$Conj:$ Suppose that a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable at $x=0$, satisfies $f(a + b) = f(a)f(b)$ for all $a,b,\in\mathbb{R}$, and is not identically zero ($\exists ~x$ such that $f(x) \neq 0$). Then $f$ is differentiable and $f'(x) = f'(0)f(x)$.

I think I need to show that $f(0) = 1$, $f(x)\neq 0$ for all $x$ and that $f$ is continuous if continuous at $f(0)$. Here's what I got so far:

$Proof:$ $f$ not identically zero implies that there exists $x \in \mathbb{R}$ such that $f(x) \neq 0$. Then:

\begin{align} f(x) &= f(x +a - a) & \text{(where $a \in \mathbb{R}$).}\\ & = f(x)f(a)f(-a) \\ & \neq 0 \end{align}

Thus, $f(x)\neq 0$ implies that $f(a)\neq 0$ for all $a \in \mathbb{R}$. Furthermore,

\begin{align} f(0) & = f(0 + 0)\\ & = f(0)^2 \end{align}

Thus, $f(0)$ must be either 0 or 1. We just showed that $f(x)\neq 0$ for all $x$, so $f(0)=1$.

Now I need the continuity piece (I think) but I'm stuck.

Thanks in advance!


I'm not sure why you are specifically looking to prove continuity. It's possible just to go directly to differentiability at $x$. Use the definition of the derivative at $x$:

$$\begin{align} f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\to0}\frac{f(x)\,f(h)-f(x)}{h}\\ &=f(x)\lim_{h\to0}\frac{f(h)-1}{h}\\ \end{align}$$

You have shown $f(0)=1$, so

$$\begin{align} f'(x)&=f(x)\lim_{h\to0}\frac{f(0+h)-f(0)}{h}\\ \end{align}$$

  • $\begingroup$ Awesome! I wasn't sure why I was doing continuity either. It was something to prove in a related problem that I thought might help. $\endgroup$ – k-dubs Nov 14 '14 at 16:59

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.