Is Laplace transform of piecewise continuous injective?

Question

Let $(\mathcal{L}f)(s)$ be the Laplace transform of a piecewise continuous function $f(t)$ defined for $t\geq 0$. If $(\mathcal{L}f)(s)=0 $ for all $s\in\mathbb{R^+}$ does this imply that $f(t)= 0$ for all $t\geq 0$ ?

Answer 1

You didn't assume anything on the convergence of the integral or integrability of $f$.

If $$F(s_0)=\lim_{T\to\infty} \int_0^T f(t)e^{-s_0 t}dt$$ converges for some $s_0$ then $$g(T)=\int_0^T f(t)e^{-s_0 t}dt$$ is bounded and for all $\Re(s)>\Re(s_0)$ (integration by parts) $$F(s)=\lim_{T\to\infty} g(T) e^{-(s-s_0)T}+\int_0^T g(t)(s-s_0)e^{-(s-s_0)t}dt=\int_0^\infty g(t)(s-s_0)e^{-(s-s_0)t}dt$$ converges absolutely.

Thus $F$ is analytic for $\Re(s)>\Re(s_0)$.

If $f$ is not identically zero then $g(a) \ne 0$ for some $a>0$ thus for $n$ and $b$ large enough $$2^{-n}\sum_{k=0}^n {n\choose k} \frac{F(s_0+1-i(2k-n)/b)}{1-i(2k-n)/b} e^{i(2k-n)a/b}=\int_0^\infty g(t) \cos(t/b-a/b)^n e^{-t}dt \ne 0$$ so $F$ is a non-zero analytic function whence it can't vanish on a real segment.