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Given a function $f$ satisfying the following two conditions for all $x$ and $y$:
(a) $f(x+y)=f(x)\cdot f(y)$,
(b) $f(x)=1+xg(x)$, where $\displaystyle \lim_{x\rightarrow 0}g(x)=1$.
Prove that $f'(x)=f(x)$.

The only thing I know is that $f'(x)=f(x)$ is true for $x=\{0,1\}$ , but how do we know that it's true for all $x$?

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Hint: $\frac{f(x+h) - f(x)}h = f(x) \cdot \frac{f(h) - 1}h$ (now use what you now about $f'(0)$). – martini May 22 '12 at 15:10
Where you wrote $x=\{0,1\}$, did you mean $x\in\{0,1\}$? You can write $x=0,1$ (with no braces) and it means the same thing. But "$x=\{0,1\}$" means $x$ is a set with two members: $0$ and $1$. – Michael Hardy May 22 '12 at 15:20
Note that $ f'=f $ then implies $ f(x)=A\exp(x) $, but from (a) $A=A*A$ and (b) rules out $A=0$ so $f$ is exactly $\exp $. – Mark Hurd May 23 '12 at 2:32

Let $x_0$ be an arbitrary real number.

Note that you have $f(0)=1$ from the second condition.

You have $$f'(x_0)=\lim\limits_{y\to 0} \frac{f(x_0+y)-f(x_0)}y = \lim\limits_{y\to 0} \frac{f(x_0)(f(y)-1)}y = f(x_0) \lim\limits_{y\to 0} \frac{f(y)-1}y = f(x_0)f'(0).$$

Since you wrote you have already proved this for $0$ and $1$, you know that $f'(0)=f(0)=1$. Thus the above equation is the same as $f'(x_0)=f(x_0)$.

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In fact, what happens at $0$ isn't even necessarily to comment separately since $\frac{f(y)-1}{y}=\frac{1+y\,g(y)-1}{y}=g(y)\to 1$ as $y\to 0$. – T. Eskin May 22 '12 at 15:20
That's even better, nice observation. (And it is essentially TonyK's solutions.) – Martin Sleziak May 22 '12 at 15:22

$$ \begin{align} f(x+h)-f(x) & = f(x)\cdot f(h) - f(x), \quad\text{from (a)} \\[8pt] & = f(x)(f(h)-1) \\[8pt] & = f(x)\cdot hg(h),\quad \text{from (b)} \end{align} $$

So $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}f(x)g(h)=f(x)$

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Assuming $f$ is differentiable:

You have that


Fix $x$ and differentiate wrt $y$.


Let $y=0$, and you get


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Maybe it is worth noticing that in this solution we have to assume that $f$ is differentiable. Or, we should read $f'(x+y)=f(x)f'(y)$ as: If $f'(y)$ exists then $f'(x+y)$ exists and it is equal to $f(x)f'(y)$. Similarly for the last equation. – Martin Sleziak May 22 '12 at 16:50
@MartinSleziak That is true. Since many aldready dealt with the quotient, I wanted to give another look to the problem. – Pedro Tamaroff May 22 '12 at 16:56

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