Uniqueness of Laplace transform

Statement: Let $L_f(t)={\int }_0^{\mathrm{\infty }}f(x)e^{-tx}dx$. Then if $f(x)=o(e^{ax})$ for some $a>0$, $L_f=L_g$ implies $f=g$.

Proof: By linearity of the integral, this is equivalent to showing $L_f=0$ implies $f=0$. Perform the substitution $x=-log(u)$ to obtain

$$\begin{array}{rcl}{L}_f(t)& =& {\int }_0^{\mathrm{\infty }}f(x)e^{-tx}dx\\ & =& {\int }_0^1{u}^{t-1}f(-log(u))du\end{array}$$

where for $t\ge a+1$, ${\displaystyle \underset{u\to 0}{lim}{u}^{t-1}f((-log(u))=\underset{x\to \mathrm{\infty }}{lim}{e}^{-(t-1)x}f(x)=0}$, so this function is continuous at 0 and the above integral makes sense. Thus, if $L_f=0$, then ${\int }_0^1{u}^{n}f(-log(u))du=0$ for all sufficiently large integers $n\ge N$. Hence by linearity ${\int }_0^{1}p(u)f(-log(u))du=0$ if $p$ is a polynomial which contains only terms of size $N$ or greater.

Since $f(-log(u))$ is continuous on [0, 1], by Stone-Weierstrass we can choose a sequence of polynomials $p_k$ (having terms only of size $N$ or greater) which uniformly converges to $f(-log(u))$. That is, for any $ϵ\ge 0$, $|f(-log(u))–p_k(u)|\le ϵ$ for all $u$ when $k$ is sufficiently large. Thus, on the one hand, $k$ sufficiently large implies

$$\begin{array}{rcl}|{\int }_0^1(f(-log(u))–p_k(u))f(-log(u))du|& \le & {\int }_0^1ϵ|f(-log(u))|du\\ & =& ϵ{\int }_0^1|f(-log(u))|du\end{array}$$

which can be made arbitrarily small. But on the other hand,

$$\begin{array}{rcl}{\int }_0^1(f(-log(u)–p_k(u))f(-log(u))du& =& {\int }_0^{1}f(-log(u){)}^{2}du+{\int }_0^1{p}_k(u)f(-log(u))du\\ & =& {\int }_0^{1}f(-log(u){)}^{2}du+0.\end{array}$$

But $f(-log(u){)}^2$ is non-negative, and the whole integral goes to zero with $n$ (since it’s smaller than any positive $ϵ$). The only way this is possible is if $f(-log(u))=0$ for all $u$ in $[0,1]$, i.e., $f(x)=0$ for all $x$ in $[0,\mathrm{\infty })$. $\mathrm{<img\; draggable="false"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/11/svg/25fc.svg">}$

Notes: This can also be used to prove the uniqueness of the Fourier transform.

We needed to be able to approximate an arbitrary continuous function $f$ on $[0,1]$ such that $f(0)=0$ uniformly with polynomials with terms only of degree greater than or equal to some K. To do this, it suffices to be able to approximate the curve $y=x$ with those polynomials, since then you can use the approximation to $y=x$ to build up other polynomials and exploit Weierstrass approximation. To do this, set $f_n(x)=x(1–(1–x{)}^n{)}^K$.