Equality of laplace transform Assuming that Laplace Transforms of two functions  $f$ and $g$ are equal, is it true that $f=g$?
There is one-to-one correspondence between functions and their Laplace Transforms, so it seems to me that it is true.

This is a theorem knows as Lerchs theorem. I will give a standard proof of this below. I will restrict myself to continuous functions, but the generalization to integrable functions (continuous almost everywhere) is straight forward (replace equal with equal almost everywhere).
We first start with a lemma:

Lemma 1: If $g$ is a continuous function on $[0,1]$ and $\int_0^1 g(x) x^n dx = 0$ for $n=0,1,2,\ldots$ then $g(x) \equiv 0$ for $x\in[0,1]$.

Proof: Fix $\epsilon > 0$. By the Weierstrass approximation theorem we can find a polynomial $P(\epsilon,x)$ s.t. $|P(\epsilon,x) - g(x)|_\infty < \epsilon$. The hypothesis on $g$ implies $\int_0^1 g(x)P(\epsilon,x)dx = 0$. Taking $\epsilon \to 0$ we get $\int_0^1 g^2(x) dx = 0$ which implies $g(x) \equiv 0$.
We can now prove a second lemma:

Lemma 2: If $g$ is a continuous function and if $\mathcal{L}(g)(s) = 0$ for all $s \geq a$ then $g(x) \equiv 0$.

Proof: Fix $s_* \geq a$. At $s=s_* + n + 1$ the condition $\mathcal{L}(g)(s) = 0$ implies that
$$0 = \mathcal{L}(g)(s) = \int_0^\infty g(x) e^{-xs} dx = \int_0^1 x^{n} \left(x^{s_*}g(-\log(x))\right)dx$$
for all $n=0,1,2,\ldots$ where we have made the substitution $e^{-x}\to x$ in the last equality. By the above lemma it follows that $x^{s_*}g(-\log(x)) \equiv 0$ which implies $g(x) \equiv 0$.
We are now ready to prove the theorem:

Lerch theorem: If $g,f$ are continuous functions and if $\mathcal{L}(f) = \mathcal{L}(g)$ for all $s\geq s_*$ then $g\equiv f$.

By the linearity of the Laplace transform we have
$$0 = \mathcal{L}(f) - \mathcal{L}(g) = \int_0^\infty (f(x)-g(x)) e^{-sx}dx$$
for all $s\geq s_*$. By the second lemma above it follows that $f(x)-g(x) \equiv 0$.
