Growth order of function.

  • #1
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Main Question or Discussion Point

let be the integral equation

[tex] h(x)= \int_{0}^{\infty} \frac{dy}{y}K(y/x)f(y) [/tex]

here the kernel is always a positive , then if [tex] h(x)=O(x^{a}) [/tex] and the integral


[tex] \int_{0}^{\infty} \frac{dy}{y}K(y)y^{a} [/tex] exists and is a positive real number then also [tex] f(x)= O(x^{a}) [/tex]
 

Answers and Replies

  • #2
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let be the integral equation

[tex] h(x)= \int_{0}^{\infty} \frac{dy}{y}K(y/x)f(y) [/tex]

here the kernel is always a positive , then if [tex] h(x)=O(x^{a}) [/tex] and the integral


[tex] \int_{0}^{\infty} \frac{dy}{y}K(y)y^{a} [/tex] exists and is a positive real number then also [tex] f(x)= O(x^{a}) [/tex]
I'll restrict to positive x as the negative case can be tackled similarly.
With a change of variable y/x =z , we get
[tex] \frac{h(x)}{x^{a}} = \int_{0}^{\inftz} \frac{dz}{z}K(z)z^{a}F(xz)[/tex]

where F(t) =f(t) /t^a. As [tex] h(x)=O(x^{a}) [/tex] , the right hand side is bounded,O(1).
I don't see why F =O(1) . F could tend to infinity much slower than
[tex] K(z)z^{a -1}[/tex] & the first equation could still hold.
 

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