# Prove that if $f$ is differentiable at $x=0$, then $f$ is differentiable on $\mathbb{R}$.

$Conj:$ 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.

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}

• 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. – k-dubs Nov 14 '14 at 16:59