The [Fourier transform](/page/Fourier%20Transform) on $L^1(\mathbb{R}^n)$ is defined by the integral $\hat{f}(\xi) = \int f(x) e^{-ix \cdot \xi}\,d\mathcal{L}^n(x)$. This integral makes sense because $|f(x)e^{-ix\cdot\xi}| = |f(x)|$ is integrable. For a function $f \in L^2(\mathbb{R}^n) \setminus L^1(\mathbb{R}^n)$, however, the integral may diverge for every $\xi$, and the pointwise definition breaks down entirely. Nevertheless, the Fourier transform extends to all of $L^2$ as a bounded linear operator — not by an integral formula, but by a density argument using the [Plancherel theorem](/theorems/247). This page explains the construction, what the resulting object $\hat{f}$ is, and what operations on it are legitimate. ## Why the Integral Fails Consider $f(x) = (1 + |x|)^{-(n+1)/2}$ for $x \in \mathbb{R}^n$. One checks that $|f|^2 \sim |x|^{-(n+1)}$, which is integrable at infinity in $\mathbb{R}^n$, so $f \in L^2(\mathbb{R}^n)$. But $|f| \sim |x|^{-(n+1)/2}$, and the integral $\int_{|x| > 1} |x|^{-(n+1)/2}\,d\mathcal{L}^n(x)$ diverges (since $(n+1)/2 \le n$ for all $n \ge 1$), so $f \notin L^1(\mathbb{R}^n)$. For such [functions](/page/Function), the expression $\int f(x) e^{-ix\cdot\xi}\,d\mathcal{L}^n(x)$ is not absolutely convergent for any $\xi$, and the integral definition of $\hat{f}$ is meaningless. The solution is to avoid the integral formula and instead define $\hat{f}$ as an $L^2$-limit. ## The Construction The extension proceeds in three steps. **Step 1: The isometry on $L^1 \cap L^2$.** For functions in $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$, the integral definition applies and produces a bounded continuous function $\hat{f} \in C_0(\mathbb{R}^n)$ by the [Riemann–Lebesgue lemma](/theorems/245). Moreover, $\hat{f} \in L^2(\mathbb{R}^n)$, and the [Plancherel theorem](/theorems/247) gives the isometry \begin{align*} \|\hat{f}\|_{L^2}^2 = (2\pi)^n\,\|f\|_{L^2}^2. \end{align*} This identity is first verified on the [Schwartz space](/page/Schwartz%20Space) $\mathcal{S}(\mathbb{R}^n) \subset L^1 \cap L^2$ (see [Theorem 247](/theorems/247) for the full proof). **Step 2: Extension by density.** The space $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$ is dense in $L^2(\mathbb{R}^n)$. (For instance, given $f \in L^2$, the truncations $f_m = f \cdot \mathbb{1}_{B(0,m)}$ lie in $L^1 \cap L^2$ and converge to $f$ in $L^2$ by dominated convergence.) For any $f \in L^2$, choose a [sequence](/page/Sequence) $\{f_m\} \subset L^1 \cap L^2$ with $f_m \to f$ in $L^2$. By the isometry from Step 1, \begin{align*} \|\hat{f}_m - \hat{f}_k\|_{L^2} = (2\pi)^{n/2}\,\|f_m - f_k\|_{L^2} \to 0 \quad \text{as } m,k \to \infty, \end{align*} so $\{\hat{f}_m\}$ is Cauchy in $L^2(\mathbb{R}^n)$. Since $L^2$ is complete, the sequence converges to some $F \in L^2(\mathbb{R}^n)$. [definition: Fourier Transform on L²] Let $f \in L^2(\mathbb{R}^n)$ and let $\{f_m\} \subset L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$ be any sequence with $f_m \to f$ in $L^2$. The **Fourier transform of $f$** is the $L^2$ function \begin{align*} \hat{f} := \lim_{m \to \infty} \hat{f}_m \quad \text{in } L^2(\mathbb{R}^n), \end{align*} where each $\hat{f}_m$ is defined by the integral formula. The limit is independent of the choice of approximating sequence. [/definition] **Well-definedness.** If $\{f_m'\}$ is another approximating sequence with $f_m' \to f$ in $L^2$, then $\|\hat{f}_m - \hat{f}_m'\|_{L^2} = (2\pi)^{n/2}\|f_m - f_m'\|_{L^2} \to 0$, so both sequences have the same $L^2$ [limit](/page/Limit). The definition depends only on $f$, not on the approximation. **Step 3: The isometry extends.** By [continuity](/page/Continuity) of the $L^2$ norm, \begin{align*} \|\hat{f}\|_{L^2} = \lim_{m \to \infty} \|\hat{f}_m\|_{L^2} = \lim_{m \to \infty} (2\pi)^{n/2}\|f_m\|_{L^2} = (2\pi)^{n/2}\|f\|_{L^2}. \end{align*} The rescaled operator $(2\pi)^{-n/2}\mathcal{F}$ is therefore an isometry of $L^2(\mathbb{R}^n)$ into itself. It is in fact **surjective** — hence a unitary isomorphism — because its range is closed and contains the dense subspace $\mathcal{F}(\mathcal{S}(\mathbb{R}^n)) = \mathcal{S}(\mathbb{R}^n)$. [quotetheorem:247] [quotetheorem:529] ## What $\hat{f}$ Is (and Is Not) For a general $f \in L^2$, the Fourier transform $\hat{f}$ is: - An **$L^2$ function**, defined up to modification on a [set](/page/Set) of $\mathcal{L}^n$-measure zero. It is not defined pointwise. - The **$L^2$-limit** of the sequence $\hat{f}_m = \int f_m(x) e^{-ix\cdot\xi}\,d\mathcal{L}^n(x)$ for any approximation $f_m \to f$ in $L^2$. In particular, $\hat{f}$ is **not** given by the integral $\int f(x) e^{-ix\cdot\xi}\,d\mathcal{L}^n(x)$, which may not converge. - **Consistent** with the integral definition when $f \in L^1 \cap L^2$: in that case the $L^2$-limit agrees $\mathcal{L}^n$-a.e. with the function defined by the integral. Because $\hat{f}$ is a genuine $L^2$ function (not a distribution), all $L^2$ operations on it are legitimate. In particular: - **Pointwise multiplication by measurable functions:** If $g: \mathbb{R}^n \to \mathbb{C}$ is measurable, the product $g \cdot \hat{f}$ is a well-defined measurable function. Expressions like $|\xi|^s \hat{f}(\xi)$ are meaningful, and the condition $|\xi|^s \hat{f} \in L^2(\mathbb{R}^n)$ is a well-posed [integrability](/page/Integral) condition. This is used in the definition of the [homogeneous Sobolev space](/page/Homogeneous%20Sobolev%20Space). - **Restriction to subsets:** The restriction $\mathbb{1}_E \hat{f}$ for any measurable $E \subset \mathbb{R}^n$ is in $L^2$, and by the Plancherel isomorphism there is a unique $L^2$ function whose Fourier transform is this restriction. This is the basis of the frequency decomposition used, for instance, in [Theorem 634](/theorems/634) (Gagliardo–Nirenberg–Sobolev interpolation). ## Frequency Truncation One of the most useful consequences of the $L^2$ theory is the ability to decompose a function into frequency bands. Given $f \in L^2(\mathbb{R}^n)$ and $N > 0$, define \begin{align*} \hat{f}_{\le N} := \mathbb{1}_{B(0,N)}\,\hat{f}, \qquad \hat{f}_{>N} := \mathbb{1}_{B(0,N)^c}\,\hat{f}. \end{align*} Both are in $L^2$, so by the Plancherel isomorphism there exist unique $f_{\le N}, f_{>N} \in L^2(\mathbb{R}^n)$ with these as their Fourier transforms. Since $\hat{f} = \hat{f}_{\le N} + \hat{f}_{>N}$ and Plancherel is injective, $f = f_{\le N} + f_{>N}$ in $L^2$. The low-frequency part $f_{\le N}$ has a compactly supported Fourier transform. Since $\hat{f}_{\le N}$ is supported in $B(0,N)$ and is in $L^2$, it is also in $L^1$ (by Cauchy–Schwarz, since $B(0,N)$ has finite measure). The [Fourier inversion formula](/theorems/528) therefore applies to $f_{\le N}$: \begin{align*} f_{\le N}(x) = \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} \hat{f}_{\le N}(\xi)\,e^{ix\cdot\xi}\,d\mathcal{L}^n(\xi), \end{align*} and $f_{\le N}$ is a bounded continuous function ($\|f_{\le N}\|_{L^\infty} \le (2\pi)^{-n}\|\hat{f}_{\le N}\|_{L^1} < \infty$). This recovers pointwise control from $L^2$ data — a key step in interpolation arguments. ## Relationship to Other Definitions The $L^2$ Fourier transform sits in a chain of successively larger domains: \begin{align*} \mathcal{S}(\mathbb{R}^n) \subset L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n) \subset L^2(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n). \end{align*} At each level, the Fourier transform is defined by a different method (integral, $L^2$-limit, duality), and each extension is consistent with the previous one. The $L^2$ definition described on this page is the middle extension. For the [distributional](/page/Distribution) extension to $\mathcal{S}'(\mathbb{R}^n)$ via the transposition formula $\hat{u}(\phi) = u(\hat{\phi})$, see the [Tempered Distributions](/page/Tempered%20Distributions) page and [Theorem 230](/theorems/230). ## References - Grafakos, L. *Classical Fourier Analysis*. 3rd ed. Springer, 2014. Chapter 2. - Stein, E. M. and Shakarchi, R. *Fourier Analysis: An Introduction*. Princeton University Press, 2003. Chapter 5. - Rudin, W. *Real and Complex Analysis*. 3rd ed. McGraw-Hill, 1987. Chapter 9.