)}\qquad \|J^{-k}f\|_{L^p(\mathbb{R}^n)} &\le C'_{n,k,p}\,\sum_{|\alpha| \le k}\|D^\alpha f\|_{L^p(\mathbb{R}^n)}. \end{align*} Inequality (A), summed over $|\alpha| \le k$ and $L^p$-norms raised to the $p$-th power, gives one direction of the equivalence. Inequality (B), interpreted directly, gives the reverse direction. For each, we identify a Fourier multiplier whose $L^p$-boundedness is the assertion. We work first on Schwartz functions $f \in \mathcal{S}(\mathbb{R}^n)$ and then extend by density. [/step]

[step:Verify (A) — bound $\|D^\alpha f\|_{L^p}$ by $\|J^{-k}f\|_{L^p}$ via the Mihlin multiplier $\xi^\alpha\,\langle\xi\rangle^{-k}$] Fix a multi-index $\alpha$ with $|\alpha| \le k$. For $f \in \mathcal{S}(\mathbb{R}^n)$, \begin{align*} \widehat{D^\alpha f}(\xi) = (i\xi)^\alpha\,\hat f(\xi) = i^{|\alpha|}\,\xi^\alpha\,\hat f(\xi), \end{align*} where $\xi^\alpha := \xi_1^{\alpha_1}\cdots\xi_n^{\alpha_n}$. From the definition of $J^{-k}$, $\widehat{J^{-k}f}(\xi) = \langle\xi\rangle^k\,\hat f(\xi)$, so $\hat f(\xi) = \langle\xi\rangle^{-k}\,\widehat{J^{-k}f}(\xi)$. Substituting, \begin{align*} \widehat{D^\alpha f}(\xi) = i^{|\alpha|}\,\xi^\alpha\,\langle\xi\rangle^{-k}\,\widehat{J^{-k}f}(\xi) = i^{|\alpha|}\,m_\alpha(\xi)\,\widehat{J^{-k}f}(\xi), \end{align*} where \begin{align*} m_\alpha : \mathbb{R}^n &\to \mathbb{R} \\ \xi &\mapsto \xi^\alpha\,\langle\xi\rangle^{-k} = \frac{\xi^\alpha}{(1 + |\xi|^2)^{k/2}}. \end{align*} Thus $D^\alpha f = i^{|\alpha|}\,T_{m_\alpha}(J^{-k}f)$. [claim:$m_\alpha = \xi^\alpha\,\langle\xi\rangle^{-k}$ satisfies the Mihlin condition: $|D^\delta m_\alpha(\xi)| \le C_{n,k,\alpha,\delta}\,|\xi|^{-|\delta|}$ for all multi-indices $\delta$ with $|\delta| \le \lfloor n/2\rfloor + 1$ and all $\xi \in \mathbb{R}^n \setminus \{0\}$] [proof] We verify the bound directly. By the Leibniz rule applied to $m_\alpha = \xi^\alpha \cdot \langle\xi\rangle^{-k}$, \begin{align*} D^\delta m_\alpha(\xi) = \sum_{\beta + \gamma = \delta}\binom{\delta}{\beta}\,D^\beta(\xi^\alpha)\,D^\gamma(\langle\xi\rangle^{-k}). \end{align*} \textit{Bound on $D^\beta(\xi^\alpha)$.} For any multi-indices $\alpha, \beta$, $D^\beta(\xi^\alpha) = \alpha!/(\alpha - \beta)!\,\xi^{\alpha - \beta}$ if $\beta \le \alpha$ component-wise, and $0$ otherwise. Hence \begin{align*} |D^\beta(\xi^\alpha)| \le c_{\alpha, \beta}\,|\xi|^{|\alpha| - |\beta|}, \end{align*} where $c_{\alpha, \beta} := \alpha!/(\alpha - \beta)!$ when $\beta \le \alpha$ and $c_{\alpha, \beta} := 0$ otherwise, using $|\xi^{\alpha - \beta}| \le |\xi|^{|\alpha| - |\beta|}$. \textit{Bound on $D^\gamma(\langle\xi\rangle^{-k})$.} By induction on $|\gamma|$ we show \begin{align*} |D^\gamma(\langle\xi\rangle^{-k})| \le C_{n, k, \gamma}\,\langle\xi\rangle^{-k - |\gamma|} \qquad \text{for all } \xi \in \mathbb{R}^n. \end{align*} Base case $|\gamma| = 0$: $|\langle\xi\rangle^{-k}| = \langle\xi\rangle^{-k}$. Inductive step: $\partial_i\langle\xi\rangle^{-k} = -k\,\xi_i\,\langle\xi\rangle^{-k-2}$. Since $|\xi_i| \le \langle\xi\rangle$, \begin{align*} |\partial_i\langle\xi\rangle^{-k}| \le k\,\langle\xi\rangle\,\langle\xi\rangle^{-k-2} = k\,\langle\xi\rangle^{-k-1}. \end{align*} For higher $|\gamma|$, expand using Leibniz on the product $\xi_i\,\langle\xi\rangle^{-k-2}$ and apply induction. \textit{Combining.} Substituting both bounds, \begin{align*} |D^\delta m_\alpha(\xi)| \le \sum_{\beta + \gamma = \delta}\binom{\delta}{\beta}\,c_{\alpha,\beta}\,|\xi|^{|\alpha| - |\beta|}\,C_{n, k, \gamma}\,\langle\xi\rangle^{-k - |\gamma|}. \end{align*} Since $|\alpha| \le k$, on $|\xi| \ge 1$ we have $\langle\xi\rangle \asymp |\xi|$, so $\langle\xi\rangle^{-k - |\gamma|} \le 2^{(k + |\gamma|)/2}\,|\xi|^{-k - |\gamma|}$, and \begin{align*} |\xi|^{|\alpha| - |\beta|}\,\langle\xi\rangle^{-k - |\gamma|} \le C\,|\xi|^{|\alpha| - |\beta| - k - |\gamma|} = C\,|\xi|^{(|\alpha| - k) - |\delta|} \le C\,|\xi|^{-|\delta|}, \end{align*} using $|\alpha| \le k$ and $|\beta| + |\gamma| = |\delta|$. On $|\xi| \le 1$, $\langle\xi\rangle \asymp 1$, so $\langle\xi\rangle^{-k-|\gamma|} \le 1$ and $|\xi|^{|\alpha| - |\beta|}$ contributes $|\xi|^{|\alpha| - |\beta|}$. Each summand requires $\beta \le \alpha$ to be nonzero, so $|\beta| \le |\alpha| \le k$ and the exponent $|\alpha| - |\beta| \ge 0$. We have to verify $|\xi|^{|\alpha| - |\beta|} \le C\,|\xi|^{-|\delta|}$ for $|\xi| \le 1$. This holds when $|\alpha| - |\beta| + |\delta| \ge 0$, which is automatic. More carefully, since $|\xi| \le 1$ implies $|\xi|^t \le |\xi|^{-|\delta|}$ for $t \ge -|\delta|$, and $|\alpha| - |\beta| \ge 0 \ge -|\delta|$, the bound holds with constant $1$. (Note: the Mihlin condition only requires the bound on $\mathbb{R}^n \setminus\{0\}$, not at $|\xi| = 0$.) Combining both regimes, $|D^\delta m_\alpha(\xi)| \le C_{n, k, \alpha, \delta}\,|\xi|^{-|\delta|}$ on $\mathbb{R}^n \setminus \{0\}$, completing the verification. [/proof] [/claim] By the [Mihlin Multiplier Theorem](/theorems/3189) applied to $m_\alpha$ with the constant $A_\alpha := C_{n, k, \alpha} := \max_{|\delta| \le \lfloor n/2\rfloor + 1} C_{n, k, \alpha, \delta}$, the operator $T_{m_\alpha}: L^p(\mathbb{R}^n) \to L^p(\mathbb{R}^n)$ is bounded with norm $\le C_{n, p}\,A_\alpha$ depending only on $n$, $k$, $p$, $\alpha$. Hence \begin{align*} \|D^\alpha f\|_{L^p(\mathbb{R}^n)} = \|T_{m_\alpha}(J^{-k}f)\|_{L^p(\mathbb{R}^n)} \le C_{n, k, p, \alpha}\,\|J^{-k}f\|_{L^p(\mathbb{R}^n)}, \end{align*} which is inequality (A). Setting $C_{n, k, p} := \max_{|\alpha| \le k}C_{n, k, p, \alpha}$ gives the bound uniformly in $|\alpha| \le k$. [/step]

[step:Verify (B) — bound $\|J^{-k}f\|_{L^p}$ by $\sum_{|\alpha| \le k}\|D^\alpha f\|_{L^p}$ via the multinomial expansion of $\langle\xi\rangle^k$] For $k = 0$, inequality (B) is elementary: $J^0 = \mathrm{Id}$ and the sum on the right contains the term $\|f\|_{L^p}$. Assume $k \ge 1$. We use the multinomial-type expansion: for any $\xi \in \mathbb{R}^n$, set $r := |\xi|$ and write \begin{align*} \langle\xi\rangle^{2k} = (1 + r^2)^k = \sum_{j=0}^k \binom{k}{j}\,r^{2j} = \sum_{j=0}^k \binom{k}{j}\,\bigl(\xi_1^2 + \cdots + \xi_n^2\bigr)^j. \end{align*} By the multinomial theorem applied to $(\xi_1^2 + \cdots + \xi_n^2)^j = \sum_{|\beta| = j} \frac{j!}{\beta!}\,\xi^{2\beta}$ where $\beta \in \mathbb{N}_0^n$, $|\beta| = j$, and $\xi^{2\beta} = \prod \xi_i^{2\beta_i}$, \begin{align*} \langle\xi\rangle^{2k} = \sum_{|\beta| \le k} \binom{k}{|\beta|}\,\frac{|\beta|!}{\beta!}\,\xi^{2\beta} = \sum_{|\beta| \le k} c_\beta\,\xi^{2\beta}, \end{align*} where $c_\beta := \binom{k}{|\beta|}\,|\beta|!/\beta!$ is a positive integer depending only on $n$ and $k$. For $|\beta| \le k$ define \begin{align*} \tilde n_\beta : \mathbb{R}^n &\to \mathbb{R} \\ \xi &\mapsto \frac{\xi^\beta\,\langle\xi\rangle^k}{Q(\xi)}, \qquad Q(\xi) := \sum_{|\gamma| \le k} (\xi^\gamma)^2 = \sum_{|\gamma| \le k} \xi^{2\gamma}. \end{align*} Note $Q(\xi) \ge 1$ (since the term $\gamma = 0$ contributes $1$), so $\tilde n_\beta$ is well-defined as a real-valued function on $\mathbb{R}^n$. [claim:Lower bound $Q(\xi) \ge c_0\,\langle\xi\rangle^{2k}$ for some $c_0 > 0$ depending only on $n$ and $k$] [proof] We establish the key bound \begin{align*} \sum_{|\gamma| \le k}\xi^{2\gamma} \ge n^{-(k - 1)}\,|\xi|^{2k} \qquad \text{for all } \xi \in \mathbb{R}^n, \end{align*} which combined with the constant term ($\gamma = 0$ contributing $\xi^0 = 1$) gives \begin{align*} Q(\xi) = \sum_{|\gamma| \le k}\xi^{2\gamma} \ge \max\bigl(1,\ n^{-(k-1)}\,|\xi|^{2k}\bigr). \end{align*} \textbf{Step 1: the sub-monomial bound $\sum_{j=1}^n \xi_j^{2k} \ge n^{-(k-1)}\,|\xi|^{2k}$.} We prove this from the power-mean (or equivalently Cauchy–Schwarz / AM–GM) inequality \begin{align*} \Bigl(\frac{1}{n}\sum_{j=1}^n a_j\Bigr)^k \le \frac{1}{n}\sum_{j=1}^n a_j^k \qquad \text{for all } a_j \ge 0,\ k \ge 1, \end{align*} applied with $a_j := \xi_j^2 \ge 0$. The left-hand side equals $n^{-k}|\xi|^{2k}$ (since $\sum \xi_j^2 = |\xi|^2$); the right-hand side equals $n^{-1}\sum_j \xi_j^{2k}$. Multiplying through by $n$, \begin{align*} n^{-(k-1)}|\xi|^{2k} \le \sum_{j=1}^n \xi_j^{2k}. \end{align*} The diagonal multi-indices $\gamma = k\,e_j$ (where $e_j$ is the $j$-th standard basis vector and $|\gamma| = k$) contribute $\xi_j^{2k}$ each to the sum $\sum_{|\gamma| \le k}\xi^{2\gamma}$. Hence \begin{align*} \sum_{|\gamma| \le k}\xi^{2\gamma} \ge \sum_{j=1}^n \xi_j^{2k} \ge n^{-(k-1)}\,|\xi|^{2k}. \end{align*} \textbf{Step 2: derive $Q(\xi) \ge c_0\,\langle\xi\rangle^{2k}$.} For $|\xi| \le 1$, $\langle\xi\rangle^{2k} = (1 + |\xi|^2)^k \le 2^k$, and $Q(\xi) \ge 1$, so $Q(\xi) \ge 2^{-k}\,\langle\xi\rangle^{2k}$. For $|\xi| \ge 1$, $\langle\xi\rangle^{2k} \le (2|\xi|^2)^k = 2^k |\xi|^{2k}$, and $Q(\xi) \ge n^{-(k-1)}\,|\xi|^{2k}$ from Step 1, so $Q(\xi) \ge 2^{-k}\,n^{-(k-1)}\,\langle\xi\rangle^{2k}$. Taking $c_0 := 2^{-k}\,\min(1, n^{-(k-1)})$, $Q(\xi) \ge c_0\,\langle\xi\rangle^{2k}$ for all $\xi \in \mathbb{R}^n$. [/proof] [/claim] Then $|\tilde n_\beta(\xi)| \le |\xi|^{|\beta|}\,\langle\xi\rangle^k\,/\,c_0\,\langle\xi\rangle^{2k} = |\xi|^{|\beta|}\,/\,(c_0\,\langle\xi\rangle^k) \le 1/c_0$ when $|\beta| \le k$. [claim:$\tilde n_\beta$ satisfies the Mihlin condition: $|D^\delta \tilde n_\beta(\xi)| \le C_{n, k, \beta, \delta}\,|\xi|^{-|\delta|}$ for $|\delta| \le \lfloor n/2 \rfloor + 1$ and $\xi \ne 0$] [proof] Write $\tilde n_\beta = (\xi^\beta\,\langle\xi\rangle^{-k}) \cdot (\langle\xi\rangle^{2k}/Q(\xi))$, i.e., $\tilde n_\beta = m_\beta \cdot h$ where $m_\beta(\xi) := \xi^\beta\,\langle\xi\rangle^{-k}$ and $h(\xi) := \langle\xi\rangle^{2k}/Q(\xi)$. \textbf{Bound on $D^\delta m_\beta$.} By the claim in Step 2 (applied with multi-index $\beta$ in place of $\alpha$, which is valid since $|\beta| \le k$), \begin{align*} |D^\delta m_\beta(\xi)| \le C_{n, k, \beta, \delta}\,|\xi|^{-|\delta|}, \qquad \xi \ne 0. \end{align*} \textbf{Bound on $D^\delta h$.} We prove \begin{align*} |D^\delta h(\xi)| \le C_{n, k, \delta}\,\langle\xi\rangle^{-|\delta|} \qquad \text{for all } \xi \in \mathbb{R}^n,\ |\delta| \ge 0. \end{align*} By the Leibniz rule on $h = \langle\xi\rangle^{2k} \cdot Q(\xi)^{-1}$, \begin{align*} D^\delta h = \sum_{\delta' + \delta'' = \delta}\binom{\delta}{\delta'}\,D^{\delta'}\langle\xi\rangle^{2k}\,D^{\delta''}(Q^{-1}). \end{align*} The bound $|D^{\delta'}\langle\xi\rangle^{2k}| \le C_{n, k, \delta'}\,\langle\xi\rangle^{2k - |\delta'|}$ follows from the same induction as in the previous claim's proof (with $-k$ replaced by $2k$). It remains to bound $D^{\delta''}(Q^{-1})$. \textbf{Sub-claim:} $|D^{\delta''}(Q(\xi)^{-1})| \le C_{n, k, \delta''}\,Q(\xi)^{-1}\,\langle\xi\rangle^{-|\delta''|}$ for all $\delta''$ and all $\xi \in \mathbb{R}^n$. \textit{Proof of sub-claim.} Apply Faà di Bruno's formula to the composition $Q^{-1} = F \circ Q$ where $F(t) := 1/t$ for $t \ge 1$ (note $Q \ge 1$ everywhere). Faà di Bruno gives \begin{align*} D^{\delta''}(F \circ Q) = \sum_{\pi}c_\pi\,F^{(|\pi|)}(Q)\,\prod_{B \in \pi} D^{\delta''_B}Q, \end{align*} where the sum is over set partitions $\pi$ of $\delta''$, $|\pi|$ is the number of blocks, $\delta''_B$ is the multi-index assigned to block $B$, and $c_\pi$ are combinatorial constants depending only on $\delta''$. Since $F^{(j)}(t) = (-1)^j j!\,t^{-(j+1)}$, $|F^{(|\pi|)}(Q)| = |\pi|!\,Q^{-(|\pi|+1)}$. For the polynomial $Q(\xi) = \sum_{|\gamma| \le k}\xi^{2\gamma}$ of degree $2k$, each derivative $D^{\delta''_B}Q$ is a polynomial of degree $\le 2k - |\delta''_B|$ (taking the convention that polynomials of negative degree are zero), so \begin{align*} |D^{\delta''_B}Q(\xi)| \le C_{n, k, \delta''_B}\,(1 + |\xi|)^{2k - |\delta''_B|} \le C\,\langle\xi\rangle^{2k - |\delta''_B|}. \end{align*} We have $\sum_{B \in \pi}|\delta''_B| = |\delta''|$, so \begin{align*} \prod_{B \in \pi}|D^{\delta''_B}Q(\xi)| \le C\,\prod_{B \in \pi}\langle\xi\rangle^{2k - |\delta''_B|} = C\,\langle\xi\rangle^{2k|\pi| - |\delta''|}. \end{align*} Combining with the bound on $F^{(|\pi|)}(Q)$ and using $Q \ge c_0\,\langle\xi\rangle^{2k}$ (from the claim in this step), \begin{align*} |F^{(|\pi|)}(Q)|\,\prod_{B \in \pi}|D^{\delta''_B}Q| \le |\pi|!\,Q^{-(|\pi|+1)}\,C\,\langle\xi\rangle^{2k|\pi| - |\delta''|} \le C'\,Q^{-1}\,(c_0^{-1}\langle\xi\rangle^{-2k})^{|\pi|}\langle\xi\rangle^{2k|\pi| - |\delta''|} = C''\,Q^{-1}\,\langle\xi\rangle^{-|\delta''|}. \end{align*} Summing over partitions $\pi$ (finitely many), \begin{align*} |D^{\delta''}(Q^{-1})| \le C_{n, k, \delta''}\,Q^{-1}\,\langle\xi\rangle^{-|\delta''|}, \end{align*} proving the sub-claim. Combining with $Q^{-1} \le c_0^{-1}\,\langle\xi\rangle^{-2k}$ from the lower bound on $Q$, \begin{align*} |D^{\delta''}(Q^{-1})| \le C\,\langle\xi\rangle^{-2k - |\delta''|}. \end{align*} Substituting into the Leibniz expansion, \begin{align*} |D^\delta h(\xi)| \le \sum_{\delta' + \delta'' = \delta}\binom{\delta}{\delta'}\,C_{n,k,\delta'}\,\langle\xi\rangle^{2k - |\delta'|}\,C_{n,k,\delta''}\,\langle\xi\rangle^{-2k - |\delta''|} = C_{n, k, \delta}\,\langle\xi\rangle^{-|\delta|}. \end{align*} This establishes the bound on $D^\delta h$. \textbf{Combining.} By the Leibniz rule applied to $\tilde n_\beta = m_\beta \cdot h$, \begin{align*} |D^\delta\tilde n_\beta(\xi)| \le \sum_{\delta_1 + \delta_2 = \delta}\binom{\delta}{\delta_1}\,|D^{\delta_1}m_\beta|\,|D^{\delta_2}h| \le C_{n, k, \beta, \delta}\,|\xi|^{-|\delta_1|}\,\langle\xi\rangle^{-|\delta_2|}. \end{align*} For $|\xi| \ge 1$, $\langle\xi\rangle \asymp |\xi|$, so $|\xi|^{-|\delta_1|}\,\langle\xi\rangle^{-|\delta_2|} \le C\,|\xi|^{-(|\delta_1| + |\delta_2|)} = C\,|\xi|^{-|\delta|}$. For $0 < |\xi| \le 1$, we use that $\langle\xi\rangle^{-|\delta_2|} \le 1$ and bound $|\xi|^{-|\delta_1|} \le |\xi|^{-|\delta|}$ when $|\delta_1| \le |\delta|$ (which always holds). Thus $|D^\delta\tilde n_\beta(\xi)| \le C_{n, k, \beta, \delta}\,|\xi|^{-|\delta|}$ on all of $\mathbb{R}^n \setminus \{0\}$. [/proof] [/claim] The pointwise identity \begin{align*} \langle\xi\rangle^k = \langle\xi\rangle^k\,\frac{Q(\xi)}{Q(\xi)} = \frac{1}{Q(\xi)}\sum_{|\beta| \le k}\xi^{2\beta}\,\langle\xi\rangle^k = \sum_{|\beta| \le k}\xi^\beta\,\tilde n_\beta(\xi). \end{align*} Therefore on the Fourier side, \begin{align*} \widehat{J^{-k}f}(\xi) = \langle\xi\rangle^k\,\hat f(\xi) = \sum_{|\beta| \le k}\xi^\beta\,\tilde n_\beta(\xi)\,\hat f(\xi) = \sum_{|\beta| \le k} (-i)^{|\beta|}\,\tilde n_\beta(\xi)\,\widehat{D^\beta f}(\xi), \end{align*} using $\widehat{D^\beta f} = (i\xi)^\beta\,\hat f$ so $\xi^\beta\,\hat f = (-i)^{|\beta|}\,\widehat{D^\beta f}$. Inverting the Fourier transform, \begin{align*} J^{-k}f = \sum_{|\beta| \le k} (-i)^{|\beta|}\,T_{\tilde n_\beta}(D^\beta f). \end{align*} By the [Mihlin Multiplier Theorem](/theorems/3189) applied to each $\tilde n_\beta$ — hypotheses verified in the claim above — each $T_{\tilde n_\beta}$ is bounded on $L^p(\mathbb{R}^n)$ with norm $\le \tilde C^{(\beta)}_{n,k,p}$. Taking the $L^p$ norm and applying the triangle inequality and Mihlin bounds, \begin{align*} \|J^{-k}f\|_{L^p(\mathbb{R}^n)} \le \sum_{|\beta| \le k}\|T_{\tilde n_\beta}(D^\beta f)\|_{L^p(\mathbb{R}^n)} \le \sum_{|\beta| \le k} \tilde C^{(\beta)}_{n,k,p}\,\|D^\beta f\|_{L^p(\mathbb{R}^n)}. \end{align*} Setting $C'_{n,k,p} := \max_{|\beta| \le k}\tilde C^{(\beta)}_{n,k,p}$, this gives inequality (B). [/step]

[step:Establish equality of underlying spaces and extend by density] For Schwartz functions $f \in \mathcal{S}(\mathbb{R}^n)$, both inequalities (A) and (B) hold. Inequality (A) gives, after summing in $\alpha$ and raising to $p$, \begin{align*} \|f\|_{W^{k,p}}^p = \sum_{|\alpha| \le k}\|D^\alpha f\|_{L^p}^p \le \sum_{|\alpha| \le k} C_{n,k,p}^p\,\|J^{-k}f\|_{L^p}^p = (k+n \text{ choose terms})\,C_{n,k,p}^p\,\|f\|_{H^{k,p}}^p, \end{align*} i.e., $\|f\|_{W^{k,p}} \le \tilde C_{n,k,p}\,\|f\|_{H^{k,p}}$. Inequality (B) gives $\|f\|_{H^{k,p}} \le C'_{n,k,p}\,\sum_\beta\|D^\beta f\|_{L^p}$, and by Hölder applied to the inequality $\sum_\beta a_\beta \le N^{1-1/p}(\sum_\beta a_\beta^p)^{1/p}$ with $N$ the number of multi-indices, $\sum_\beta\|D^\beta f\|_{L^p} \le N^{1-1/p}\,\|f\|_{W^{k,p}}$, giving $\|f\|_{H^{k,p}} \le C'_{n,k,p}\,N^{1-1/p}\,\|f\|_{W^{k,p}} =: \tilde C'_{n,k,p}\,\|f\|_{W^{k,p}}$. The two inequalities show that for $f \in \mathcal{S}(\mathbb{R}^n)$, the norms $\|f\|_{W^{k,p}}$ and $\|f\|_{H^{k,p}}$ are equivalent. To extend to $W^{k,p}(\mathbb{R}^n)$: $\mathcal{S}(\mathbb{R}^n)$ is dense in $W^{k,p}(\mathbb{R}^n)$ for $1 \le p < \infty$ (a standard fact, established by mollification and cutoff). Given $f \in W^{k,p}(\mathbb{R}^n)$, choose $(f_m)_{m \in \mathbb{N}} \subset \mathcal{S}(\mathbb{R}^n)$ with $f_m \to f$ in $W^{k,p}$. The equivalence of norms on Schwartz space gives $\|f_m - f_\ell\|_{H^{k,p}} \le \tilde C'_{n,k,p}\,\|f_m - f_\ell\|_{W^{k,p}} \to 0$ as $m, \ell \to \infty$, so $(f_m)$ is Cauchy in $H^{k,p}$. Since $H^{k,p}$ is complete (it is the image of $L^p$ under the Banach-isomorphism $J^k$), there is a limit $g \in H^{k,p}(\mathbb{R}^n)$ with $f_m \to g$ in $H^{k,p}$. Convergence in $H^{k,p}$ implies convergence in $\mathcal{S}'$ (since the embedding $H^{k,p} \hookrightarrow \mathcal{S}'$ is continuous), as does convergence in $W^{k,p}$, so $g = f$ in $\mathcal{S}'$. Hence $f \in H^{k,p}(\mathbb{R}^n)$ and $\|f\|_{H^{k,p}} = \lim_m \|f_m\|_{H^{k,p}} \le \tilde C'_{n,k,p}\,\lim_m\|f_m\|_{W^{k,p}} = \tilde C'_{n,k,p}\,\|f\|_{W^{k,p}}$. The reverse direction $W^{k,p} \subseteq H^{k,p}$ is established symmetrically using inequality (A): given $f \in H^{k,p}(\mathbb{R}^n)$, $\mathcal{S}(\mathbb{R}^n)$ is dense in $H^{k,p}$ (since it is dense in $L^p$ and $J^k$ is an isomorphism $L^p \cong H^{k,p}$), and the same Cauchy-completeness argument yields $f \in W^{k,p}$ with $\|f\|_{W^{k,p}} \le \tilde C_{n,k,p}\,\|f\|_{H^{k,p}}$. Together, $H^{k,p}(\mathbb{R}^n) = W^{k,p}(\mathbb{R}^n)$ as sets and the norms are equivalent with constants $c_{n,k,p} := 1/\tilde C_{n,k,p}$ and $C_{n,k,p} := \tilde C'_{n,k,p}$ depending only on $n$, $k$, $p$. This completes the proof. [/step]