An exercise in the book stated "Find $2$ functions $f_1 \not =f_2$ such that $L\{f_1\}=L\{f_2\}$. Now I think I know the answer ($f_1=f_2$ except at any number of discrete points) but previously I was so convinced that this can't happen that I wanted to see what happens if I try to prove that the Laplace transformation is one-to-one. Here is what I came up with:
Let $f'(t)$ and $g'(t)$ be functions such that $L(f'(t))=L(g'(t))$. In order to prove that the LaPlace transform is one-to one (at least for most "nice" functions), we need to prove that $f'(t)=g'(t)$
Let $F(s)=L(f(t))$ and $G(s)=L(g(t))$
$L(f'(t))=sF(s)-f(0)=L(g'(t))=sG(s)-g(0)$
$F(s)-G(s)=\dfrac {f(0)-g(0)}{s}$
$L^{-1}(F(s)-G(s))=L^{-1}(\dfrac {f(0)-g(0)}{s})$
$f(t)-g(t)=f(0)-g(0)$
$\dfrac {d}{dt} [f(t)-g(t)]=\dfrac {d}{dt} [f(0)-g(0)]$
$f'(t)-g'(t)=0$
$f'(t)=g'(t)$
So if it is correct, what does this proof imply exactly? Of course the Laplace transformation is not one-to-one, but what restrictions does this proof impose on the two functions other than $\int {f'dt}$ and $\int g' dt$ exist and $L(f)$ and $L(g)$ exist?