$\frac{f(t+T)}{f(T)} = f(t)$ then $f$ is an exponential function If $f$ is a function such that for all $(T, t) \in \mathbb{R}^2_+$ we have : $$\frac{f(t+T)}{f(T)} = f(t)$$
Then how can I prove that $f$ is an exponential function ? 
 A: If you want to prove this function is exponential, you need to explicitly tell that function is differentiable
Given $$f(t+T)=f(t)f(T)$$
Now take derivative w.r.t. $t$.
$$f'(t+T)=f'(t)f(T)$$
Now put $t=0$, to get 
$$f(T)f(0)=f'(T)$$
Let $y=f(T)$ and now solve differential equation.
A: You need some smoothness for $f$, i.e., the $f$ is for example continuous, otherwise, using the Axiom of Choice one can define a discontinuous, nowhere bounded $f$ satisfying the condition in the OP, in the same way that one can define a discontinuous $g$ satisfying $g(x+y)=g(x)+g(y)$.
Assume that $f$ is continuous. Then set 
$$
g(x)=1+\int_0^x f(t)\,dt.
$$
Hence $g'=f$ and $g(0)=1.$
Then $g$ is continuously differentiable and
$$
g(x)=1+\lim_{n\to\infty}\frac{x}{n}\sum_{k=0}^{n-1} f(kx/n)
=1+\lim_{n\to\infty}\frac{x}{n}\sum_{k=0}^{n-1} \big(f(x/n)\big)^k
=1+\lim_{n\to\infty}\frac{x}{n}\frac{f(x)-1}{f(x/n)-1}
=1+\big(f(x)-1\big)\lim_{n\to\infty}\frac{1}{\frac{f(x/n)-1}{x/n}},
$$
which shows that $f$ is also differentiable.
Ince $f$ is differentiable, obtain 
$$
f'(t)\leftarrow\frac{f(t+h)-f(t)}{h}=f(t)\frac{f(h)-f(0)}{h}\to f'(0)f(t)
$$
and hence
$$
\mathrm{e}^{-f'(0)t}\big(f'(t)-f(0)f(t)\big)=0\quad\Longleftrightarrow\quad
\left(\mathrm{e}^{-f'(0)t}f(t)\right)'=0\quad\Longleftrightarrow\quad \mathrm{e}^{-f'(0)t}f(t)=c\quad\Longleftrightarrow\quad f(t)=c\mathrm{e}^{f'(0)t}
$$
But $f(0)=1$ implies that $c=1$ and finally $f(t)=\mathrm{e}^{f'(0)t}$.
