I am trying to construct the exponential function on $\mathbb R$ by first finding all functions $f$ such that $f = f'$ (which should be all the constant multiples of $\exp$), then characterizing $\exp$ by the initial condition $f(0) = 1$.
I intend to use only the definitions and properties of derivatives and integrals, and the fundamental theorem(s) of calculus; in particular, I am trying to avoid Taylor series and/or the notion of uniform convergence. Of course, these restrictions are by no means precise, and I really don't know if it's possible to come up with a reasonably "simple" construction using only this set of tools, but I figured I'd give it a try.
So far I've only managed to prove uniqueness in the sense that if $f = f'$ and $c \in \mathbb R$, then $f$ is uniquely determined by $f(c)$. With this comes the corollary that if $f(c) = 0$ for any $c \in \mathbb R$, then $f = 0$. Existence seems pretty difficult to get to without any of the heavy machinery.
Here is the proof of uniqueness as described above:
Lemma. If $f = f'$, $g = g'$, and $c \in \mathbb R$, then $f + c \cdot g = f' + c \cdot g' = f' + (c \cdot g)' = (f + c \cdot g)'$.
Lemma. If $f = f'$ and $f(c) = 0$ for some $c \in \mathbb R$, then $f(x) = 0$ for all $x \leq c$. (I had trouble proving it for $x > c$.)
Proof. By the FTC, we have $\int_a^b f = f(b) - f(a)$. If $f(x) > 0$ for $c - \varepsilon \leq x \leq c$, then $\int_{c - \varepsilon}^c f > 0$, but $f(c) - f(c - \varepsilon) < 0$, a contradiction. Similar contradiction for $f(x) < 0$.
Theorem. If $f = f'$, $g = g'$, and $f(c) = g(c)$ for some $c \in \mathbb R$, then $f = g$.
Proof. Suppose $f(d) \neq g(d)$ for some $d \in \mathbb R$. WLOG, assume $g(d) \neq 0$. If $d < c$, then $(f - g)(d) \neq 0 = (f - g)(c)$; if $c < d$, then $\left( f - \frac{f(d)}{g(d)} \cdot g \right)(c) \neq 0 = \left( f - \frac{f(d)}{g(d)} \cdot g \right)(d).$ In any case, we contradict the second lemma.
Corollary. If $f = f'$ and $f(c) = 0$ for some $c \in \mathbb R$, then $f = 0$.
Proof. $0 = 0'$.
Now I want to prove the existence of $f$ for any initial condition $f(x) = y$, but I don't know how.