The strategy is to work entirely in Fourier space. For each $\xi \neq 0$, the matrix $P(\xi) = I - \frac{\xi \otimes \xi}{|\xi|^2} \in \mathbb{R}^{n \times n}$ is the orthogonal projection onto $\xi^\perp$, and its complement $Q(\xi) = \frac{\xi \otimes \xi}{|\xi|^2} \in \mathbb{R}^{n \times n}$ projects onto $\operatorname{span}(\xi)$. These pointwise matrices induce bounded operators $\mathbb{P}, \mathbb{Q}$ on $L^2(\mathbb{R}^n; \mathbb{R}^n)$ via the Fourier multiplier $\widehat{\mathbb{P}f}(\xi) := P(\xi)\hat{f}(\xi)$, and similarly for $\mathbb{Q}$. The divergence-free condition $\nabla \cdot v = 0$ translates to $\xi \cdot \hat{v}(\xi) = 0$ a.e., which is precisely the condition $\hat{v}(\xi) \in \xi^\perp$, i.e., $P(\xi)\hat{v}(\xi) = \hat{v}(\xi)$. The proof proceeds as follows: Step 1 establishes the algebraic properties of $P$ and $Q$; Step 2 shows $L^2_\sigma$ is closed; Step 3 constructs the decomposition and proves orthogonality; Step 4 identifies the orthogonal complement as the space of $L^2$ gradient fields. **Step 1: Algebraic properties of the multipliers.** For $\xi \in \mathbb{R}^n \setminus \{0\}$, define $P(\xi) := I - \frac{\xi \otimes \xi}{|\xi|^2}$ and $Q(\xi) := \frac{\xi \otimes \xi}{|\xi|^2}$. Both are real symmetric $n \times n$ matrices. [claim:Complementary Orthogonal Projections] For each $\xi \neq 0$: (i) $P(\xi)^2 = P(\xi)$ and $Q(\xi)^2 = Q(\xi)$; (ii) $P(\xi) + Q(\xi) = I$; (iii) $P(\xi)Q(\xi) = Q(\xi)P(\xi) = 0$; (iv) $\|P(\xi)\|_{\mathrm{op}} \le 1$ and $\|Q(\xi)\|_{\mathrm{op}} \le 1$. [/claim] [proof] Property (ii) holds by definition. For (i), compute \begin{align*} (Q(\xi))^2 &= \frac{\xi \otimes \xi}{|\xi|^2} \cdot \frac{\xi \otimes \xi}{|\xi|^2} = \frac{(\xi \otimes \xi)(\xi \otimes \xi)}{|\xi|^4}. \end{align*} To evaluate the numerator, note that $(\xi \otimes \xi)(\xi \otimes \xi) = (\xi\xi^\top)(\xi\xi^\top) = \xi(\xi^\top \xi)\xi^\top = |\xi|^2\,\xi\xi^\top = |\xi|^2\,(\xi \otimes \xi)$. Hence \begin{align*} (Q(\xi))^2 &= \frac{|\xi|^2\,(\xi \otimes \xi)}{|\xi|^4} = \frac{\xi \otimes \xi}{|\xi|^2} = Q(\xi). \end{align*} Then \begin{align*} P^2 &= (I - Q)^2 = I - 2Q + Q^2 = I - 2Q + Q = I - Q = P. \end{align*} Property (iii) follows: \begin{align*} PQ &= (I - Q)Q = Q - Q^2 = 0, \\ QP &= Q(I - Q) = Q - Q^2 = 0. \end{align*} For (iv), fix $v \in \mathbb{R}^n$. Since $P$ is symmetric ($P^\top = P$) and idempotent ($P^2 = P$), \begin{align*} |P(\xi)v|^2 &= (P(\xi)v)^\top(P(\xi)v) = v^\top P(\xi)^\top P(\xi) v = v^\top P(\xi)^2 v = v^\top P(\xi) v. \end{align*} By the [Cauchy–Schwarz Inequality](/theorems/432) applied in $\mathbb{R}^n$, \begin{align*} v^\top P(\xi) v &= v \cdot P(\xi)v \le |v|\,|P(\xi)v|. \end{align*} Hence $|P(\xi)v|^2 \le |v|\,|P(\xi)v|$, giving $|P(\xi)v| \le |v|$, so $\|P(\xi)\|_{\mathrm{op}} \le 1$. The same argument applies to $Q$. [/proof] **Step 2: $L^2_\sigma$ is a closed subspace.** [claim:Closedness Of The Divergence Free Subspace] $L^2_\sigma(\mathbb{R}^n)$ is a closed subspace of $L^2(\mathbb{R}^n; \mathbb{R}^n)$. [/claim] [proof] $L^2_\sigma$ is clearly a linear subspace. To show closedness, let $(v_k)_{k \ge 1} \subseteq L^2_\sigma$ converge to $v$ in $L^2$. For any $\varphi \in C^\infty_c(\mathbb{R}^n)$, the [Cauchy–Schwarz Inequality](/theorems/432) gives \begin{align*} \left|\int_{\mathbb{R}^n} v \cdot \nabla \varphi\,d\mathcal{L}^n\right| &= \left|\int_{\mathbb{R}^n} (v - v_k) \cdot \nabla\varphi\,d\mathcal{L}^n\right| \le \|v - v_k\|_{L^2}\,\|\nabla\varphi\|_{L^2} \to 0 \end{align*} as $k \to \infty$, where we used $\int v_k \cdot \nabla\varphi = 0$ since $v_k \in L^2_\sigma$. Hence $\int v \cdot \nabla\varphi = 0$ for all $\varphi \in C^\infty_c$, so $v \in L^2_\sigma$. [/proof] Since $L^2_\sigma$ is a closed subspace of the [Hilbert space](/page/Hilbert%20Space) $L^2(\mathbb{R}^n; \mathbb{R}^n)$, the orthogonal decomposition $L^2 = L^2_\sigma \oplus L^2_{\sigma^\perp}$ exists by the [Orthogonal Decomposition Theorem](/theorems/241). **Step 3: Construction of the decomposition and the Leray projector.** The matrices $P(\xi), Q(\xi) \in \mathbb{R}^{n \times n}$ from Step 1 act on vectors in $\mathbb{R}^n$ for each fixed $\xi$. We now use them to construct operators on $L^2(\mathbb{R}^n; \mathbb{R}^n)$. For $f \in L^2(\mathbb{R}^n; \mathbb{R}^n)$, define the [functions](/page/Function) $g, h: \mathbb{R}^n \to \mathbb{R}^n$ by \begin{align*} g(\xi) &:= P(\xi)\,\hat{f}(\xi), \quad h(\xi) := Q(\xi)\,\hat{f}(\xi). \end{align*} Since $\|P(\xi)\|_{\mathrm{op}} \le 1$ by Step 1(iv), we have $|g(\xi)| \le |\hat{f}(\xi)|$ pointwise, so $g \in L^2(\mathbb{R}^n; \mathbb{R}^n)$. The same argument gives $h \in L^2(\mathbb{R}^n; \mathbb{R}^n)$. By the [Plancherel Identity](/theorems/529), the [Fourier transform](/page/Fourier%20Transform) $f \mapsto \hat{f}$ (in the convention $\hat{f}(\xi) = \int f(x)\,e^{-i\xi \cdot x}\,d\mathcal{L}^n(x)$) is a bijection from $L^2(\mathbb{R}^n; \mathbb{R}^n)$ to itself: injectivity holds because $\|\hat{f}\|_{L^2} = (2\pi)^{n/2}\|f\|_{L^2}$, and surjectivity because the rescaled map $(2\pi)^{-n/2}\hat{\phantom{f}}$ is unitary and hence surjective. Since $g, h \in L^2$, there exist unique $\mathbb{P}f, \mathbb{Q}f \in L^2(\mathbb{R}^n; \mathbb{R}^n)$ satisfying \begin{align*} \widehat{\mathbb{P}f}(\xi) &= g(\xi) = P(\xi)\,\hat{f}(\xi), \quad \widehat{\mathbb{Q}f}(\xi) = h(\xi) = Q(\xi)\,\hat{f}(\xi). \end{align*} Since $P(\xi) + Q(\xi) = I$ pointwise by Step 1(ii), we have $g(\xi) + h(\xi) = \hat{f}(\xi)$, and the injectivity of the Fourier transform gives $\mathbb{P}f + \mathbb{Q}f = f$. [claim:Fourier Characterisation Of Divergence Free Fields] $v \in L^2_\sigma(\mathbb{R}^n)$ if and only if $\xi \cdot \hat{v}(\xi) = 0$ for $\mathcal{L}^n$-a.e. $\xi \in \mathbb{R}^n$. [/claim] [proof] The [distributional](/page/Distribution) condition $\nabla \cdot v = 0$ means $\int_{\mathbb{R}^n} v \cdot \nabla\varphi\,d\mathcal{L}^n = 0$ for all $\varphi \in C^\infty_c(\mathbb{R}^n)$. By the [Parseval Identity](/theorems/248), this equals $\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} \hat{v}(\xi) \cdot \overline{\widehat{\nabla\varphi}(\xi)}\,d\mathcal{L}^n(\xi)$. Since $\widehat{\nabla\varphi}(\xi) = i\xi\,\hat\varphi(\xi)$ by the [Fourier Differentiation Rule](/theorems/249), this becomes $\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} \hat{v}(\xi) \cdot \overline{i\xi\,\hat\varphi(\xi)}\,d\mathcal{L}^n(\xi) = \frac{-i}{(2\pi)^n}\int_{\mathbb{R}^n} (\xi \cdot \hat{v}(\xi))\,\overline{\hat\varphi(\xi)}\,d\mathcal{L}^n(\xi)$. Since this vanishes for all $\varphi \in C^\infty_c$ and the Fourier transform maps $C^\infty_c$ onto a dense subset of $L^2$, we conclude $\xi \cdot \hat{v}(\xi) = 0$ for $\mathcal{L}^n$-a.e. $\xi$. Conversely, if $\xi \cdot \hat{v}(\xi) = 0$ a.e., then the same computation shows $\int v \cdot \nabla\varphi = 0$ for all $\varphi \in C^\infty_c$. [/proof] Write $v := \mathbb{P}f$ and $w := \mathbb{Q}f$ for brevity. To verify $v \in L^2_\sigma$, the Fourier characterisation gives: \begin{align*} \xi \cdot \hat{v}(\xi) &= \xi \cdot P(\xi)\hat{f}(\xi) = \xi \cdot \hat{f}(\xi) - \xi \cdot \frac{\xi(\xi \cdot \hat{f}(\xi))}{|\xi|^2} = \xi \cdot \hat{f}(\xi) - \frac{|\xi|^2(\xi \cdot \hat{f}(\xi))}{|\xi|^2} = 0. \end{align*} To verify orthogonality, by the [Parseval Identity](/theorems/248): \begin{align*} \int_{\mathbb{R}^n} v \cdot w\,d\mathcal{L}^n &= \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} \hat{v}(\xi) \cdot \overline{\hat{w}(\xi)}\,d\mathcal{L}^n(\xi) = \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} \overline{\hat{f}(\xi)}^T P(\xi)^T Q(\xi)\,\hat{f}(\xi)\,d\mathcal{L}^n(\xi) = 0, \end{align*} where the last equality uses $P(\xi)^T Q(\xi) = P(\xi)Q(\xi) = 0$ from the claim in Step 1, and we used that $P$ and $Q$ are real symmetric to write $\hat{v} \cdot \overline{\hat{w}} = (P\hat{f}) \cdot \overline{(Q\hat{f})} = \overline{\hat{f}}^T P Q \hat{f}$. Since $f = \mathbb{P}f + \mathbb{Q}f$ with $\mathbb{P}f \in L^2_\sigma$ and $\mathbb{P}f \perp \mathbb{Q}f$, the operator $\mathbb{P}$ defined at the start of this step is the orthogonal projection of $L^2(\mathbb{R}^n; \mathbb{R}^n)$ onto $L^2_\sigma$. **Step 4: Identification of the orthogonal complement.** [claim:Orthogonal Complement Is The Gradient Space] A function $w \in L^2(\mathbb{R}^n; \mathbb{R}^n)$ belongs to $L^2_\sigma(\mathbb{R}^n)^\perp$ if and only if there exists $\phi \in L^2_{\mathrm{loc}}(\mathbb{R}^n)$ with $\nabla\phi \in L^2(\mathbb{R}^n; \mathbb{R}^n)$ such that $w = \nabla\phi$. [/claim] [proof] ($\supseteq$) Suppose $w = \nabla\phi$ with $\nabla\phi \in L^2$. By the [Fourier Differentiation Rule](/theorems/249), $\hat{w}(\xi) = i\xi\,\hat\phi(\xi)$, so $\hat{w}(\xi)$ is a scalar multiple of $\xi$ for $\mathcal{L}^n$-a.e. $\xi$. For any $v \in L^2_\sigma$, the Fourier characterisation from Step 3 gives $\xi \cdot \hat{v}(\xi) = 0$ a.e., so $\hat{v}(\xi) \perp \xi$ and hence $\hat{v}(\xi) \cdot \overline{\hat{w}(\xi)} = \hat{v}(\xi) \cdot \overline{i\xi\,\hat\phi(\xi)} = -i\,\overline{\hat\phi(\xi)}\,(\xi \cdot \hat{v}(\xi)) = 0$ for $\mathcal{L}^n$-a.e. $\xi$. By the [Parseval Identity](/theorems/248), \begin{align*} \int_{\mathbb{R}^n} v \cdot w\,d\mathcal{L}^n &= \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} \hat{v}(\xi) \cdot \overline{\hat{w}(\xi)}\,d\mathcal{L}^n(\xi) = 0. \end{align*} Hence $w \in {L^2_\sigma}^\perp$. ($\subseteq$) Let $w \in {L^2_\sigma}^\perp$. By Step 3, $w = \mathbb{Q}f$ for some $f \in L^2$, so $\hat{w}(\xi) = Q(\xi)\hat{f}(\xi) = \frac{\xi(\xi \cdot \hat{f}(\xi))}{|\xi|^2}$. Define $\hat\phi: \mathbb{R}^n \setminus \{0\} \to \mathbb{C}$ by \begin{align*} \hat\phi(\xi) &:= \frac{-i\,\xi \cdot \hat{f}(\xi)}{|\xi|^2}. \end{align*} Then $\widehat{\nabla\phi}(\xi) = i\xi\,\hat\phi(\xi)$ by the [Fourier Differentiation Rule](/theorems/249), so $\widehat{\nabla\phi}(\xi) = \frac{\xi(\xi \cdot \hat{f}(\xi))}{|\xi|^2} = \hat{w}(\xi)$, so $w = \nabla\phi$ in the distributional sense. Since $\hat{w} \in L^2$, we have $\nabla\phi \in L^2$ by the [Plancherel Identity](/theorems/529). Finally, $\hat\phi$ is locally [integrable](/page/Integral) (since $|\hat\phi(\xi)| \le |\hat{f}(\xi)|/|\xi|$ and $1/|\xi| \in L^2_{\mathrm{loc}}(\mathbb{R}^n)$ for $n \ge 2$), so $\phi \in L^2_{\mathrm{loc}}(\mathbb{R}^n)$. [/proof]