Fourier Transform

Example: Gaussian Function

Consider the Gaussian function: \[ f(x) = e^{-\frac{x^2}{2\sigma^2}}. \] Its Fourier transform (using the convention of this site) is: \[ \begin{align*} \hat{f}(\xi) &= \int_{-\infty}^{\infty} e^{-\frac{x^2}{2\sigma^2}}e^{i\xi x} \, dx \\\\ &= \int_{-\infty}^{\infty} e^{-\frac{x^2}{2\sigma^2} + i\xi x} \, dx. \end{align*} \] Complete the square in the exponent: \[ \begin{align*} -\frac{x^2}{2\sigma^2} + i\xi x &= -\frac{1}{2\sigma^2}(x^2 - 2i\sigma^2\xi x) \\\\ &= -\frac{1}{2\sigma^2}((x - i\sigma^2\xi)^2 + \sigma^4\xi^2). \end{align*} \] Therefore: \[ \hat{f}(\xi) = e^{-\frac{\sigma^2\xi^2}{2}} \int_{-\infty}^{\infty} e^{-\frac{(x-i\sigma^2\xi)^2}{2\sigma^2}} \, dx. \] The integrand is entire and decays exponentially as \(|\mathrm{Re}\, z| \to \infty\), uniformly on horizontal strips of bounded imaginary part. Cauchy's theorem applied to a rectangular contour therefore justifies shifting the contour from \(\mathbb{R}\) to \(\mathbb{R} + i\sigma^2\xi\) without changing the value, since the vertical-side contributions vanish in the limit: \[ \begin{align*} \int_{-\infty}^{\infty} e^{-\frac{(x-i\sigma^2\xi)^2}{2\sigma^2}} \, dx &= \int_{-\infty}^{\infty} e^{-\frac{u^2}{2\sigma^2}} \, du \\\\ &= \sigma\sqrt{2\pi}. \end{align*} \] Thus: \[ \hat{f}(\xi) = \sigma\sqrt{2\pi} \cdot e^{-\frac{\sigma^2\xi^2}{2}}. \]

The transform of a Gaussian is again a Gaussian, with the width inverted. A function narrow in space has a wide transform, and conversely. The whole family is therefore preserved by the Fourier transform, and the single choice \(\sigma = 1\) gives a genuine eigenfunction, since \(e^{-x^2/2}\) transforms to \(\sqrt{2\pi}\, e^{-\xi^2/2}\). Normalizing by \(\sqrt{2\pi}\) therefore fixes \(e^{-x^2/2}\) outright, and the same unitary rescaling reappears in the discrete setting, where dividing the Fourier matrix by the square root of its size puts its eigenvalues on the unit circle. This behavior makes Gaussians fundamental in uncertainty principles, quantum mechanics, and signal processing.

Proof (the four properties):

Linearity is the linearity of the integral. The translation property follows from the substitution \(u = x - a\): \[ \int_{-\infty}^{\infty} f(x - a)\, e^{ix\xi}\, dx = \int_{-\infty}^{\infty} f(u)\, e^{i(u + a)\xi}\, du = e^{ia\xi} \hat{f}(\xi). \] Scaling follows from the substitution \(u = ax\). When \(a \lt 0\) this reverses the orientation of the line, and the reversal combines with the Jacobian into the single factor \(1/|a|\): \[ \int_{-\infty}^{\infty} f(ax)\, e^{ix\xi}\, dx = \frac{1}{|a|} \int_{-\infty}^{\infty} f(u)\, e^{iu\xi/a}\, du = \frac{1}{|a|} \hat{f}\left(\frac{\xi}{a}\right). \] Differentiation follows from integration by parts. The boundary term vanishes because \(f(x) \to 0\) as \(|x| \to \infty\), which leaves \[ \int_{-\infty}^{\infty} f'(x)\, e^{ix\xi}\, dx = -\int_{-\infty}^{\infty} f(x)\, i\xi\, e^{ix\xi}\, dx = -i\xi\, \hat{f}(\xi). \] Every derivative of a Schwartz function is again a Schwartz function, so the same step applies with \(f'\) in place of \(f\), and induction on \(n\) gives \(\mathcal{F}\{f^{(n)}\}(\xi) = (-i\xi)^n \hat{f}(\xi)\).

Proof:

Fix integers \(m, n \geq 0\). Because \(x \mapsto x^k f(x)\) lies in \(\mathcal{S}(\mathbb{R})\) and hence in \(L^1(\mathbb{R})\) for every integer \(k \geq 0\), the integral defining \(\hat{f}\) may be differentiated under the integral sign any number of times. This makes \(\hat{f}\) smooth and gives \[ \frac{d^n}{d\xi^n} \hat{f}(\xi) = \widehat{(ix)^n f}(\xi). \] Write \(h(x) = (ix)^n f(x)\), so that \(h \in \mathcal{S}(\mathbb{R})\) and \(\frac{d^n}{d\xi^n} \hat{f} = \hat{h}\). Applying the differentiation property to \(h\) gives \(\widehat{h^{(m)}}(\xi) = (-i\xi)^m \hat{h}(\xi)\), and therefore \[ \left| \xi^m \frac{d^n}{d\xi^n} \hat{f}(\xi) \right| = \bigl| \xi^m \hat{h}(\xi) \bigr| = \bigl| \widehat{h^{(m)}}(\xi) \bigr| \leq \| h^{(m)} \|_{L^1}. \] The right-hand side is finite and independent of \(\xi\), because \(h^{(m)}\) again lies in \(\mathcal{S}(\mathbb{R}) \subset L^1(\mathbb{R})\). Since \(m\) and \(n\) were arbitrary, \(\hat{f}\) satisfies every bound in the definition of the Schwartz class.

Proof (Schwartz class):

The theorem statement gives the inversion formula for the class \(\{f \in L^1(\mathbb{R}) : \hat f \in L^1(\mathbb{R})\}\). The proof we present here is restricted to the smaller Schwartz class \(\mathcal{S}(\mathbb{R}) \subset \{f \in L^1 : \hat f \in L^1\}\). The extension from \(\mathcal{S}(\mathbb{R})\) to the full \(L^1\) class indicated in the statement is discussed after the proof. The Schwartz-class technique is Gaussian regularization. We insert a Gaussian damping factor that makes Fubini applicable, then remove it in the limit.

Step 1: regularized inverse.
For \(\epsilon \gt 0\), define \[ I_\epsilon(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}(\xi)\, e^{-\epsilon \xi^2/2}\, e^{-ix\xi}\, d\xi. \] Since \(\hat f \in \mathcal{S}(\mathbb{R})\) by the invariance of the Schwartz class, the integrand is absolutely integrable for every \(\epsilon \geq 0\). In particular the integral converges and \(I_\epsilon(x)\) is well-defined.

Step 2: Fubini interchange.
We substitute \(\hat f(\xi) = \int f(y) e^{iy\xi}\, dy\). The resulting double integrand satisfies \[ \bigl|f(y)\, e^{iy\xi}\, e^{-\epsilon\xi^2/2}\, e^{-ix\xi}\bigr| = |f(y)|\, e^{-\epsilon\xi^2/2} \in L^1(\mathbb{R} \times \mathbb{R}), \] the product of an \(L^1\) function in \(y\) with an \(L^1\) function in \(\xi\). Therefore Fubini's theorem, applied to the real and imaginary parts of the integrand separately, permits the order swap: \[ \begin{align*} I_\epsilon(x) &= \int_{-\infty}^{\infty} f(y)\, K_\epsilon(y - x)\, dy, \\\\ K_\epsilon(z) &= \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{iz\xi}\, e^{-\epsilon\xi^2/2}\, d\xi. \end{align*} \] The kernel \(K_\epsilon\) is the forward Fourier transform of the Gaussian \(\xi \mapsto e^{-\epsilon\xi^2/2}\), evaluated at \(z\), divided by \(2\pi\). We reuse the Gaussian-transform calculation given earlier, with the roles of the space and frequency variables exchanged. Apply the formula \[ \widehat{e^{-x^2/(2\sigma^2)}}(\xi) = \sigma\sqrt{2\pi}\, e^{-\sigma^2 \xi^2/2} \] to the variable \(\xi\) in the role of \(x\), with \(\sigma^2 = 1/\epsilon\). This gives \[ \int e^{iz\xi}\, e^{-\epsilon\xi^2/2}\, d\xi = \sqrt{2\pi/\epsilon}\, e^{-z^2/(2\epsilon)}. \] Thus \[ K_\epsilon(z) = \frac{1}{\sqrt{2\pi\epsilon}}\, e^{-z^2/(2\epsilon)}, \] a Gaussian of unit mass: \(\int K_\epsilon(z)\, dz = 1\).

Step 3: convolution limit.
By the substitution \(z = y - x\), we have \(I_\epsilon(x) = \int f(x + z)\, K_\epsilon(z)\, dz\). Using \(\int K_\epsilon = 1\) to write \(f(x) = \int f(x)\, K_\epsilon(z)\, dz\), we obtain \[ I_\epsilon(x) - f(x) = \int_{-\infty}^{\infty} \bigl[ f(x + z) - f(x) \bigr]\, K_\epsilon(z)\, dz. \] Fix \(\eta \gt 0\). Every derivative of a Schwartz function is bounded, so \(|f(x + z) - f(x)| \leq \|f'\|_\infty\, |z|\) for all \(x\) and \(z\). Taking \(\delta = \eta / (1 + \|f'\|_\infty)\) therefore gives \(|f(x + z) - f(x)| \lt \eta\) whenever \(|z| \leq \delta\), uniformly in \(x\). Split the integral accordingly. On the inner region \(|z| \leq \delta\), positivity of \(K_\epsilon\) and unit mass give \[ \left| \int_{|z| \leq \delta} [f(x+z) - f(x)]\, K_\epsilon(z)\, dz \right| \leq \eta. \] On the outer region \(|z| \gt \delta\), the substitution \(u = z/\sqrt{\epsilon}\) shows \[ \int_{|z| \gt \delta} K_\epsilon(z)\, dz = \int_{|u| \gt \delta/\sqrt{\epsilon}} \frac{1}{\sqrt{2\pi}}\, e^{-u^2/2}\, du \longrightarrow 0 \quad \text{as } \epsilon \to 0, \] since the lower bound \(\delta/\sqrt{\epsilon} \to \infty\) and the integrand is the standard-normal density. Combined with the trivial bound \(|f(x+z) - f(x)| \leq 2 \sup_{\mathbb{R}} |f|\) on the outer region, \[ \limsup_{\epsilon \to 0}\, |I_\epsilon(x) - f(x)| \leq \eta. \] Since \(\eta \gt 0\) was arbitrary, \(\lim_{\epsilon \to 0} I_\epsilon(x) = f(x)\).

Step 4: spectral-side limit.
We return to the definition of \(I_\epsilon\). The integrand \[ \hat f(\xi)\, e^{-\epsilon \xi^2/2}\, e^{-ix\xi} \] is dominated by \(|\hat f(\xi)| \in L^1(\mathbb{R})\) uniformly in \(\epsilon\) and converges pointwise to \(\hat f(\xi)\, e^{-ix\xi}\) as \(\epsilon \to 0\). The dominated convergence theorem applies along any sequence \(\epsilon_j \to 0\), and the resulting value does not depend on the sequence chosen, so \[ \lim_{\epsilon \to 0} I_\epsilon(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \hat f(\xi)\, e^{-ix\xi}\, d\xi. \] Equating the two limits established in Steps 3 and 4 yields the inversion formula for every \(f \in \mathcal{S}(\mathbb{R})\).

Proof:

For a function \(h\) on \(\mathbb{R}\) and \(y \in \mathbb{R}\), write \((\tau_y h)(x) = h(x + y)\) for its translate. Translation invariance gives \(\|\tau_y h\|_p = \|h\|_p\). A step function is a finite linear combination of indicators of bounded intervals.

Step 1: step functions are dense in \(L^p\).
Let \(g \in L^p(\mathbb{R})\). Writing \(g = (\Re g)^+ - (\Re g)^- + i\,(\Im g)^+ - i\,(\Im g)^-\), each of the four parts is non-negative, Borel, and dominated by \(|g|\), so it suffices to treat \(0 \leq g \in L^p\). First, \(g\, \mathbf{1}_{[-M, M]} \to g\) in \(L^p\) as \(M \to \infty\) through the integers, by the dominated convergence theorem with dominator \(|g|^p\), so we may assume that \(g\) vanishes outside \([-M, M]\). Next, the staircase functions \[ s_n = \min\bigl( 2^{-n} \lfloor 2^n g \rfloor,\ n \bigr) \] are simple, satisfy \(0 \leq s_n \leq g\), and obey \(|s_n - g| \leq 2^{-n}\) wherever \(g \lt n\). Since \(g\) is finite almost everywhere, \(s_n \to g\) almost everywhere, and the dominated convergence theorem with dominator \(g^p\) gives \(\|s_n - g\|_p \to 0\). Each \(s_n\) is a finite combination \(\sum_j \alpha_j \mathbf{1}_{E_j}\) of indicators of Borel sets \(E_j \subseteq [-M, M]\).

It remains to approximate \(\mathbf{1}_E\) for a Borel set \(E \subseteq [-M, M]\). Fix \(\eta \gt 0\). Outer regularity gives an open \(U \supseteq E\) with \(|U \setminus E| \lt \eta\), and \(W = U \cap (-M - 1, M + 1)\) is open, contains \(E\), and has \(|W \setminus E| \lt \eta\) and \(|W| \leq 2M + 2\). Every open subset of \(\mathbb{R}\) is the disjoint union of its connected components, which are open intervals, and there are at most countably many of them because each contains a rational number. Write \(W = \bigsqcup_k I_k\) accordingly. By countable additivity \(\sum_k |I_k| = |W| \lt \infty\), so there is \(K\) with \(\sum_{k \gt K} |I_k| \lt \eta\). The set \(V = I_1 \sqcup \cdots \sqcup I_K\) satisfies \(V \triangle E \subseteq (W \setminus E) \cup (W \setminus V)\), hence \(|V \triangle E| \lt 2\eta\), and \(\mathbf{1}_V = \sum_{k \leq K} \mathbf{1}_{I_k}\) is a step function with \[ \|\mathbf{1}_V - \mathbf{1}_E\|_p = |V \triangle E|^{1/p} \lt (2\eta)^{1/p}. \] Replacing each \(\mathbf{1}_{E_j}\) in \(s_n\) this way and chaining the approximations proves Step 1.

Step 2: translation is continuous in \(L^p\).
We claim \(\|\tau_y g - g\|_p \to 0\) as \(y \to 0\) for every \(g \in L^p(\mathbb{R})\). For the indicator of a bounded interval \(I\) with endpoints \(a \lt b\), the function \(\tau_y \mathbf{1}_I - \mathbf{1}_I\) takes the values \(0\) and \(\pm 1\), the latter exactly on \((I - y) \triangle I\), a set of measure \(2 \min(|y|, b - a) \leq 2|y|\). Hence \(\|\tau_y \mathbf{1}_I - \mathbf{1}_I\|_p \leq (2|y|)^{1/p} \to 0\), and by the triangle inequality the claim holds for every step function. For general \(g\), fix \(\eta \gt 0\) and a step function \(s\) with \(\|g - s\|_p \lt \eta\) by Step 1. Then \[ \begin{align*} \|\tau_y g - g\|_p &\leq \|\tau_y (g - s)\|_p + \|\tau_y s - s\|_p + \|s - g\|_p \\\\ &\lt 2\eta + \|\tau_y s - s\|_p, \end{align*} \] so \(\limsup_{y \to 0} \|\tau_y g - g\|_p \leq 2\eta\) for every \(\eta \gt 0\).

Step 3: Gaussian smoothing converges in \(L^p\).
For \(\epsilon \gt 0\), let \[ K_\epsilon(z) = \frac{1}{\sqrt{2\pi\epsilon}}\, e^{-z^2/(2\epsilon)} \] be the kernel of the inversion proof, positive, even, and of unit mass. Let \(g \in L^p\) vanish outside \([-M, M]\), and assume in addition \(g \in L^1\). Define \[ \begin{align*} g_\epsilon(x) &= \int_{-\infty}^{\infty} g(y)\, K_\epsilon(x - y)\, dy \\\\ &= \int_{-\infty}^{\infty} g(x + z)\, K_\epsilon(z)\, dz, \end{align*} \] the integral converging absolutely for every \(x\) because \(g \in L^1\) and \(K_\epsilon \leq (2\pi\epsilon)^{-1/2}\). The second form comes from the translation \(y = x + z\) and the evenness of \(K_\epsilon\). Using unit mass to write \(g(x) = \int g(x)\, K_\epsilon(z)\, dz\), \[ g_\epsilon(x) - g(x) = \int_{-\infty}^{\infty} \bigl[ g(x + z) - g(x) \bigr]\, K_\epsilon(z)\, dz. \] For \(p = 1\) the triangle inequality for integrals, and for \(p = 2\) the Cauchy-Schwarz inequality applied to the factors \([g(x + z) - g(x)] \sqrt{K_\epsilon(z)}\) and \(\sqrt{K_\epsilon(z)}\), give in both cases \[ |g_\epsilon(x) - g(x)|^p \leq \int_{-\infty}^{\infty} |g(x + z) - g(x)|^p\, K_\epsilon(z)\, dz. \]

The function \((x, z) \mapsto |g(x + z) - g(x)|^p K_\epsilon(z)\) is non-negative and Borel on \(\mathbb{R}^2\), being built from \(g\) and continuous maps. The Borel sets of \(\mathbb{R}^2\) lie in the product \(\sigma\)-algebra of the Borel sets of \(\mathbb{R}\), since every open subset of \(\mathbb{R}^2\) is a countable union of open rectangles with rational corners. Integrating in \(x\) and applying, Lebesgue measure being \(\sigma\)-finite, Tonelli's theorem, \[ \begin{align*} \|g_\epsilon - g\|_p^p &\leq \int_{-\infty}^{\infty} \omega(z)\, K_\epsilon(z)\, dz, \\\\ \omega(z) &= \|\tau_z g - g\|_p^p. \end{align*} \]

Here \(\omega(z) \leq (\|\tau_z g\|_p + \|g\|_p)^p = 2^p \|g\|_p^p\) for every \(z\), and \(\omega(z) \to 0\) as \(z \to 0\) by Step 2. Fix \(\eta \gt 0\) and choose \(\delta \gt 0\) with \(\omega(z) \lt \eta\) for \(|z| \leq \delta\). Splitting the integral as in the inversion proof, \[ \|g_\epsilon - g\|_p^p \leq \eta + 2^p \|g\|_p^p \int_{|z| \gt \delta} K_\epsilon(z)\, dz, \] and the tail integral tends to \(0\) as \(\epsilon \to 0\), as computed in Step 3 of that proof. Hence \(\limsup_{\epsilon \to 0} \|g_\epsilon - g\|_p^p \leq \eta\) for every \(\eta \gt 0\), that is, \(g_\epsilon \to g\) in \(L^p\).

Step 4: the smoothed function is a Schwartz function.
Keep \(g \in L^1\) vanishing outside \([-M, M]\). The kernel \(K_\epsilon\) is a constant multiple of the Gaussian \(e^{-z^2/(2\sigma^2)}\) of the Gaussian example above with \(\sigma^2 = \epsilon\), hence a member of \(\mathcal{S}(\mathbb{R})\). Every derivative \(K_\epsilon^{(n)}\) is therefore bounded, and every quantity \(C_{m,n} = \sup_{u} |u|^m |K_\epsilon^{(n)}(u)|\) is finite. We claim \[ g_\epsilon^{(n)}(x) = \int_{-M}^{M} g(y)\, K_\epsilon^{(n)}(x - y)\, dy \quad \text{for all } n \geq 0. \] Assuming it for \(n\), the difference quotients of the integrand in \(x\) are bounded, by the mean value theorem, by \(|g(y)|\, \|K_\epsilon^{(n+1)}\|_\infty\), an integrable function of \(y\) independent of \(x\). The dominated convergence theorem, applied along any sequence of increments tending to zero, then differentiates under the integral sign and gives the case \(n + 1\). So \(g_\epsilon\) is smooth.

For the decay, let \(|x| \geq 2M\). Every \(y\) in the range of integration has \(|y| \leq M\), so \(|x - y| \geq |x| - |y| \geq |x|/2\), and \[ \begin{align*} |x^m g_\epsilon^{(n)}(x)| &\leq \int_{-M}^{M} |g(y)|\, 2^m |x - y|^m\, |K_\epsilon^{(n)}(x - y)|\, dy \\\\ &\leq 2^m C_{m,n}\, \|g\|_1. \end{align*} \] For \(|x| \lt 2M\) the same integral is at most \((2M)^m \|K_\epsilon^{(n)}\|_\infty \|g\|_1\). Thus \(\sup_x |x^m g_\epsilon^{(n)}(x)| \lt \infty\) for all \(m, n \geq 0\), and \(g_\epsilon \in \mathcal{S}(\mathbb{R})\).

Step 5: assembly.
Let \(f \in L^p(\mathbb{R})\) and set \(f_M = f\, \mathbf{1}_{[-M, M]}\). As in Step 1, \(\|f - f_M\|_p \to 0\) as \(M \to \infty\). Each \(f_M\) lies in \(L^1\). For \(p = 2\) this follows from the Cauchy-Schwarz inequality, which gives \(\int |f_M| \leq (2M)^{1/2} \|f\|_2\). Steps 3 and 4 therefore apply to \(g = f_M\), and \((f_M)_\epsilon \in \mathcal{S}(\mathbb{R})\) converges to \(f_M\) in \(L^p\) as \(\epsilon \to 0\). Choose \(M_n\) with \(\|f - f_{M_n}\|_p \lt 1/n\), then \(\epsilon_n\) with \(\|(f_{M_n})_{\epsilon_n} - f_{M_n}\|_p \lt 1/n\), and set \(f_n = (f_{M_n})_{\epsilon_n}\). Then \(\|f_n - f\|_p \lt 2/n\).

The construction \(f \mapsto (f_M)_\epsilon\) does not depend on \(p\). If \(f \in L^1 \cap L^2(\mathbb{R})\), both requirements on \(M_n\) hold for all large \(M\), and both requirements on \(\epsilon_n\) hold for all small \(\epsilon\), so \(M_n\) and \(\epsilon_n\) can be chosen to satisfy the \(L^1\) and \(L^2\) conditions at once. The resulting sequence converges to \(f\) in both norms simultaneously.

Proof:

We prove the identity in two stages: first on the Schwartz class \(\mathcal{S}(\mathbb{R})\), then by density on all of \(L^1 \cap L^2(\mathbb{R})\). It suffices to prove the polarized form. The special case \(f = g\) follows by setting them equal.

Stage 1: identity on the Schwartz class.
Fix \(f, g \in \mathcal{S}(\mathbb{R})\). By the invariance of the Schwartz class, \(\hat f, \hat g \in \mathcal{S}(\mathbb{R})\) as well. Apply the Fourier inversion theorem (proved above) to \(g\): \[ g(x) = \frac{1}{2\pi}\int \hat g(\xi)\, e^{-ix\xi}\, d\xi. \] Taking the complex conjugate of both sides, and using \(\overline{e^{-ix\xi}} = e^{+ix\xi}\), \[ \overline{g(x)} = \frac{1}{2\pi}\int \overline{\hat g(\xi)}\, e^{ix\xi}\, d\xi. \] Multiplying by \(f(x)\) and integrating in \(x\), \[ \int f(x)\overline{g(x)}\, dx = \frac{1}{2\pi}\iint f(x)\, e^{ix\xi}\, \overline{\hat g(\xi)}\, dx\, d\xi. \] To justify the order swap in the double integral, we check the integrability hypothesis of Fubini's theorem. The absolute value of the integrand is \(|f(x)| \cdot |\hat g(\xi)|\), and Tonelli's theorem applied to this non-negative measurable function gives \[ \iint |f(x)| \cdot |\hat g(\xi)|\, dx\, d\xi = \|f\|_1 \cdot \|\hat g\|_1 \lt \infty, \] since both \(f\) and \(\hat g\) are Schwartz, hence in \(L^1(\mathbb{R})\). Therefore the integrand is in \(L^1(\mathbb{R} \times \mathbb{R})\), and Fubini's theorem, again applied to the real and imaginary parts separately, permits the order swap: \[ \begin{align*} \frac{1}{2\pi}\iint f(x)\, e^{ix\xi}\, \overline{\hat g(\xi)}\, dx\, d\xi &= \frac{1}{2\pi}\int \overline{\hat g(\xi)} \left( \int f(x)\, e^{ix\xi}\, dx \right) d\xi \\\\ &= \frac{1}{2\pi}\int \hat f(\xi)\, \overline{\hat g(\xi)}\, d\xi, \end{align*} \] where the inner integral is \(\hat f(\xi)\) by definition. This establishes the polarized identity on \(\mathcal{S}(\mathbb{R})\).

Stage 2: extension to \(L^1 \cap L^2(\mathbb{R})\).
Equip \(L^1 \cap L^2(\mathbb{R})\) with the \(L^2\) norm. Then \(\mathcal{S}(\mathbb{R})\) is a dense subspace, by the density lemma proved above. The lemma gives more, and the argument uses it. For a fixed \(f \in L^1 \cap L^2\), it produces an approximating sequence \(\{f_n\} \subset \mathcal{S}(\mathbb{R})\) converging to \(f\) simultaneously in both \(L^1\) and \(L^2\). We fix such a sequence for the remainder of the proof.

Stage 1 specialized to \(f = g\) gives \[ \|f\|_2^2 = \frac{1}{2\pi}\|\hat f\|_2^2 \] for every \(f \in \mathcal{S}(\mathbb{R})\). Equivalently, the linear map \(T : \mathcal{S}(\mathbb{R}) \to L^2(\mathbb{R})\), \(Tf = \hat f / \sqrt{2\pi}\), is an isometry from \((\mathcal{S}(\mathbb{R}), \|\cdot\|_2)\) into \(L^2(\mathbb{R})\). From this, two observations about our sequence \(\{f_n\}\) follow:

• The sequence \(\hat{f_n}\) is Cauchy in \(L^2\).
  By Stage 1 applied to the differences \(f_n - f_m \in \mathcal{S}(\mathbb{R})\), \(\|\hat{f_n} - \hat{f_m}\|_2 = \sqrt{2\pi}\, \|f_n - f_m\|_2 \to 0\). Hence by completeness of \(L^2\), \(\hat{f_n}\) converges to some \(F \in L^2(\mathbb{R})\).
• The limit \(F\) coincides with the classical Fourier transform \(\hat f\).
  Since \(f \in L^1\), the integral \(\hat f(\xi) = \int f(x)\, e^{ix\xi}\, dx\) is absolutely convergent and defines \(\hat f \in L^\infty(\mathbb{R})\). The \(L^1\)-convergence \(f_n \to f\) (which our sequence provides) gives \(\hat{f_n} \to \hat f\) uniformly on \(\mathbb{R}\), by the elementary estimate \[ \begin{align*} \|\hat{f_n} - \hat f\|_\infty &= \sup_\xi \left|\int (f_n - f)(x)\, e^{ix\xi}\, dx\right| \\\\ &\leq \|f_n - f\|_1 \to 0. \end{align*} \] Combined with \(\hat{f_n} \to F\) in \(L^2\), passing to a subsequence (using subsequential a.e. convergence for \(L^2\) convergence) yields pointwise a.e. convergence to both \(F\) and \(\hat f\), so \(F = \hat f\) a.e.

The \(L^2\) norm is continuous, so we may take the limit \(n \to \infty\) in the Stage 1 identity \(\|f_n\|_2^2 = \tfrac{1}{2\pi}\|\hat{f_n}\|_2^2\). This yields \[ \|f\|_2^2 = \tfrac{1}{2\pi}\|\hat f\|_2^2. \] The polarized form follows by the same argument applied to a second sequence \(\{g_n\} \subset \mathcal{S}(\mathbb{R})\) approximating \(g \in L^1 \cap L^2\) simultaneously in \(L^1\) and \(L^2\). The bilinear identity \[ \int f_n \overline{g_n}\, dx = \tfrac{1}{2\pi}\int \hat{f_n}\, \overline{\hat{g_n}}\, d\xi \] passes to the limit on both sides by continuity of the \(L^2\) inner product. Here \(f_n \to f\) and \(g_n \to g\) in \(L^2\), and the same argument used above for \(\hat f\) gives \(\hat{f_n} \to \hat f\) and \(\hat{g_n} \to \hat g\) in \(L^2\).

Proof (convolution to product):

Using the convention of this site with \(\hat{f}(\xi) = \int f(x)e^{ix\xi} \, dx\): \[ \begin{align*} \mathcal{F}\{f * g\}(\xi) &= \int_{-\infty}^{\infty} (f * g)(x)e^{ix\xi} \, dx \\\\ &= \int_{-\infty}^{\infty} \left(\int_{-\infty}^{\infty} f(y)g(x-y) \, dy\right) e^{ix\xi} \, dx \\\\ &= \int_{-\infty}^{\infty} f(y) \left(\int_{-\infty}^{\infty} g(x-y)e^{ix\xi} \, dx\right) dy. \end{align*} \] Substituting \(u = x - y\) (so \(x = u + y\) and \(dx = du\)): \[ \begin{align*} &= \int_{-\infty}^{\infty} f(y) \left(\int_{-\infty}^{\infty} g(u)e^{i(u+y)\xi} \, du\right) dy \\\\ &= \int_{-\infty}^{\infty} f(y)e^{iy\xi} \left(\int_{-\infty}^{\infty} g(u)e^{iu\xi} \, du\right) dy \\\\ &= \left(\int_{-\infty}^{\infty} f(y)e^{iy\xi} \, dy\right) \cdot \left(\int_{-\infty}^{\infty} g(u)e^{iu\xi} \, du\right) \\\\ &= \hat{f}(\xi) \cdot \hat{g}(\xi). \end{align*} \] The interchange of integration order is justified by Fubini's theorem when \(f, g \in L^1(\mathbb{R})\), applied to the real and imaginary parts of the integrand separately.

Proof (product to convolution):

Under sufficient regularity (for instance, \(f, g \in L^1\) with \(\hat f \in L^1\)), the inversion formula gives \[ f(x) = \frac{1}{2\pi}\int \hat f(\eta) e^{-ix\eta}\, d\eta. \] Substituting into the Fourier transform of \(fg\) and applying Fubini's theorem to the real and imaginary parts separately: \[ \begin{align*} \mathcal{F}\{f \cdot g\}(\xi) &= \int_{-\infty}^{\infty} \left[\frac{1}{2\pi}\int_{-\infty}^{\infty} \hat f(\eta)\, e^{-ix\eta}\, d\eta\right] g(x)\, e^{ix\xi}\, dx \\\\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty} \hat f(\eta)\left[\int_{-\infty}^{\infty} g(x)\, e^{ix(\xi - \eta)}\, dx\right] d\eta \\\\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty} \hat f(\eta)\, \hat g(\xi - \eta)\, d\eta = \frac{1}{2\pi}(\hat f * \hat g)(\xi). \end{align*} \]

FNO Architecture Sketch

An FNO performs the following steps, repeating steps 2 to 4 in each Fourier layer:

1. Lift input to higher-dimensional channel space
2. Apply spectral convolution: \(\mathcal{F}^{-1}(R \cdot \mathcal{F}(v))\) where \(R\) is a learnable weight matrix
3. Add the pointwise linear term \(Wv\)
4. Apply non-linearity (activation function), for example GELU
5. Project to output

The spectral convolution step is where the convolution theorem provides the key efficiency gain.