Hilbert transform and singular integrals
Use principal-value cancellation and a signed frequency multiplier while respecting endpoint failures and periodic zero modes.
On this page
The question for today
What you’ll understand
- Explain why a principal value differs from an absolute integral.
- Compute the Hilbert transform on elementary periodic modes.
- Distinguish real-line L² identities from the circle's constant-mode correction.
Some useful operators have kernels too singular for an ordinary absolute convolution bound. The Hilbert transform is a central example: its kernel has a reciprocal singularity, but opposing sides cancel when the truncation is symmetric. Its Fourier multiplier rotates positive and negative frequencies in opposite directions, linking real-variable estimates with complex phase. The operator is bounded on many Lebesgue spaces but fails at the strong endpoints. We will keep its sign convention explicit and distinguish the real line from the periodic circle, where the constant frequency needs separate treatment.
Cancellation replaces absolute integrability
On the line, for a Schwartz function define by the limit of as decreases to zero. Near the singularity, pair positive and negative y to obtain a numerator that vanishes linearly. Smoothness controls that difference, while rapid decay handles the far tail. The singular kernel itself is not locally absolutely integrable, so applying Young's L¹-kernel bound is not legitimate.
The principal value specifies symmetric exclusion around the singularity. It is not the same as assigning separate values to two divergent one-sided integrals and then subtracting them. Changing a truncation convention without compensating terms can alter the operator. For rough inputs, bounded extension and convergence of truncated formulas require additional theorems; the transparent Schwartz calculation is a safe initial domain. Cancellation is the structural reason the operator can act stably despite the pointwise singularity of its kernel.
Frequency rotation gives an L² identity
The multiplier has magnitude one away from zero, so Plancherel extends H as an isometry of . Its square is minus the identity there, since the single frequency zero has Lebesgue measure zero and does not affect an L² multiplier. On the period-two-pi circle, define the multiplier at integer n with zero value at n=0. The constant mode is now an actual basis vector rather than a negligible frequency point.
Consequently the periodic transform annihilates constants and its square equals minus the identity plus projection onto the mean. Positive-frequency cosines become sines, while positive-frequency sines become negative cosines. This phase rule preserves the energy of the mean-zero part, not the total energy of an arbitrary periodic input. Keeping the zero mode visible prevents a common mistake: applying the real-line square identity to a constant periodic function and claiming its transform can recover a constant that was removed.
The interior exponent range matters
Singular-integral theory gives bounded extensions on real-line Lᵖ for exponents strictly between one and infinity, together with a weak type (1,1) endpoint estimate. The transform is not generally bounded from L¹ to L¹ or from L∞ to L∞. These are not small technical omissions: simple interval inputs produce reciprocal tails and logarithmic singularities. Their local and distant behavior explain why a bounded input can generate an unbounded output and an integrable input can generate a nonintegrable absolute tail.
More general singular-integral operators require hypotheses on kernel size, regularity away from the diagonal, and a boundedness or cancellation structure that makes the initial operator meaningful. A reciprocal-looking formula alone does not imply the full Calderón–Zygmund conclusion. In proofs, decomposing an input into controlled and exceptional pieces ties these operators to maximal weak bounds. In applications, check the operator's actual domain and truncation, then identify whether the desired output estimate is strong, weak, or in a replacement endpoint space.
Work it through
A periodic phase transform with a mean
On the normalized circle let f=5+3cos(x)−2sin(2x). Find its periodic Hilbert transform, the energy of that transform, and H squared applied to f.
The constant five is annihilated. Applying the positive-frequency mode rules gives Hf=3sin(x)+2cos(2x), with the second sign changing because H of a sine is its negative cosine.
Orthogonality yields transformed energy 9/2+4/2=13/2. The original energy is 25+13/2=63/2, so the missing energy is precisely the squared mean 25.
Apply H again to obtain −3cos(x)+2sin(2x). This is −f+5, agreeing with the periodic identity that restores the mean after negating the input.
The periodic Hilbert transform is an isometry on mean-zero L² functions, and the constant-mode correction is necessary on all periodic inputs.
The living notebook
Explore the model
Interactive lab · waves
Inspect a quadrature phase shift
Pause and Reset, then set both controls to phase difference −90° and amplitude ratio 1. At time zero the first component is sin(2x) and the second is −cos(2x), the Hilbert-transform phase rule for that sine. Reset and set +90° and ratio 1; the second component becomes +cos(2x), the opposite sign. Inspect the two components, not just their sum. The scene illustrates a mode's phase relation and does not evaluate a singular integral on arbitrary input.
- Sum at x = 0
- 0.866
- Resultant amplitude
- 1.732
- Phase difference
- 60°
Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). 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 sign in the multiplier corresponds to a definite phase direction; matching amplitudes alone cannot determine it.
Pause & reason
Check your understanding
Your turn
Try it yourself
For f the indicator of [0,1] on the real line, calculate Hf(x) for x outside the endpoints using principal value when necessary. Evaluate at x=2 and x=1/2, and identify one endpoint failure of a strong norm bound.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use the equivalent integral (1/pi) times the integral from zero to one of 1/(x−t). Keep absolute values in the logarithms.
Compare with the solution
Integration gives Hf(x)=(1/pi)log|x/(x−1)| for x not equal to zero or one; symmetric principal value gives the same formula inside the interval. At two the value is log(2)/pi, and at one half it is zero. Near either endpoint its magnitude grows logarithmically, so this bounded input has an output not in L∞. Far away the transform behaves like 1/(pi x), whose absolute integral diverges, also demonstrating failure of a strong L¹-to-L¹ bound. These failures coexist with interior Lᵖ boundedness.
Keep the idea
Take it with you
- Principal values encode cancellation rather than absolute integrability.
- The Fourier multiplier fixes the Hilbert transform's phase signs.
- Endpoint and zero-frequency distinctions are essential to correct operator identities.
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