Modulus of smoothness: Difference between revisions

Revision as of 06:46, 11 April 2018

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==Moduli of smoothness==

The modulus of smoothness of order <math>n</math> <ref>DeVore, Ronald A., Lorentz, George G., Constructive approximation, Springer-Verlag, 1993.</ref>

of a function <math>f\in C[a,b]</math> is the function <math>\omega_n:[0,\infty)\to\mathbb{R}</math> defined by

<ref>DeVore, Ronald A., Lorentz, George G., Constructive approximation, Springer-Verlag, 1993.</ref>

of a function <math>f\in C[a,b]</math> is the function <math>\omega_n:[0,\infty)\to\mathbb{R}</math> defined by

:<math>\omega_n(t,f,[a,b])=\sup_{h\in[0,t]}\sup_{x\in[a,b-nh]} \left |\Delta_h^n(f,x) \right | \qquad \text{ for } \quad 0\le t\le \frac{b-a} n,</math>

and

:<math>\omega_n(t,f,[a,b])=\omega_n\left(\frac{b-a}{n},f,[a,b]\right), \qquad \text{ for } \quad t>\frac{b-a}{n},</math>

where the [[finite difference]] (''n''-th order forward difference) is defined as

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==Properties==

1. *<math>\omega_n(0)=0,</math> <math>\omega_n(0+)=0.</math>

2. *<math>\omega_n</math> is non-decreasing on <math>[0,\infty).</math>

3. *<math>\omega_n</math> is continuous on <math>[0,\infty).</math>

4. <math>\omega_n(mt)\leq m^n\omega_n(t)</math>,*For <math>m\in\mathbb{N}</math>, <math>t\geq0.geq 0</math> we have:

::<math>\omega_n(mt)\leq m^n\omega_n(t).</math>

5. *<math>\omega_n(f,\lambda t)\leq (\lambda +1)^n\omega_n(f,t),</math>, for <math>\lambda>0.</math>

6. *For <math>r\in \N</math>, denote bylet <math>W^r</math> denote the space of continuous function on <math>[-1,1]</math> that have <math>(r-1)</math>-st absolutely continuous derivative on <math>[-1,1]</math> and
::<math>\left \|f^{(r)} \right \|_{L_{\infty}[-1,1]}<+\infty.</math>
:If <math>f\in W^r,</math>, then
::<math>\omega_r(t,f,[-1,1])\leq t^r \left \|f^{(r)} \right \|_{L_{\infty}[-1,1]}, t\geq 0,</math>
:where <math>\|g(x)\|_{L_{\infty}[-1,1]}={\mathrm{ess} \sup}_{x\in [-1,1]}|g(x)|.</math>

==Applications==

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For every natural number <math>n</math>, if <math>f</math> is <math>2\pi</math>-periodic continuous function, there exists a [[trigonometric polynomial]] <math>T_n</math> of degree <math>\le n</math> such that

:<math>\left |f(x)-T_n(x \right )|\leq c(k)\omega_k\left(\frac{1}{n},f\right),\quad x\in[0,2\pi],</math>

where the constant <math>c(k)</math> depends on <math>k\in\mathbb{N}.</math>

==References==