# Growth order of function.

• zetafunction

#### zetafunction

let be the integral equation

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

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

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

zetafunction said:
let be the integral equation

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

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

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

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