[proofplan] We adopt the Fourier-coefficient convention $\hat{u}(k) := (2\pi)^{-n} \int_{\mathbb{T}^n} u(x)\, e^{-i k \cdot x}\, d\mathcal{L}^n(x)$ for $u \in L^2(\mathbb{T}^n)$, $k \in \mathbb{Z}^n$; with this normalisation Parseval's identity reads $\|u\|_{L^2(\mathbb{T}^n)}^2 = (2\pi)^n \sum_{k \in \mathbb{Z}^n} |\hat{u}(k)|^2$. The bridge between the two definitions is the polynomial-equivalence inequality $c_m \sum_{|\alpha| \le m} |k^\alpha|^2 \le (1 + |k|^2)^m \le C_m \sum_{|\alpha| \le m} |k^\alpha|^2$ on $\mathbb{Z}^n$, which translates between the Fourier-side weight and the sum of monomials produced on the weak-derivative side. Combined with Parseval applied to each $D^\alpha f$, this inequality yields equivalence of norms once we establish that the two notions of derivative agree. The proof has four parts: (1) prove the polynomial-equivalence inequality via the multinomial expansion; (2) Fourier $\Rightarrow$ weak-derivative — define $D^\alpha f$ via the Fourier multiplier $(ik)^\alpha$ and verify the integration-by-parts characterisation; (3) weak-derivative $\Rightarrow$ Fourier — test against plane waves and apply Parseval to each $D^\alpha f \in L^2$; (4) collect the bounds to conclude equivalence of norms. [/proofplan] [step:Establish the polynomial-equivalence inequality on $\mathbb{Z}^n$] [claim:For every $m \in \mathbb{N} \cup \{0\}$ there exist constants $c_m, C_m > 0$ depending only on $m$ and $n$ such that for every $k \in \mathbb{Z}^n$,] \begin{align*} c_m \sum_{|\alpha| \le m} |k^\alpha|^2 \le (1 + |k|^2)^m \le C_m \sum_{|\alpha| \le m} |k^\alpha|^2, \end{align*} where $\alpha = (\alpha_1, \dots, \alpha_n) \in (\mathbb{N} \cup \{0\})^n$ is a multi-index, $|\alpha| = \alpha_1 + \dots + \alpha_n$, and $k^\alpha = k_1^{\alpha_1} \cdots k_n^{\alpha_n}$. [/claim] [proof] The case $m = 0$ is the identity $1 \le 1 \le 1$. Assume $m \ge 1$. By the multinomial theorem, \begin{align*} (1 + |k|^2)^m = (1 + k_1^2 + \dots + k_n^2)^m = \sum_{\beta_0 + \beta_1 + \dots + \beta_n = m} \binom{m}{\beta_0, \beta_1, \dots, \beta_n}\, k_1^{2\beta_1} \cdots k_n^{2\beta_n}. \end{align*} Setting $\alpha := (\beta_1, \dots, \beta_n)$ and $\beta_0 = m - |\alpha|$, this rewrites as \begin{align*} (1 + |k|^2)^m = \sum_{|\alpha| \le m} \binom{m}{m - |\alpha|, \alpha_1, \dots, \alpha_n}\, (k^\alpha)^2 = \sum_{|\alpha| \le m} a_\alpha (k^\alpha)^2, \end{align*} where $a_\alpha = \frac{m!}{(m - |\alpha|)!\, \alpha_1! \cdots \alpha_n!} > 0$ for every multi-index $\alpha$ with $|\alpha| \le m$. Since the multi-indices with $|\alpha| \le m$ form a finite set, define \begin{align*} c_m := \min_{|\alpha| \le m} a_\alpha > 0, \qquad C_m := \max_{|\alpha| \le m} a_\alpha > 0. \end{align*} Then for every $k \in \mathbb{Z}^n$, \begin{align*} c_m \sum_{|\alpha| \le m} (k^\alpha)^2 \le \sum_{|\alpha| \le m} a_\alpha (k^\alpha)^2 = (1 + |k|^2)^m \le C_m \sum_{|\alpha| \le m} (k^\alpha)^2, \end{align*} which is the desired inequality (since $|k^\alpha|^2 = (k^\alpha)^2$ for $k \in \mathbb{Z}^n$ and $\alpha$ a multi-index). [/proof] [/step] [step:Show that Fourier $H^m$ implies weak-derivative $H^m$, with controlled norm] Suppose $f \in H^m(\mathbb{T}^n)$ in the Fourier sense, i.e., $f \in L^2(\mathbb{T}^n)$ and $\sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^m |\hat{f}(k)|^2 < \infty$, where $\hat{f}(k) = (2\pi)^{-n}\int_{\mathbb{T}^n} f(x)\, e^{-i k \cdot x}\, d\mathcal{L}^n(x)$. Fix a multi-index $\alpha$ with $|\alpha| \le m$. By the polynomial-equivalence inequality (Step 1), \begin{align*} |k^\alpha|^2 \le \sum_{|\beta| \le m} |k^\beta|^2 \le \frac{1}{c_m} (1 + |k|^2)^m, \end{align*} so \begin{align*} \sum_{k \in \mathbb{Z}^n} |k^\alpha|^2\, |\hat{f}(k)|^2 \le \frac{1}{c_m} \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^m |\hat{f}(k)|^2 < \infty. \end{align*} Hence the sequence $\big( (ik)^\alpha\, \hat{f}(k) \big)_{k \in \mathbb{Z}^n}$ lies in $\ell^2(\mathbb{Z}^n)$, and by the Riesz-Fischer theorem on $\mathbb{T}^n$ it is the sequence of Fourier coefficients of a unique function $g_\alpha \in L^2(\mathbb{T}^n)$, namely \begin{align*} g_\alpha = \sum_{k \in \mathbb{Z}^n} (ik)^\alpha\, \hat{f}(k)\, e^{i k \cdot x}, \end{align*} where the series converges in $L^2(\mathbb{T}^n)$, and $\hat{g_\alpha}(k) = (ik)^\alpha\, \hat{f}(k)$ for every $k \in \mathbb{Z}^n$. By Parseval's identity (with the present convention, $\|g_\alpha\|_{L^2(\mathbb{T}^n)}^2 = (2\pi)^n \sum_k |\hat{g_\alpha}(k)|^2$), \begin{align*} \|g_\alpha\|_{L^2(\mathbb{T}^n)}^2 = (2\pi)^n \sum_{k \in \mathbb{Z}^n} |k^\alpha|^2\, |\hat{f}(k)|^2 < \infty. \end{align*} We claim that $g_\alpha$ is the weak $\alpha$-th derivative of $f$, i.e., that for every test function $\varphi \in C^\infty(\mathbb{T}^n)$, \begin{align*} \int_{\mathbb{T}^n} g_\alpha\, \varphi\, d\mathcal{L}^n = (-1)^{|\alpha|} \int_{\mathbb{T}^n} f\, \partial^\alpha \varphi\, d\mathcal{L}^n. \end{align*} Both integrals are well-defined: by Hölder's inequality with $p = q = 2$ on the bounded measure space $\mathbb{T}^n$, the products $g_\alpha \varphi$ and $f\, \partial^\alpha \varphi$ are in $L^1(\mathbb{T}^n)$. To verify the identity, expand $\varphi$ in its Fourier series. Since $\varphi \in C^\infty(\mathbb{T}^n)$, the decay of Fourier coefficients of smooth periodic functions gives $\hat{\varphi}(k) = O(|k|^{-N})$ as $|k| \to \infty$ for every $N \in \mathbb{N}$, hence $\sum_k |\hat{\varphi}(k)|^2 (1 + |k|^2)^M < \infty$ for every $M$. By Parseval's identity for the $L^2$ pairing on $\mathbb{T}^n$, \begin{align*} \int_{\mathbb{T}^n} g_\alpha\, \overline{\varphi}\, d\mathcal{L}^n = (2\pi)^n \sum_{k \in \mathbb{Z}^n} \hat{g_\alpha}(k)\, \overline{\hat{\varphi}(k)} = (2\pi)^n \sum_{k} (ik)^\alpha\, \hat{f}(k)\, \overline{\hat{\varphi}(k)}. \end{align*} On the other hand, differentiating the Fourier series of $\varphi$ termwise (justified by the rapid decay) gives the identity for Fourier coefficients of derivatives, $\widehat{\partial^\alpha \varphi}(k) = (ik)^\alpha\, \hat{\varphi}(k)$, so \begin{align*} \int_{\mathbb{T}^n} f\, \overline{\partial^\alpha \varphi}\, d\mathcal{L}^n = (2\pi)^n \sum_{k \in \mathbb{Z}^n} \hat{f}(k)\, \overline{(ik)^\alpha\, \hat{\varphi}(k)} = (2\pi)^n \sum_{k} \hat{f}(k)\, \overline{(ik)^\alpha}\, \overline{\hat{\varphi}(k)}. \end{align*} Since $k \in \mathbb{Z}^n$ has real components, $\overline{ik_j} = -ik_j$ for each $j$, and therefore \begin{align*} \overline{(ik)^\alpha} = \prod_{j=1}^n \overline{(ik_j)^{\alpha_j}} = \prod_{j=1}^n (-ik_j)^{\alpha_j} = \prod_{j=1}^n (-1)^{\alpha_j} (ik_j)^{\alpha_j} = (-1)^{|\alpha|}\, (ik)^\alpha. \end{align*} Substituting, \begin{align*} \int_{\mathbb{T}^n} f\, \overline{\partial^\alpha \varphi}\, d\mathcal{L}^n = (-1)^{|\alpha|} (2\pi)^n \sum_{k} (ik)^\alpha\, \hat{f}(k)\, \overline{\hat{\varphi}(k)} = (-1)^{|\alpha|} \int_{\mathbb{T}^n} g_\alpha\, \overline{\varphi}\, d\mathcal{L}^n. \end{align*} Replacing $\overline{\varphi}$ by $\varphi$ (the identity holds for all complex test functions, in particular real-valued ones, and complex extension by linearity preserves the identity) yields the weak-derivative identity. Hence $D^\alpha f = g_\alpha$ in the weak sense, with $\|D^\alpha f\|_{L^2(\mathbb{T}^n)}^2 = (2\pi)^n \sum_k |k^\alpha|^2 |\hat{f}(k)|^2$. Summing over $|\alpha| \le m$, exchanging the order of summation (justified by Tonelli's theorem applied to counting measure on $\{|\alpha| \le m\} \times \mathbb{Z}^n$, since all terms are non-negative), and using the upper bound from Step 1, \begin{align*} \sum_{|\alpha| \le m} \|D^\alpha f\|_{L^2(\mathbb{T}^n)}^2 = (2\pi)^n \sum_{k} |\hat{f}(k)|^2 \sum_{|\alpha| \le m} |k^\alpha|^2 \le \frac{(2\pi)^n}{c_m} \sum_{k} (1 + |k|^2)^m |\hat{f}(k)|^2. \end{align*} Therefore $f \in H^m(\mathbb{T}^n)$ in the weak-derivative sense, with \begin{align*} \|f\|_{H^m, \text{weak}}^2 \le \frac{(2\pi)^n}{c_m}\, \|f\|_{H^m, \text{Fourier}}^2. \end{align*} [guided] We are given $f \in L^2(\mathbb{T}^n)$ with $\sum_k (1+|k|^2)^m |\hat{f}(k)|^2 < \infty$ (Fourier-side hypothesis), where the Fourier coefficient is normalised as $\hat{f}(k) = (2\pi)^{-n}\int_{\mathbb{T}^n} f(x) e^{-i k \cdot x}\, d\mathcal{L}^n(x)$ so that Parseval reads $\|u\|_{L^2}^2 = (2\pi)^n \sum_k |\hat{u}(k)|^2$. The strategy has three substeps: define a candidate $g_\alpha$ for the weak derivative purely on the Fourier side, verify it satisfies the integration-by-parts characterisation, and compute its $L^2$ norm. **Substep a (define the candidate).** The Fourier multiplier rule for differentiation says that, formally, $\widehat{\partial^\alpha u}(k) = (ik)^\alpha \hat{u}(k)$ — each derivative produces a factor of $ik_j$. We define $g_\alpha$ to be the function in $L^2(\mathbb{T}^n)$ with these prescribed Fourier coefficients. To produce such a function we need to know the candidate sequence is square-summable; this is exactly what the polynomial-equivalence inequality from Step 1 guarantees: applied with $|\beta| = |\alpha|$, \begin{align*} |(ik)^\alpha \hat{f}(k)|^2 = |k^\alpha|^2 |\hat{f}(k)|^2 \le \frac{1}{c_m} (1+|k|^2)^m |\hat{f}(k)|^2, \end{align*} so summing over $k$ gives a finite bound by the Fourier-side hypothesis. Thus $\big((ik)^\alpha \hat{f}(k)\big)_{k \in \mathbb{Z}^n} \in \ell^2(\mathbb{Z}^n)$. By the Riesz-Fischer theorem on $\mathbb{T}^n$, there is a unique $g_\alpha \in L^2(\mathbb{T}^n)$ with $\hat{g_\alpha}(k) = (ik)^\alpha \hat{f}(k)$ for every $k \in \mathbb{Z}^n$, given by the $L^2$-convergent series $g_\alpha(x) = \sum_k (ik)^\alpha \hat{f}(k)\, e^{ik \cdot x}$. **Substep b (verify the weak-derivative property).** We must show that for every $\varphi \in C^\infty(\mathbb{T}^n)$, \begin{align*} \int_{\mathbb{T}^n} g_\alpha\, \varphi\, d\mathcal{L}^n = (-1)^{|\alpha|} \int_{\mathbb{T}^n} f\, \partial^\alpha \varphi\, d\mathcal{L}^n. \end{align*} Both sides are finite by Hölder ($p = q = 2$ on the bounded space $\mathbb{T}^n$). The verification uses two ingredients: (i) Parseval's identity translates $L^2$ pairings into $\ell^2$ sums of Fourier coefficients; (ii) smoothness of $\varphi$ gives rapid decay $\hat{\varphi}(k) = O(|k|^{-N})$ for every $N$, so all the series we manipulate converge absolutely. By Parseval applied to the pairing $(g_\alpha, \varphi)_{L^2} = \int g_\alpha\, \overline{\varphi}\, d\mathcal{L}^n$, \begin{align*} \int_{\mathbb{T}^n} g_\alpha\, \overline{\varphi}\, d\mathcal{L}^n = (2\pi)^n \sum_k (ik)^\alpha\, \hat{f}(k)\, \overline{\hat{\varphi}(k)}. \end{align*} For the right-hand side, the rapid decay of $\hat{\varphi}$ legitimises differentiating the Fourier series of $\varphi$ termwise, yielding the Fourier coefficients of derivatives identity $\widehat{\partial^\alpha \varphi}(k) = (ik)^\alpha \hat{\varphi}(k)$, and hence \begin{align*} \int_{\mathbb{T}^n} f\, \overline{\partial^\alpha \varphi}\, d\mathcal{L}^n = (2\pi)^n \sum_k \hat{f}(k)\, \overline{(ik)^\alpha\, \hat{\varphi}(k)}. \end{align*} Now we relate $\overline{(ik)^\alpha}$ to $(ik)^\alpha$. Since each $k_j$ is real, $\overline{ik_j} = -ik_j$, and so \begin{align*} \overline{(ik)^\alpha} = \prod_j \overline{(ik_j)^{\alpha_j}} = \prod_j (-ik_j)^{\alpha_j} = \prod_j (-1)^{\alpha_j}(ik_j)^{\alpha_j} = (-1)^{|\alpha|}(ik)^\alpha, \end{align*} because the exponents $\alpha_j$ sum to $|\alpha|$. Plugging in, \begin{align*} \int_{\mathbb{T}^n} f\, \overline{\partial^\alpha \varphi}\, d\mathcal{L}^n = (-1)^{|\alpha|} (2\pi)^n \sum_k (ik)^\alpha\, \hat{f}(k)\, \overline{\hat{\varphi}(k)} = (-1)^{|\alpha|} \int_{\mathbb{T}^n} g_\alpha\, \overline{\varphi}\, d\mathcal{L}^n. \end{align*} Replacing $\overline{\varphi}$ by $\varphi$ (the identity holds for all complex test functions, and the substitution preserves linearity) gives the weak-derivative identity. Hence $D^\alpha f = g_\alpha$ in the weak sense. **Substep c (norm bound).** Parseval's identity gives $\|D^\alpha f\|_{L^2}^2 = \|g_\alpha\|_{L^2}^2 = (2\pi)^n \sum_k |k^\alpha|^2 |\hat{f}(k)|^2$. Summing over $|\alpha| \le m$ and exchanging the order (Tonelli's theorem applies because all terms are non-negative) yields \begin{align*} \sum_{|\alpha| \le m} \|D^\alpha f\|_{L^2}^2 = (2\pi)^n \sum_k |\hat{f}(k)|^2 \sum_{|\alpha| \le m} |k^\alpha|^2. \end{align*} Applying the upper bound from Step 1, $\sum_{|\alpha| \le m} |k^\alpha|^2 \le c_m^{-1}(1+|k|^2)^m$, this is at most $(2\pi)^n c_m^{-1} \|f\|_{H^m, \text{Fourier}}^2$. So $\|f\|_{H^m, \text{weak}}^2 \le (2\pi)^n c_m^{-1} \|f\|_{H^m, \text{Fourier}}^2$, completing the forward direction. [/guided] [/step] [step:Show that weak-derivative $H^m$ implies Fourier $H^m$, with controlled norm] Suppose $f \in L^2(\mathbb{T}^n)$ has weak derivatives $D^\alpha f \in L^2(\mathbb{T}^n)$ for every multi-index $\alpha$ with $|\alpha| \le m$. We use the convention $\hat{u}(k) = (2\pi)^{-n}\int_{\mathbb{T}^n} u(x)\, e^{-i k \cdot x}\, d\mathcal{L}^n(x)$, under which Parseval reads $\|u\|_{L^2(\mathbb{T}^n)}^2 = (2\pi)^n \sum_k |\hat{u}(k)|^2$. By the integration-by-parts characterisation of the weak derivative, for every multi-index $\alpha$ with $|\alpha| \le m$ and every $k \in \mathbb{Z}^n$, the function $\varphi_k(x) := e^{i k \cdot x}$ belongs to $C^\infty(\mathbb{T}^n)$ and satisfies $\partial^\alpha \varphi_k(x) = (ik)^\alpha e^{i k \cdot x}$, hence $\overline{\partial^\alpha \varphi_k}(x) = \overline{(ik)^\alpha}\, e^{-i k \cdot x}$. Therefore \begin{align*} \int_{\mathbb{T}^n} D^\alpha f(x)\, e^{-i k \cdot x}\, d\mathcal{L}^n(x) = (-1)^{|\alpha|} \int_{\mathbb{T}^n} f(x)\, \overline{\partial^\alpha \varphi_k(x)}\, d\mathcal{L}^n(x) = (-1)^{|\alpha|}\, \overline{(ik)^\alpha} \int_{\mathbb{T}^n} f(x)\, e^{-i k \cdot x}\, d\mathcal{L}^n(x). \end{align*} Dividing both sides by $(2\pi)^n$ and recalling the convention for $\hat{u}(k)$, \begin{align*} \widehat{D^\alpha f}(k) = (-1)^{|\alpha|}\, \overline{(ik)^\alpha}\, \hat{f}(k). \end{align*} We compute $\overline{(ik)^\alpha}$ explicitly. Since each $k_j \in \mathbb{Z}$ is real, $\overline{ik_j} = -ik_j$, so \begin{align*} \overline{(ik)^\alpha} = \prod_{j=1}^n \overline{(ik_j)^{\alpha_j}} = \prod_{j=1}^n (-ik_j)^{\alpha_j} = \prod_{j=1}^n (-1)^{\alpha_j}(ik_j)^{\alpha_j} = (-1)^{|\alpha|}\, (ik)^\alpha. \end{align*} Substituting, \begin{align*} \widehat{D^\alpha f}(k) = (-1)^{|\alpha|} \cdot (-1)^{|\alpha|} (ik)^\alpha\, \hat{f}(k) = (ik)^\alpha\, \hat{f}(k). \end{align*} By Parseval's identity applied to $D^\alpha f \in L^2(\mathbb{T}^n)$, \begin{align*} \|D^\alpha f\|_{L^2(\mathbb{T}^n)}^2 = (2\pi)^n \sum_{k \in \mathbb{Z}^n} |\widehat{D^\alpha f}(k)|^2 = (2\pi)^n \sum_{k} |k^\alpha|^2\, |\hat{f}(k)|^2. \end{align*} Summing over multi-indices $|\alpha| \le m$, exchanging the order (terms are non-negative; Tonelli's theorem on counting measure justifies the swap), and using the polynomial-equivalence inequality (Step 1, lower bound), \begin{align*} \sum_{|\alpha| \le m} \|D^\alpha f\|_{L^2(\mathbb{T}^n)}^2 &= (2\pi)^n \sum_{k \in \mathbb{Z}^n} |\hat{f}(k)|^2 \sum_{|\alpha| \le m} |k^\alpha|^2 \\ &\ge (2\pi)^n \cdot \frac{1}{C_m} \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^m |\hat{f}(k)|^2 \\ &= \frac{(2\pi)^n}{C_m}\, \|f\|_{H^m, \text{Fourier}}^2. \end{align*} Since the left-hand side is finite by hypothesis, $\sum_k (1 + |k|^2)^m |\hat{f}(k)|^2 < \infty$, so $f \in H^m(\mathbb{T}^n)$ in the Fourier sense. Rearranging, \begin{align*} \|f\|_{H^m, \text{Fourier}}^2 \le \frac{C_m}{(2\pi)^n}\, \|f\|_{H^m, \text{weak}}^2. \end{align*} [guided] We are given that $f \in L^2(\mathbb{T}^n)$ has weak derivatives $D^\alpha f \in L^2(\mathbb{T}^n)$ for every $|\alpha| \le m$, and we want to deduce the Fourier-side estimate. The plan is: identify the Fourier coefficients of $D^\alpha f$ as $(ik)^\alpha \hat{f}(k)$, then apply Parseval to convert $\|D^\alpha f\|_{L^2}^2$ into a sum on the Fourier side, and finally invoke Step 1's lower bound. **Substep a (test against plane waves).** The integration-by-parts characterisation of the weak derivative says that for every $\varphi \in C^\infty(\mathbb{T}^n)$, \begin{align*} \int_{\mathbb{T}^n} D^\alpha f\, \varphi\, d\mathcal{L}^n = (-1)^{|\alpha|} \int_{\mathbb{T}^n} f\, \partial^\alpha \varphi\, d\mathcal{L}^n. \end{align*} Why test against plane waves? With our convention $\hat{u}(k) = (2\pi)^{-n} \int u(x) e^{-i k \cdot x}\, d\mathcal{L}^n(x)$, the Fourier coefficient $\widehat{D^\alpha f}(k)$ is essentially the pairing of $D^\alpha f$ against $e^{i k \cdot x}$. So we choose $\varphi_k(x) := e^{i k \cdot x}$, which is smooth on $\mathbb{T}^n$ and satisfies $\partial^\alpha \varphi_k(x) = (ik)^\alpha e^{i k \cdot x}$ — derivatives of plane waves act as multiplication by $(ik)^\alpha$. The integration-by-parts identity, applied with $\overline{\varphi_k}$ in place of $\varphi$ (a real linear combination of $\varphi_k$ and $\overline{\varphi_k}$ would also work; this avoids that), gives \begin{align*} \int_{\mathbb{T}^n} D^\alpha f(x)\, e^{-i k \cdot x}\, d\mathcal{L}^n(x) = (-1)^{|\alpha|}\, \overline{(ik)^\alpha}\, \int_{\mathbb{T}^n} f(x)\, e^{-i k \cdot x}\, d\mathcal{L}^n(x). \end{align*} Dividing by $(2\pi)^n$, $\widehat{D^\alpha f}(k) = (-1)^{|\alpha|}\, \overline{(ik)^\alpha}\, \hat{f}(k)$. **Substep b (simplify $\overline{(ik)^\alpha}$).** Each component $k_j$ is real (an integer), so $\overline{ik_j} = -ik_j$. We compute \begin{align*} \overline{(ik)^\alpha} = \prod_{j=1}^n \overline{(ik_j)^{\alpha_j}} = \prod_{j=1}^n (-ik_j)^{\alpha_j} = \prod_{j=1}^n (-1)^{\alpha_j}(ik_j)^{\alpha_j} = (-1)^{|\alpha|}(ik)^\alpha, \end{align*} since the exponents $\alpha_j$ sum to $|\alpha|$. The two factors of $(-1)^{|\alpha|}$ in $\widehat{D^\alpha f}(k)$ cancel, yielding the clean Fourier-multiplier identity \begin{align*} \widehat{D^\alpha f}(k) = (ik)^\alpha\, \hat{f}(k). \end{align*} This says: in this convention, taking a weak $\alpha$-derivative on the spatial side corresponds to multiplying Fourier coefficients by $(ik)^\alpha$. **Substep c (apply Parseval and sum).** With our convention Parseval reads $\|u\|_{L^2}^2 = (2\pi)^n \sum_k |\hat{u}(k)|^2$. Applied to $D^\alpha f$, \begin{align*} \|D^\alpha f\|_{L^2(\mathbb{T}^n)}^2 = (2\pi)^n \sum_k |k^\alpha|^2 |\hat{f}(k)|^2. \end{align*} Summing over $|\alpha| \le m$ and exchanging the order (Tonelli's theorem applies because the summands are non-negative) yields \begin{align*} \sum_{|\alpha| \le m} \|D^\alpha f\|_{L^2(\mathbb{T}^n)}^2 = (2\pi)^n \sum_k |\hat{f}(k)|^2 \sum_{|\alpha| \le m} |k^\alpha|^2. \end{align*} Applying the lower bound of the polynomial-equivalence inequality, $\sum_{|\alpha| \le m} |k^\alpha|^2 \ge C_m^{-1} (1+|k|^2)^m$, \begin{align*} \sum_{|\alpha| \le m} \|D^\alpha f\|_{L^2(\mathbb{T}^n)}^2 \ge \frac{(2\pi)^n}{C_m} \sum_k (1+|k|^2)^m |\hat{f}(k)|^2 = \frac{(2\pi)^n}{C_m}\, \|f\|_{H^m, \text{Fourier}}^2. \end{align*} Since the left-hand side is finite by hypothesis, the Fourier-side norm is finite, so $f \in H^m(\mathbb{T}^n)$ in the Fourier sense. Rearranging gives the bound $\|f\|_{H^m, \text{Fourier}}^2 \le C_m (2\pi)^{-n} \|f\|_{H^m, \text{weak}}^2$. [/guided] [/step] [step:Combine the bounds to conclude norm equivalence and complete the proof] Steps 2 and 3 together establish: (a) If $f \in L^2(\mathbb{T}^n)$ satisfies the Fourier-side condition $\sum_k (1+|k|^2)^m |\hat{f}(k)|^2 < \infty$, then all weak derivatives $D^\alpha f$ for $|\alpha| \le m$ exist and lie in $L^2(\mathbb{T}^n)$, with $\|f\|_{H^m, \text{weak}}^2 \le (2\pi)^n c_m^{-1} \|f\|_{H^m, \text{Fourier}}^2$. (b) If $f \in L^2(\mathbb{T}^n)$ has weak derivatives $D^\alpha f \in L^2(\mathbb{T}^n)$ for all $|\alpha| \le m$, then $f$ satisfies the Fourier-side condition, with $\|f\|_{H^m, \text{Fourier}}^2 \le C_m (2\pi)^{-n} \|f\|_{H^m, \text{weak}}^2$. Hence the two definitions of $H^m(\mathbb{T}^n)$ produce the same set of functions, and the two norms satisfy \begin{align*} \frac{(2\pi)^n}{C_m}\, \|f\|_{H^m, \text{Fourier}}^2 \le \|f\|_{H^m, \text{weak}}^2 \le \frac{(2\pi)^n}{c_m}\, \|f\|_{H^m, \text{Fourier}}^2, \end{align*} which is the equivalence of norms (multiplicative constants depending on $m$ and $n$ alone). This completes the proof. [/step]