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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 ?

J. OK
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2 Answers2

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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.

Jaideep Khare
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  • This assumes the function is differentiable around 0, but OP only asks for continuous. – gt6989b May 10 '17 at 20:03
  • @gt6989b Thanks, I have added that in my answer. – Jaideep Khare May 10 '17 at 20:07
  • Actually, continuity is enough, but a different argument is needed. – Wojowu May 10 '17 at 20:14
  • Yes, but that is not always easy to assume. There is a proof without that assumption – gt6989b May 10 '17 at 20:14
  • @Wojowu If you know, then please add that as an answer, I would like to see that. – Jaideep Khare May 10 '17 at 20:24
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    If you assume continuity as one point $c$ then via $f(x + h) = f(x - c)f(c + h)$ we can see that $f$ is continuous everywhere. If $f$ vanishes at one point it vanishes everywhere. So let's assume $f$ to be non-zero and then $f(x) = (f(x/2))^{2} > 0$ so that $f$ is positive. Now integrate $f(x)f(t)=f(x + t)$ between $0$ and $1$ with respect to $t$ and you get $f(x)F(1) = F(x + 1) - F(x)$ where $F(y) = \int_{0}^{y}f(t),dt$ and since $F$ is differentiable, $f$ is also differentiable. – Paramanand Singh May 17 '17 at 10:48
  • @ParamanandSingh It is awesome, can I add that to my answer? – Jaideep Khare May 17 '17 at 12:16
  • You can add in your answer. I have no issues. – Paramanand Singh May 17 '17 at 14:51
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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 $f(t)=\mathrm{e}^{f'(0)t}$.

Yiorgos S. Smyrlis
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