Exponential of Sum/Real Numbers/Proof 6

Proof
Fix $a \in \R$ and define the function $f_a : \R \to \R$ by:


 * $\map {f_a} x = \map \exp {a - x} \exp x$

for all $x \in \R$.

We aim to establish that:


 * $\map {f_a} x = \map \exp {a - x} \exp x = \exp a$

for all $a, x \in \R$.

Then, we can fix $x, y \in \R$ and set $a = x + y$ to obtain:


 * $\map {f_a} x = \exp y \exp x = \map \exp {x + y}$

which is the claim.

Note that $f_a$ is differentiable and we have:

From Zero Derivative implies Constant Function, $f_a$ is constant.

That is, there exists $C \in \R$ such that:


 * $\map {f_a} x = C$

for all $x \in \R$.

We have:

So:


 * $\map {f_a} x = \exp a$

for all $a, x \in \R$.

That is:


 * $\map \exp {a - x} \map \exp x = \exp a$

for all $a, x \in \R$.

Hence the claim.