# Functional Equation Solved Using Differentiation

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?

• You could also simply use $f'(x+y)=\frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$. Commented Jan 5, 2019 at 23:53
• 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 : math.stackexchange.com/a/1885860/72031 Commented Jan 6, 2019 at 17:00

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 $$f(x)=e^{kx}=e^{f'(0)x}.$$

You can guess the function to be:

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

since

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

• But is there any way to solve the questions using differentiation? Commented Jan 4, 2019 at 18:13

$$f(0)=1$$ 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 $$a)$$.