0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You could also simply use $f'(x+y)=\frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$. $\endgroup$ – LutzL Jan 5 at 23:53
  • 1
    $\begingroup$ 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 $\endgroup$ – Paramanand Singh Jan 6 at 17:00
4
$\begingroup$

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

$\endgroup$
0
$\begingroup$

You can guess the function to be:

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

since

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

$\endgroup$
  • $\begingroup$ But is there any way to solve the questions using differentiation? $\endgroup$ – Matthew Tan Jan 4 at 18:13
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.