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The proof I followed is under the following link

Proof of Lerchs theorem

The problem is described as follow:

  1. By steps listed in the proof the first lemma illustrated: If $g$ is a continuous function on $[0,1]$ and $\int_0^1 g(x)x^ndx=0$ for $n=0,1,2,…$ then $g(x)=0$ for $x\in[0,1]$.
  2. The lemma 2 is to prove that if $\mathcal{L}\{f(x)\}=0$ than $f(x)=0,x\in[0,\infty)$. By definition of laplace transform: \begin{aligned} \mathcal{L}\{f(x)\}&=\int_0^\infty e^{-st}f(t)dt\\ &=\int_1^0 u^{s}f(-\ln(u))\left(-\frac{1}{u}\right)du\\ &=\int_0^1 u^n u^{s-1-n}f(-\ln(u))du\\ &=0\\ \end{aligned}
  3. However the domain of $u$ is $(0,1]$ which means that it is an improper integral instead of a proper integral with the domain of $[0,1]$. Which mean that I cannot apply the result from lemma 1 to prove it.

Hence, if anyone can help me on proving the improper integral version of lemma 1 or, provide another prove. Thanks so much!!

David C. Ullrich
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SHORE SHEN
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  • Adding a question. In some other proof they trying to illustrate that $u^{s-1-n}f(-\ln(u))$ will be continues at u=0. But the thing is the function has to satisfy any n that is a positive integer, which means that $s-1-n$ will be a negative number and $\lim_{u\to 0}u^{s-n-1}=\lim_{t\to\infty}e^{-t(s-1-n)}=\lim_{t\to\infty}e^{(n+1-s)t}=\infty$ if n is big enough. Which make the proof does not make any sense. – SHORE SHEN Aug 25 '18 at 14:59
  • There is another problem that assuming $\exists a\in R\Rightarrow \lim_{t\to\infty}e^{-at}f(t)=0$ so that the lemma 2 can be proved at the interval [0,1]. How can I prove that $\exists a\in R\Rightarrow \lim_{t\to\infty}e^{-at}f(t)=0$ – SHORE SHEN Aug 25 '18 at 16:43

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