growth order of function.

[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})

[Eynstone]

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.