How can I derive this function?

Question

Let $f(x)$ be a function satifying conditions:

• $f(x+y)=f(x)f(y)$.
• $f$ is differentiable.
• $f$ is continuous.

How can I derive $f(x)$?

Answer 1

From $f(0)=f(0+0)=f(0)\cdot f(0)$ we conclude that $f(0)$ is either $0$ or $1$. Now if $f(0)=0$, then $$f(x)=f(x+0)=f(x)\cdot f(0)=0$$. Now suppose $f(0)=1$, then we can see $$f(2\cdot x)=f(x)^2, f(3\cdot x)=f(x)^3 \cdots $$ From which we can guess that $f(x)=e^{ax}$.

Answer 2

First, universal observations: $$f(0+0) = f(0)^2 \implies f(0) = 0 \text{ or } f(0) = 1 \\ f(2x) = f(x)^2, f(3x) = f(x)^3, ..., f(nx) = f(x)^n, n \in \mathbb{N}, n>0 \\ f({x \over m}) = f(x)^{1 \over m}, m \in \mathbb{N}, f(qx) = f(x)^q, q \in \mathbb{Q}, q > 0$$ For case $f(0) = 0$ we have $f(x) = f(x+0) = f(x)f(0) = 0$, in both (a) and (b); easy to see that $f(x) = 0$ really is a solution.

For case $f(0) = 1$ first we note that $f(0) = f(x)f(-x)$ and thus $f(-qx) = fx)^{-q}, q \in \mathbb{Q}, q > 0$. After that we can take $a = f(1)$ (obviously $a > 0$, since $a = f({1 \over 2})^2$). Then $\forall x \in \mathbb{R}, \forall \varepsilon > 0, \exists q \in \mathbb{Q} : |q-x| < \varepsilon$. Since $f(q) = a^q$, from continuity we can conclude $f(x) = a^x$. (Since being differentiable implies continuity, the same holds for (a)). So $f$ can't be anything other than $a^x$ in this case, but since $a^x$ is continuous and differentiable and $a^{x+y} = a^xa^y$ for $\forall a > 0$, it's really a solution.

End result: $f(x) = 0$ or $f(x) = a^x, a > 0$.