Fourier Transform

Consider the Gaussian function: \[ f(x) = e^{-\frac{x^2}{2\sigma^2}}. \] Its Fourier transform (using the mathematical convention) 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.

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 \(\mathcal{F}\{f^{(n)}\}(\xi) = (-i\xi)^n \hat{f}(\xi)\).

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.

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 requires additional machinery and 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 \(f \in \mathcal{S}(\mathbb{R})\).

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. Indeed, \(C_c^\infty(\mathbb{R}) \subset \mathcal{S}(\mathbb{R})\), and \(C_c^\infty(\mathbb{R})\) is dense in \(L^1 \cap L^2\) by the standard truncation-and-mollification argument (cut off outside a large ball, then convolve with a smooth bump). The same construction, applied to a fixed \(f \in L^1 \cap L^2\), produces an approximating sequence \(\{f_n\} \subset \mathcal{S}(\mathbb{R})\) converging to \(f\) simultaneously in both \(L^1\) and \(L^2\). That simultaneous convergence follows from the dominated convergence theorem applied to the cutoffs, combined with the standard \(L^p\)-continuity of mollification for \(1 \leq p \lt \infty\). 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 \(L^2\).

Using the mathematical convention 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.

Under sufficient regularity (for instance, \(f, g \in L^1\) with \(\hat f, \hat g \in L^1\)), 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: \[ \begin{align*} \mathcal{F}\{f \cdot g\}(\xi) &= \int_{-\infty}^{\infty} \left[\frac{1}{2\pi}\int_{-\infty}^{\infty} \hat f(\eta)\, e^{-ix\eta}\, d\eta\right] g(x)\, e^{ix\xi}\, dx \\\\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty} \hat f(\eta)\left[\int_{-\infty}^{\infty} g(x)\, e^{ix(\xi - \eta)}\, dx\right] d\eta \\\\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty} \hat f(\eta)\, \hat g(\xi - \eta)\, d\eta = \frac{1}{2\pi}(\hat f * \hat g)(\xi). \end{align*} \]

Each FNO layer performs:

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 local residual connection \(Wv\)
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.