advanced35 min studyLesson 374 in Mathematics

Fourier transform on Schwartz functions

Choose Euclidean Fourier normalization, justify differentiation and integration by parts, and calculate Gaussian transforms.

On this page

What you’ll understand

  • Define Schwartz decay through weighted derivatives.
  • Derive Fourier differentiation and modulation identities with correct signs.
  • Calculate the transform of a scaled Gaussian.

On the real line, frequencies form a continuum rather than an integer list. The Fourier transform expresses this change while keeping the same idea of comparison with oscillatory characters. To justify the operations cleanly, begin with functions whose derivatives all decay faster than any polynomial. This Schwartz class is small enough to permit rigorous manipulation and large enough to be dense in many useful spaces. We now use ordinary Lebesgue measure on Euclidean space and place two pi in the exponential; circle formulas in earlier lessons used a different frequency coordinate.

Rapid decay means every weighted derivative

A function belongs to S(Rd)\mathcal S(\mathbb R^d) when it is smooth and every quantity sup⁡x∣xα∂βf(x)∣\sup_x|x^\alpha\partial^\beta f(x)| is finite for multiindices α,β\alpha,\beta. This controls not only the function's tail but all derivative tails. Compactly supported smooth functions and Gaussians belong to the class. The function 1/(1+x2)1/(1+x^2) is smooth and integrable on the line but not Schwartz, because sufficiently large polynomial weights overwhelm its decay. Smoothness alone and rapid decay of the function alone are each weaker requirements.

These weighted bounds imply absolute integrability of all derivatives multiplied by any fixed polynomial. They allow differentiation under the Fourier integral and repeated integration by parts, with boundary terms vanishing. They also supply a natural topology: convergence in Schwartz space means convergence of every weighted-derivative supremum, not merely pointwise convergence. This stronger structure will later let distributions act continuously on test functions and makes the transform a continuous invertible map of the same test-function space.

f^(ξ)=∫Rdf(x)e−2πix⋅ξ dx,sup⁡x∣xα∂βf(x)∣<∞(α,β arbitrary)\widehat f(\xi)=\int_{\mathbb R^d}f(x)e^{-2\pi i x\cdot\xi}\,dx,\qquad\sup_x|x^\alpha\partial^\beta f(x)|<\infty\quad(\alpha,\beta\text{ arbitrary})
The integral uses Euclidean Lebesgue measure and the 2π frequency convention. The seminorm condition defines the Schwartz class.↔ Scroll wide equations horizontally.

Derivatives become frequency weights

Integrating a spatial derivative by parts moves the derivative onto the exponential, producing multiplication by 2πiξj2\pi i\xi_j. Differentiating with respect to frequency instead brings down −2πixj-2\pi i x_j. Repeating these identities bounds every weighted frequency derivative in terms of integrable weighted spatial derivatives. Consequently the Fourier transform maps Schwartz space to itself. The signs are not cosmetic: they determine translation direction and the correct multiplier for a differential operator.

A spatial translation multiplies the transform by a phase, while multiplication of the function by an oscillation shifts its transform. Scaling trades spatial width for frequency width with a volume prefactor. These rules explain how localized oscillatory signals can be analyzed without redoing every integral. They must be applied consistently with the chosen exponential sign and frequency normalization; a formula written for angular frequency has different factors of two pi. Verify one simple Gaussian or translated test function whenever conventions are changed.

∂xjf^(ξ)=2πiξjf^(ξ),∂ξjf^(ξ)=−2πixjf^(ξ)\widehat{\partial_{x_j}f}(\xi)=2\pi i\xi_j\widehat f(\xi),\qquad\partial_{\xi_j}\widehat f(\xi)=-2\pi i\widehat{x_jf}(\xi)
For Schwartz input the differentiation and integration-by-parts operations are justified by integrable weighted derivatives.↔ Scroll wide equations horizontally.

The Gaussian is a normalization check

In one dimension, the Gaussian e−πx2e^{-\pi x^2} transforms to itself. One derivation differentiates its transform, uses integration by parts to obtain a first-order differential equation, and fixes the constant with the Gaussian integral. Scaling then gives the transform of e−πax2e^{-\pi a x^2} for positive aa. The result has prefactor a−1/2a^{-1/2} and exponent −πξ2/a-\pi\xi^2/a. Concentrating the spatial Gaussian therefore broadens its frequency profile while changing its total mass.

The Schwartz calculation is a safe starting point, not a claim that every signal decays rapidly. Indicator functions, power-law tails, and point masses require extensions or different theorems. A differential identity derived in the Schwartz class can later extend by continuity to a larger space or hold in a distributional sense, but that extension must be named. Formal manipulation of a divergent integral is not justified by resemblance to the Gaussian computation. Retain the test class until a separate norm or duality argument supplies the broader meaning.

Scale a Gaussian, then differentiate

Let f(x)=exp(−4pi x²). Find its Fourier transform, its value at zero frequency, and the Fourier transform of f'.

  1. Use the one-dimensional Gaussian scaling formula with a=4. The transform is (1/2)exp(−pi xi²/4). The reciprocal-width prefactor is essential; it equals the spatial integral at zero frequency.

  2. Evaluate at xi=0 to obtain one half. This agrees with substituting u=2x in the normalized Gaussian integral, so the zero-frequency mass gives an independent check.

  3. Spatial differentiation has multiplier 2pi i xi. Multiplying the transform gives pi i xi exp(−pi xi²/4). The original derivative is −8pi x exp(−4pi x²), consistent with the frequency-differentiation identity and its sign.

The result

The transforms are (1/2)e^(−pi xi²/4) and pi i xi e^(−pi xi²/4), respectively. The derivative has zero transform at zero frequency because its spatial integral is zero.

Explore the model

Interactive lab

Interactive lab · heat

A derivative's frequency scale has a visible effect

Pause and Reset, then set both controls to mode n=1 and diffusivity κ=0.50. After one Step, time is 0.5 and the sine-mode amplitude is exp(−0.25). Reset and set n=2 and κ=0.50; after one Step predict exp(−1). Compare attenuation. This periodic analytic heat mode uses angular frequency n, not the Euclidean ξ coordinate of the Gaussian transform, and it is not a numerical PDE solver.

A derivative's frequency scale has a visible effectThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forTwo spatial derivatives contribute a squared frequency; frequency conventions determine which factors accompany that square.

Check your understanding

With transform exp(−2pi i x xi), what is the transform of a Schwartz function's derivative?

Try it yourself

Let g(x)=exp(2pi i 3x)exp(−pi x²). Find its transform and the frequency at which its magnitude is largest.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Combine the modulation with the Fourier exponential inside the integral before using the standard Gaussian transform.

Compare with the solution

The two oscillatory factors combine to exp(−2pi i x(xi−3)), so the transform is exp(−pi(xi−3)²). Its magnitude has maximum one at xi=3, because the Gaussian exponent is zero there and negative elsewhere. Modulation moves the frequency center; it does not translate the spatial Gaussian envelope. The result remains a Schwartz function because multiplying a Gaussian by a fixed smooth oscillation preserves all weighted derivative bounds.

Take it with you

  • Schwartz regularity controls every polynomially weighted derivative.
  • The exponential convention fixes all derivative and modulation signs.
  • Calculations on test functions need a separate justification before extending to rough inputs.

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