Littlewood–Paley decompositions
Organize smooth frequency bands into a square function and distinguish exact L² orthogonality from Lᵖ norm equivalence.
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The question for today
What you’ll understand
- Construct a smooth dyadic partition away from zero frequency.
- State square-function norm equivalence with its exponent and cutoff hypotheses.
- Use band energies to understand cancellation and derivative weights.
A single sharp frequency cutoff asks which oscillations are retained. A Littlewood–Paley decomposition asks how a function is distributed across many overlapping frequency scales. Its square function keeps the sizes of band components without allowing them to cancel one another pointwise. The resulting norm is comparable to the original Lᵖ norm in a substantial interior exponent range. This is not an assertion that the two functions are pointwise equal, nor that arbitrary frequency bins give the same theorem. Smoothness, bounded overlap, and a partition that does not miss frequencies are part of the construction.
Build bands with controlled overlap
Choose a smooth radial nonincreasing cutoff on , equal to one on the unit ball and zero outside the ball of radius two, with values between zero and one. Define for integers j. Its support is contained in an annulus with radii and . Telescoping shows that these nonnegative symbols sum to one at every nonzero frequency. Only a bounded number overlap at any such frequency.
Let be the Fourier multiplier with symbol . For L² inputs, finite sums of the bands reconstruct the input in L² as their low and high endpoints tend to opposite infinities, by Plancherel and dominated convergence. Zero frequency is negligible for Euclidean L². On general tempered distributions, a homogeneous construction needs care with polynomials supported at zero in frequency. An inhomogeneous decomposition instead retains one low-frequency block explicitly; its reconstruction and function-space definitions avoid silently throwing away that information.
Take squares before combining scales
Define the square function . In L², integrating the square and using Plancherel turns the estimate into comparison of with one. Finite overlap and a partition summing to one bound that frequency expression above and below by positive constants. This is almost orthogonality: smooth overlapping bands give norm comparison, whereas disjoint orthogonal spectral projections can give exact energy equality.
The Littlewood–Paley theorem extends the two-sided norm comparison to every Euclidean Lᵖ space with exponent strictly between one and infinity for the fixed smooth dyadic family described here. The constants depend on p, dimension, and cutoff bounds. One proof uses uniform estimates for signed sums of the band multipliers and an inequality for random signs, together with duality. Another uses vector-valued singular-integral estimates. Neither proof is a direct pointwise identity, and neither grants the same strong bounds at both endpoints.
Differentiate one scale at a time
A spatial derivative multiplies frequency by its size, so on a band near scale two to the j it contributes roughly that factor. Weighted band energies therefore describe Sobolev regularity and distinguish smooth low-frequency behavior from expensive high-frequency variation. The word roughly is deliberate: an annulus contains a range of magnitudes, so replacing every frequency by the band's nominal scale gives comparable quantities, not exact equality. At the low-frequency end homogeneous norms require their usual interpretation and finiteness assumptions.
This framework is useful when an operator acts differently on different scales or when nonlinear products mix low and high frequencies. Before applying a square-function bound, identify the particular smooth family and verify uniform bounds after dilation. A finite plot of several sine terms can reveal cancellation and show which increment belongs to a frequency band, but it does not prove the infinite-dimensional Lᵖ theorem. Keep the exact finite calculation, the L² comparison proof, and the deeper Lᵖ extension as separate levels of evidence.
Work it through
An exact orthogonal toy square function
On the normalized circle, put frequencies |n|≤1 in B0, frequencies 2≤|n|≤3 in B1, and frequencies 4≤|n|≤7 in B2. For f=2sin(x)+sin(3x)+(1/2)cos(6x), compute its energy and compare its value with the square function at pi/2.
The three disjoint spectral projections are 2sin(x), sin(3x), and (1/2)cos(6x). Orthogonality gives total energy 4/2+1/2+(1/4)/2=21/8. Integrating their pointwise squared magnitudes gives exactly the same energy.
At pi/2, the three components equal 2, −1, and −1/2. Their ordinary sum is 1/2 because they cancel.
At that point the toy square function is the square root of 4+1+1/4, or √21/2, approximately 2.2913. It does not share the cancellation of the signed sum. These disjoint finite circle projections illustrate exact L² orthogonality; they are not the smooth Euclidean family used in the full Lᵖ theorem.
The integrated squared energies agree exactly, while pointwise values differ greatly. A square function measures scale sizes without claiming to reproduce the signal pointwise.
The living notebook
Explore the model
Interactive lab · spectral
Inspect a high-frequency increment
Pause and Reset, then set both controls to N=3 and decay p=1. At time zero predict value 2/3 at x=π/2. Reset and set N=6 and p=1; the increment from frequencies 4 through 6 has value 1/5 there, giving total 13/15. Inspect the additional small-scale oscillations. The scene sums specified modes rather than constructing smooth dyadic blocks or evaluating a square function.
- Sum at x = π/2
- 0.867
- Smallest wavelength
- 2π / 5
- Highest-mode coefficient
- 0.2
A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. 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 forA band increment can cancel or reinforce at a point, while its own squared energy remains nonnegative.
Pause & reason
Check your understanding
Your turn
Try it yourself
Using the worked example's circle bands, take f=cos(2x)+cos(5x). Find its energy, its derivative energy, and the weighted band energy obtained by weighting B1 by 2² and B2 by 4².
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Each cosine has normalized energy one half. A derivative weights the actual frequency squared; the band estimate uses a representative scale instead.
Compare with the solution
The input energy is 1/2+1/2=1. Differentiating gives −2sin(2x)−5sin(5x), whose energy is (4+25)/2=29/2. The representative-scale band expression is 4 times 1/2 plus 16 times 1/2, equal to ten. It differs from 29/2 because frequency five was represented by the band's lower scale four. Such weighted band quantities can be comparable to derivative energy under appropriate annular support bounds, without being exactly equal term by term.
Keep the idea
Take it with you
- Smooth dyadic partitions resolve scale while controlling overlap and derivatives.
- Square functions remove inter-band cancellation and give interior Lᵖ norm equivalence.
- Representative frequency scales produce comparability, not universal exact identities.
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