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Let $E:\mathbb{R} \to \mathbb{R}$ be an infinitely continuously differentiable function and $E$ is not zero function such that $$E(u+v)=E(u)E(v).$$ Show that $E(x)=e^{ax}$ for some $a\in \mathbb{R}$.

My partial answer:

The function $x\mapsto e^{ax}$ satisfies the properties immediately. For $y=0$, we have $E(x)=E(x)E(0)$. Thus, $E(0)=1$. Let $y\in \mathbb{R}$ is fixed, then $$E'(x+y)=E(y)E'(x)$$.

I don't know how to continue the answer. Could you give me some hint? Thanks.

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$E(x)=0$ also satisfies the condition, doesn't it? –  Peter Košinár Dec 19 '13 at 0:32
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So far, nothing rules out $E(x) \equiv 0$. With your $E'(x+y) = E(y)E'(x)$, set $x = 0$ and see where that takes you. –  Daniel Fischer Dec 19 '13 at 0:33
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Maybe you can take the $\log$ on both sides of the equation and rewrite it to Cauchy-equation. –  Ragnar Dec 19 '13 at 0:33
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@Ragnar I think that approach should be avoided since log itself is defined by exponential function –  tes Dec 19 '13 at 0:44
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I didn't think of taking a $\log$. Let $a = E'(0)$ and consider $f(x) = E(x)e^{-ax}$. –  Daniel Fischer Dec 19 '13 at 1:00
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2 Answers 2

up vote 2 down vote accepted

Following up on Dan's suggestion we have (d/dx)E(x+y) = E'(x)E(y). Setting x = 0 we get E'(y) = E'(0)E(y). The unique solution of this differential equation is E(y) = $Ce^{ay}$ where a = E'(0). But if E(0) = 1, C = 1.

The case of E $\equiv$ 1 comes when a = 0; E $\equiv$ 0 is from C = 0.

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Using the definition of the derivative,

$$E'(x) = \lim_{h\to 0}\frac{E(x+h)-E(x)}{h} = \ldots =\left(\lim_{h\to 0}\frac{E(h)-1}{h} \right)E(x). $$

By the functional equation, $E(0)=E(0)^2$, so $E(0)=0$ or $1$. If $E(0)=0$, the functional equation gives $E(x)=0$ for all $x$, and you don't want the zero solution. So $E(0)=1$. Therefore the last limit exists and is equal to $E'(0)$. So the derivative of $E$ is proportional to $E$. $E$ has the form $E(x) = ce^{ax}$ for some constants $c$ and $a$. Again, since $E(0)=1$, $c=1$.

This is how I would do it. Your work looks valid, but I'm not sure how to progress beyond the last equation you wrote.

I just noticed that this is pretty similar to Daniel Fischer's suggestion, which I think he made before this answer. If Daniel makes his suggestion an answer, it would not bother me if you accepted it instead of mine.

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@Stephan Smith -- good for you Stephan; I also don't care what solution is accepted. The main thing is to be helpful; also I want to learn myself. –  Betty Mock Dec 19 '13 at 17:10
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