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Let $f$ be a function with domain $R$ that satisfies the conditions: $$f(x+y)=f(x)f(y), \forall x,y $$ and $$f(0) \neq 0$$

(a) Show that $f(0)=1$

(b) Prove that $f(x) \neq 0$, for all $x\in R$

(c) Assuming that $f'(x)$ exists for all $x \in R$, use the definition of the derivative to show that $f(x)$ satisfies the equation $f'(x)=kf(x)$, where $k=f'(0)$

I've tried solving part (a) and (b) by substituting $x=y=0$ and $y=-x$ respectively, but I can't seem to solve part (c) as I can't avoid dividing by 0 when dealing with the limit. Does anyone know how to work around this?

Yiorgos S. Smyrlis
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Hector Lombard
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  • You could also simply use $f'(x+y)=\frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$. – Lutz Lehmann Jan 05 '19 at 23:53
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    You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : https://math.stackexchange.com/a/1885860/72031 – Paramanand Singh Jan 06 '19 at 17:00

3 Answers3

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Note that, as $h\to 0$, then $$ f'(x)\leftarrow\frac{f(x+h)-f(x)}{h}=\frac{f(x)f(h)-f(x)f(0)}{h}=f(x)\frac{f(h)-f(0)}{h}\to f(x)\,f'(0) $$ and hence $$ f'(x)=f'(0)\,f(x), $$ which means that $f'(x)=k\,f(x)$, where $k=f'(0)$.

Note. Clearly, as $f$ satisfies the ODE, $y'=ky$, then it is of the form $f(x)=ce^{kx}$, and since $f(0)=1$, then $$ f(x)=e^{kx}=e^{f'(0)x}. $$

Yiorgos S. Smyrlis
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You can guess the function to be:

$$f(x) = a^x$$

since

$a^{p+q} =a^pa^q $

Archer
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$$f(0)=1$$ is apparent because when $x=y=0$, we have $f(0)=f^2(0)$ which can only happen at $0$ or $1$, and $0$ is ruled out by the question.

For $b)$, notice that $f(x)=0\implies f(x+y)=0f(y)=0$ for any $y$. So in effect, if one value of $x$ yields $0$, all of them will. This is contradicted by part $a)$.

Rhys Hughes
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