Dirichlet and Fejér kernels
Express truncation and averaging as circle convolutions, and see how positivity changes stability and convergence.
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The question for today
What you’ll understand
- Derive the Dirichlet and Fejér frequency weights.
- Explain the significance of positivity and normalized kernel mass.
- Compute the difference between a partial sum and a Cesàro mean.
Keeping the first few frequencies is only one way to approximate a periodic function. Averaging several such truncations changes the approximation rule and replaces an oscillatory kernel by a positive one. Both constructions preserve the constant mode, but they behave differently near sharp features and under perturbations. Writing them as convolutions reveals the reason: unit signed mass does not control the absolute size of a kernel, whereas positivity turns unit mass into a direct averaging bound. We retain the circle measure dx divided by two pi throughout.
Truncation is convolution with Dirichlet's kernel
Define circle convolution by . For integrable , inserting the finite Fourier polynomial and interchanging a finite sum with the integral gives . Summing the geometric progression gives a sine quotient away from multiples of the period. The apparently undefined value at zero has the continuous extension , directly visible from the original finite sum.
Dirichlet's kernel integrates to one because its constant coefficient is one, but it takes both signs. Its absolute integral grows on the order of the logarithm of the cutoff, rather than remaining uniformly bounded. Thus the elementary convolution bound based on kernel absolute mass worsens with the number of frequencies. This explains a genuine obstacle to uniform control on all continuous or integrable inputs. It does not say that every particular smooth input is unstable, since smoothness and cancellation can yield much stronger information.
Averaging truncations creates positivity
The Fejér mean averages through , so a frequency with index appears in exactly of those sums when . Its coefficient weight is consequently . The corresponding kernel is a squared sine quotient divided by . Squaring makes it nonnegative, and its constant coefficient still makes its integral one. Notice the index convention: averaging through produces denominator .
For bounded inputs, a positive mass-one kernel bounds the magnitude of the output by the input supremum norm. For integrable inputs it also gives an L¹ contraction, by the integral triangle inequality and translation invariance of circle measure. These statements control differences of inputs because convolution is linear. The price is deliberate attenuation of the highest retained frequencies. A Fejér mean is generally not the orthogonal projection onto its frequency range, so improved averaging stability should not be confused with the best fixed-degree L² approximation property.
Mass concentrates without becoming a delta function
For every fixed neighborhood of zero on the circle, the Fejér kernel's integral outside that neighborhood tends to zero. The denominator in its explicit formula stays away from zero there, while the numerator is bounded and the factor in front shrinks. Positivity, unit mass, and this concentration show uniform convergence to every continuous periodic input: split the convolution error into a small neighborhood controlled by uniform continuity and a far region controlled by its shrinking mass.
This proof describes an approximate identity. It does not mean the kernels converge to an ordinary integrable function concentrated at one point; instead they approach the identity operation in specified function spaces, or a point mass in a weak measure sense. For Lᵖ inputs with 1≤p<∞, density and uniform convolution bounds provide norm convergence. Arbitrary bounded measurable functions need not have uniform convergence. A jump remains a place where the chosen averaging rule, the nearby limits, and the assigned point value must be distinguished.
Work it through
Averaging changes even a short polynomial
Let f(x)=cos(x)+cos(2x). Compare S2f and the mean sigma2f=(S0f+S1f+S2f)/3, including their values at zero.
The three partial sums are S0f=0, S1f=cos(x), and S2f=cos(x)+cos(2x). Adding them gives 2cos(x)+cos(2x).
Divide by three to get sigma2f=(2/3)cos(x)+(1/3)cos(2x). These are exactly the Fejér weights 1−1/3 and 1−2/3 for the two positive frequency magnitudes.
At zero the full degree-two projection has value two, while the averaged approximation has value one. This difference is not a numerical error: the two approximations are defined by different multipliers. For comparison the associated Fejér kernel has F2(0)=3 and F2(pi)=1/3.
S2 reproduces this degree-two polynomial exactly, whereas sigma2 still damps both modes. For this fixed polynomial the damping disappears as the averaging cutoff tends to infinity.
The living notebook
Explore the model
Interactive lab · spectral
Inspect a cutoff before averaging cutoffs
Pause and Reset, then set both controls to N=1 and decay p=1. At time zero note that only sin(x) is present. Reset and set N=2 and p=1; predict the added term sin(2x)/2. Averaging these two displayed truncations by hand would give sin(x)+sin(2x)/4. The scene displays ordinary finite sums, not a Fejér kernel or an automatic Cesàro average.
- 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 forAveraging approximation operators changes their frequency weights; it is not the same operation as simply choosing the largest cutoff.
Pause & reason
Check your understanding
Your turn
Try it yourself
For f(x)=exp(3ix), find sigma2f and sigma3f under the convention averaging S0 through SN. Compute the L² norm of f−sigma3f using normalized circle measure.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Ask how many of the averaged partial sums contain frequency three; a character has L² norm one.
Compare with the solution
None of S0, S1, or S2 contains frequency three, so sigma2f=0. Among S0 through S3 only S3 contains it, giving sigma3f=(1/4)exp(3ix). The error is (3/4)exp(3ix), whose L² norm is 3/4 and whose squared norm is 9/16. The ordinary projection S3f already equals f; the averaging rule attenuates a newly admitted frequency rather than reproducing it immediately.
Keep the idea
Take it with you
- Dirichlet convolution performs a sharp frequency truncation.
- Fejér averaging introduces positive kernels and tapered weights.
- Approximate-identity convergence must name the input class and topology.
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