How do I prove that the only possible function is $exp$?

Question

Let´s say we have a differentiable function

$f : \mathbb{R} -> \mathbb{R}$

with

$f' = f$

and

$f(0) = 1$ .

How do I show that the only possible function for this to work
$f = exp$ ?

From $f' = f $

I concluded that

$1 = \frac{1}{f} \cdot f' => (ln(f))' $

If we define

$ g' := (ln(f))$

Then we know that

$ g= exp(f)$

But I´m stuck right there. How do I go on?

Answer 1

Since $(\ln f)' = 1$, by integration, $\ln f(x)= x + C$, where $C$ is a constant. Using the condition $f(0) = 1$, we find that $C = 0$. Hence, $\ln f(x) = x$, i.e., $f(x) = \exp(x)$.

Note: Since you were not given that $f(x) > 0$ for all $x$ from the start, perhaps a safer way to approach the problem is to consider the function $g(x) := f(x) \exp(-x)$. By the chain rule, $$g'(x) = f'(x)\exp(-x) - f(x)\exp(-x) = (f'(x) - f(x))\exp(-x) = 0$$ Hence, $g(x)$ is constant. Since $f(0) = 1$, $g(0) = 1$. Therefore, $g(x) = 1$, i.e., $f(x) = \exp(x)$.

For uniqueness, suppose $f_1$ and $f_2$ satisfy the initial value problem $f' - f = 0$, $f(0) = 1$. Then $f_1 - f_2$ satisfy the differential equation with $(f_1 - f_2)(0) = 0$. Hence, by the fundamental theorem of calculus, $f_1(x) - f_2(x) = 0$, i.e., $f_1(x) = f_2(x)$.