Can Laplace Transform Prove f(t) = O(t) for a Given Integral Equation?

[mhill]

let be a function f(t) , and i want to prove that f(t)=O(t) in big-O notation.

i know that Laplace transform of f(t) is F(s) then i perform the integral

F(s)= \int_{0}^{\infty} dt f(t) e^{-st} if we assume f(t)=O(t) then

F(s)= \int_{0}^{\infty} dt f(t) e^{-st} \le \int_{0}^{\infty} dt e^{-st}t

so it would be enough that F(s) \le Cs^{-2} for a positive constant 'C'

is this enough ?

 

[HallsofIvy]

It is not clear what it is you are trying to prove. You have shown that if f(t)= O(t)[/itex] then F(t)= O(s^-2). You have NOT shown the converse: that if F(t)= O(s^-2) then f(t)= O(t).

 

[mhill]

sorry perhaps i did not explain myself well, the general idea y had is this given a function f(t) that is the solution to an integral equation

g(s)= \int_{0}^{\infty} K(s,t)f(t)

g(s) and K(s,t) are known , my question (more general than the previous one) is , would be a method to prove that f(t)= O(t) if we know that f(t) is the solution of certain integral equation ? , thanks.