Math 2.0

Harmonic analysis on Euclidean spaces Definition

Jackson kernel

Builds on Fejér kernelTrigonometric polynomial

Definition

Let n≥1n \ge 1 be an integer. The Jackson kernel of order nn is the function Jn(t)=λn(sin⁡(nt/2)sin⁡(t/2))4,t∈ℝ,J_n(t) = \lambda_n \left(\frac{\sin(nt/2)}{\sin(t/2)}\right)^4, \qquad t \in \R, with the value λnn4\lambda_n n^4 at the multiples of 2π2\pi. The constant λn>0\lambda_n > 0 is chosen so that 12π∫−ππJn(t) dt=1\frac{1}{2\pi} \int_{-\pi}^{\pi} J_n(t) \, dt = 1.

Remarks

Since (sin⁡(nt/2)sin⁡(t/2))2=nFn−1(t)\left(\frac{\sin(nt/2)}{\sin(t/2)}\right)^2 = n F_{n-1}(t), where Fn−1F_{n-1} is the Fejér kernel, JnJ_n is a constant times Fn−12F_{n-1}^2. So JnJ_n is an even, nonnegative trigonometric polynomial of degree 2n−22n - 2. Its key property is 12π∫−ππ∣t∣kJn(t) dt≤Cnk\frac{1}{2\pi} \int_{-\pi}^{\pi} \abs{t}^k J_n(t) \, dt \le \frac{C}{n^k} for k=1,2k = 1, 2, with CC independent of nn.

Example: for n=1n = 1, the quotient is 11, so J1=1J_1 = 1 and λ1=1\lambda_1 = 1.

Non-example: the Fejér kernel FnF_n is nonnegative with mean 11, but it fails the key property for k=1k = 1. Its first moment 12π∫∣t∣Fn(t) dt\frac{1}{2\pi} \int \abs{t} F_n(t) \, dt is of order log⁡nn\frac{\log n}{n}, not 1n\frac{1}{n}.

Texts differ in the indexing; some call JmJ_m with m≈n/2m \approx n/2 the kernel of degree nn. The convolution f↦f∗Jnf \mapsto f * J_n is the Jackson operator.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Jackson%27s_inequality
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