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I was reading a proof for a theorem in Erhan Cinlars intro to stochastic processes textbook that any equation with the following $f(x+s) = f(x)f(s)$, with $f(0)=1$. Then $f(x)=e^{(-ax)}$ for some $a$. I have been trying to prove this on my own. I came up with the following, but I didn't quite like the proof, and was wondering if someone could offer a proof utilizing Laplace transforms, or something of that flavor. Here are the following ways I have tried to show it.

1) $f'(t) =\lim_{h\to0} \frac{f(t+h) - f(t-h)}{h} = f(t) f'(0) = f(t)$ a Therefore by differential equations we get our desired result.

2) using laplace transform on $f(t+a) = f(t)f(a)$. First notice that $f(t)$ cannot be zero anywhere (assume its zero at $x$, then $f(t) = f(t+x-x) = f(t-x)f(x) = 0)$. Since it cannot be zero, we know $F(s)$ is not zero anywhere (integral of a strictly positive function is strictly positive function is strictly positive). Therefore, we have $F(S)e^{Sa} = F(S)f(a)$ which is equivalent to $0 = F(S) (e^{Sa}-f(a))$. Which is only true if $e^{Sa} = f(a)$ for all $S$, and a fixed. However, this is only true if $a = 0\dots$ So I am not sure what I am doing wrong here... I wanted to try and prove it using Laplace transforms. I would appreciate any help with understanding this.

BigM
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user7120528
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  • Any function with this property and which is also continuous has the property that $f(x)=e^{ax}$ for some $a$. – Thomas Andrews Nov 05 '16 at 22:54
  • Okay, and for the proof? That is what I am after... I was trying technique 2, and I just want to know another technique for this proof rather than the first technique. – user7120528 Nov 05 '16 at 22:56
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    You need some more conditions, such as continuity. – copper.hat Nov 05 '16 at 22:58
  • @ThomasAndrews I think we only need continuity in a neighbourhood of $0$. Can this be relaxed more? I mean, is the continuity needed? Even if it can be implicitly avoided by showing that no discontinuous function can meed the conditions? –  Nov 05 '16 at 23:01
  • How do you know $f'(0)=1$? And your limit is wrong - either $t+h$ or $t-h$ or, if you use both, it should be $2h$ in the denominator. – Thomas Andrews Nov 05 '16 at 23:01
  • Can you show its derivative is almost the same as the function? – Jacob Wakem Nov 05 '16 at 23:03
  • There is some measurability condition that is sufficient. Wiki https://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function says that being Lebesgue measurable is sufficient. – copper.hat Nov 05 '16 at 23:07
  • This is the "exponential" version of Cauchy's functional equation. See http://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation for a raft of resources about the Cauchy equation and the OP's equation (which is Equation (1) there) – grand_chat Nov 05 '16 at 23:11
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    If $f(k)=c$ for positive $c$ then $f(-k)=c^{-1}, f(2k)=c^2,f(3k)=c^3,\ldots$ implying $f(nk)=c^n$ and thus $f(k/n)=c^{1/n}$ and $f(qk)=c^q$ for rational $q$. If $f(1)=e^a$ then this gives $f(x)=e^{ax}$ for rational $x$, and continuity then gives $f(x)=e^{ax}$ for real $x$ – Henry Nov 05 '16 at 23:11
  • @copper.hat Thank you!! –  Nov 05 '16 at 23:13

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