Constructing $\exp$ on $\mathbb R$ 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.
 A: Supoose $u$ and $v$ are functions so that $u'=u$, $v' = v$ and so $u(0)=v(0) = 1$.
Put $\Phi(t) = u(t)v(-t)$ for $t\in\mathbb{R}$.
$$\Phi'(t) = u'(t)v(-t) - u(t)v'(-t) = u(t)v(-t) - u(t)v(-t) = 0.$$ 
The function $\Phi$ is constant.  Since $\Phi(0)  = 1$, $u(t)v(-t) = 1$ for all
$t\in{\mathbb{R}}$.
We can apply this result to $v$ and $v$ to get 
$$v(-t) = {1\over v(t)},\qquad t\in\mathbb{R}.$$
We conclude that $u(t)/v(t) = 1$ for $t\in\mathbb{R}$, so $u = v$.  
Here is one way to get existence.
Define $l(x) = \int_1^x{dt\over t}$, for $x > 0$.   This function is continuous and differentiable on $(0,\infty)$.  
Observe that
$$l(xy) = \int_1^{xy} {dt\over t} = \int_1^x {dt\over t} + \int_x^{xy}{dt\over t}
= l(x) + \int_x^{xy} {dt\over t}.$$
Using a change of variable, we get
$$\int_x^{xy} {dt\over t} = \int_1^y {x\,dt\over xt} = l(y).$$
We have $l(xy) = l(x)+l(y)$ for $x, y > 0$.  We clearly have $l(1) = 0$.
And it's not hard to show that $l(x^\alpha) = \alpha l(x)$ for $\alpha\in\mathbb{R}$ and $x > 0$.  
By the fundamental theorem of calculus we know that 
$$l'(x) = 1/x$$
for $x > 0$. This function is strictly increasing and therefore 1-1 on $(0,\infty)$.  
The inverse function to $l$ will satisfy the the initial value problem
$u'(t) = u(t)$ and $u(0) = 1$.  This gives existence.
You can use the properties of $l$ to see this is an exponential function.  The base for this exponential is $l^{-1}(1)$.
