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.
 A: 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 $f(x)-g(x) \equiv 0$.
A: This is almost exactly the same idea as the above answer, but I keep it as another formulation of the same idea.
Here is the precise formulation and proof of the theorem:

Lerch's theorem (Uniqueness of inverse of Laplacian): Let $f $ be a function defined on all nonnegative real numbers and is continuous except on a finite number of points. Suppose there exist constants $M, a$ such that $|f (x)| \leq Me^{ax} $ for all $x $. If $L[f](s)$ is zero for all $s > s_0$ for some constant $s_0$, then $f$ is identically zero except on the points of discontinuity.

Proof:
First assume that $f$ has no discontinuities. We'll prove the general case for a function having finite number of discontinuities at the end.
For any polynomial $P (x)= \sum_{k=0}^n a_kx^k $ we have that $\int_0^\infty e^{-st}P (e^{-t})f (t)dt=...=\sum a_kL[f](s+k) \equiv 0 $ for $s > s_0$. I skipped the easy calculation and the reader can check the "..." part. Letting $x = e^{-t} $ this condition the last integral becomes $$\int_0^1 x^{s-1}P (x)f (-\log x)dx \equiv 0,$$ for all $s>s_0$. Now fix $s_1 > \max\{s_0, 1, a+1\}$. Then $$x^{s_1-1}|f (-\log x)| \leq Me^{s_1-a-1} $$ so that the function $$G:=x^{s_1-1}f (-\log x), 0 < x \leq 1, $$ approaches zero as $x $ approaches $0$ from the right. Define $G(0)$ to be zero. Then $G$ is a continuous function on $[0,1]$. Also remember that the value of an integral does not change if we define or redefine a function at a point. $G $ is thus a function that is bounded and continuous on $[0,1] $ and satisfies $\int_0^1 G (x)P (x)dx  \equiv 0$ for any polynomial $P $. If we now approximate $G$ by polynomials that converge uniformly to $G$ and use the argument of Lemma 1 given by user @Winther above we have that $G \equiv 0$ and thus $f \equiv 0$.
We get a similar result if we assume that $f $ has only a finite number of discontinuities, and the result would be that $f $ is identically equal to zero except possibly on its finite number of discontinuous points. In the course of proving that we would need to break $f $, and  thus $G$, into a finite number of continuous parts, use several subintervals $[x_m, x_{m+1}] $, and use different approximating Weierstrass polynomials for each of these subintervals to get that $G $, and thus $f$, is equivalent to zero except on discontinuous points.
The majority of the proof and the precise statement of this theorem is almost directly copied from the following books's Appendix II, page 678:
D. L. Kreider, R. G. Kuller, D. R. Ostberg, F. W. Perkins, An Introduction to Linear Analysis, Addison-Wesley Publishing Co., Inc., Reading, Mass.-Don Mills, Ont. 1966.
Notice that the argument doesn't work if we do not assume $f $ to be of exponential order (i.e. having the constants $M, a $ as above). However, as a practical matter, I remember an MIT professor said that almost all continuous functions we shall encounter in real life problems will be of exponential order.
