advanced39 min studyLesson 377 in Mathematics

Distributions and Fourier multipliers

Extend Fourier operations to tempered distributions and distinguish bounded L² symbols from unbounded differentiation.

On this page

What you’ll understand

  • Define the transform of a tempered distribution by duality.
  • Calculate delta and derivative transforms without divergent ordinary integrals.
  • State L² multiplier bounds and a sufficient smoothness criterion for other Lᵖ spaces.

A point source has no ordinary integrable density, yet it is one of the simplest inputs for a differential equation. Tempered distributions extend Fourier analysis by letting such objects act on rapidly decaying test functions. Fourier multipliers then describe translation-invariant operators through their frequency action. This language turns differentiation, smoothing, and singular transforms into related constructions, but it does not make them equally bounded. The relevant domain, output space, and symbol regularity must accompany each formal multiplier formula.

Transform the test function, not a nonexistent density

A tempered distribution is a continuous linear functional on Schwartz space, using bilinear rather than conjugate-linear test pairing here. Define its Fourier transform by ⟨T^,ϕ⟩=⟨T,ϕ^⟩\langle\widehat T,\phi\rangle=\langle T,\widehat\phi\rangle with the Euclidean negative-exponent convention. The Schwartz transform's continuity makes this another tempered distribution. For an ordinary integrable function embedded as a distribution, Fubini verifies agreement with its usual transform. This agreement is the reason for the dual definition, not an assumption that every distribution possesses an ordinary density.

The delta at zero evaluates a test function at zero, so its transform acts by integrating the test function and is the constant distribution one. Conversely the constant distribution transforms to a delta. Distributional differentiation is defined by moving a derivative onto the test function with a minus sign. Combining this rule with Fourier duality gives the familiar frequency multiplier for differentiation, now without requiring a classical derivative or an absolutely convergent transform integral.

⟨T^,ϕ⟩=⟨T,ϕ^⟩,δ0^=1,∂xjT^=2πiξjT^\langle\widehat T,\phi\rangle=\langle T,\widehat\phi\rangle,\qquad\widehat{\delta_0}=1,\qquad\widehat{\partial_{x_j}T}=2\pi i\xi_j\widehat T
The test functions are Schwartz and pairing is bilinear. The equalities are distributional, not unregularized ordinary integrals.↔ Scroll wide equations horizontally.

The symbol determines an L² operator

For a bounded measurable symbol mm, define Tmf=F−1(mf^)T_mf=\mathcal F^{-1}(m\widehat f) on L². Plancherel gives norm at most ∥m∥∞\|m\|_\infty, and testing on frequency sets where the symbol nearly reaches its essential supremum proves equality of operator norms. This definition uses multiplication of L² functions by a bounded measurable function. It should not be confused with multiplying an arbitrary distribution by an arbitrary measurable symbol, which is not generally defined.

A smooth symbol whose derivatives have at most polynomial growth does multiply Schwartz test functions continuously and therefore can multiply tempered distributions by duality. Polynomial symbols encode constant-coefficient differential operators, but they are usually unbounded on all of L². Their domains require enough frequency decay, as in Sobolev spaces. By contrast the heat multiplier at positive time is bounded and damps high frequencies. Writing both operations as multipliers clarifies their relationship while retaining the different domain and boundedness requirements.

Tmf^=mf^,∥Tm∥L2→L2=∥m∥L∞(m∈L∞)\widehat{T_mf}=m\widehat f,\qquad\|T_m\|_{L^2\to L^2}=\|m\|_{L^\infty}\quad(m\in L^\infty)
Essential supremum is taken in frequency Lebesgue measure. Polynomial differential symbols do not meet the bounded-symbol hypothesis.↔ Scroll wide equations horizontally.

Beyond L², boundedness is not enough

For Euclidean dimension dd, one convenient sufficient multiplier theorem assumes m∈Cd+2(Rd∖{0})m\in C^{d+2}(\mathbb R^d\setminus\{0\}) and uniformly bounded quantities ∣ξ∣∣α∣∣∂αm(ξ)∣|\xi|^{|\alpha|}|\partial^\alpha m(\xi)| for all multiindices of order at most d+2d+2. Then the associated operator is bounded on Lᵖ for every 1<p<∞1<p<\infty, with constants depending on these bounds, dimension, and p. This is a deliberately stronger-than-minimal form of the Hörmander–Mikhlin theorem; it supplies a safe usable hypothesis rather than the sharpest derivative count.

The theorem is not an endpoint assertion for L¹ or L∞, and merely bounded symbols need not be Lᵖ multipliers for other exponents. A proof decomposes the symbol into smooth frequency annuli, estimates their kernels, and uses singular-integral theory. That proof is deeper than the elementary Plancherel bound. When solving an equation by division in frequency, also check zeros of its symbol: division through a zero can introduce nonexistence, nonuniqueness, or compatibility conditions rather than a bounded inverse.

A periodic resolvent as a diagonal operator

On the normalized period-2pi circle, solve (1−d²/dx²)u=cos(3x)+2sin(x) among periodic distributions, and calculate the normalized squared L² norm of the solution.

  1. This is the circle analogue of a Euclidean multiplier. On exp(inx), the operator 1−d²/dx² has symbol 1+n², which is positive for every integer frequency and therefore has no zero-frequency obstruction.

  2. Divide the third-frequency cosine by ten and the first-frequency sine by two. The solution is u=(1/10)cos(3x)+sin(x). Substitution verifies the equation directly.

  3. Orthogonality makes its energy (1/10)²/2+1/2=101/200. The input energy is 1/2+4/2=5/2, consistent with the inverse symbol being at most one. Dividing each distributional coefficient by the nonzero symbol also proves uniqueness among periodic distributions.

The result

The resolvent attenuates the higher-frequency component more strongly. The solution's energy is 101/200 under normalized circle measure.

Explore the model

Interactive lab

Interactive lab · heat

A different smoothing multiplier

Pause and Reset, then set both controls to mode n=1 and diffusivity κ=0.50. After two Steps predict amplitude exp(−0.5). Reset and set n=3 and κ=0.50; after two Steps predict exp(−4.5). These are the periodic heat multipliers exp(−κn²t), not the resolvent multiplier 1/(1+n²) in the worked example. Compare their shared suppression of higher frequencies without identifying the two operators.

A different smoothing multiplierThe 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 forThe exact multiplier specifies the operator; similar smoothing behavior does not make two symbols interchangeable.

Check your understanding

Which hypothesis alone guarantees a Fourier multiplier is bounded on Euclidean L²?

Try it yourself

Let delta_a evaluate a Schwartz test function at the real point a. Find the Fourier transforms of delta_a and its distributional derivative under the exp(−2pi i x xi) convention.

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

Give me a hint

Evaluate the transformed test function at a, then use the derivative rule rather than treating the point mass as an ordinary function.

Compare with the solution

The transformed delta acts by integrating exp(−2pi i a xi) times the frequency test function, so its transform is the ordinary bounded oscillation exp(−2pi i a xi), regarded as a distribution. Differentiation multiplies this by 2pi i xi, giving the transform of delta_a prime. These expressions are meaningful as tempered distributions even though neither a point mass nor its derivative is an ordinary L¹ input. The derivative's transformed expression grows linearly, so it is not a bounded L² multiplier symbol.

Take it with you

  • Tempered Fourier transforms are defined through Schwartz test functions.
  • Bounded measurable symbols characterize bounded L² Fourier multipliers.
  • Other Lᵖ bounds and inverses require additional regularity or nonvanishing hypotheses.

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