Do the continuously differentiable functions form an integral domain? Let $f$ and $g$ be two real valued continuously differentiable functions with domain $[a,b]$ where $f(x)g(x)=0$ for all $x$ in $[a,b]$. Can it be concluded that $f$ and $g$ are individually zero functions?
 A: No. Consider the functions over $[-1,1]$ 
$$
f(x)=\begin{cases}
0 & \text{for $-1\le x\le 0$}\\
e^{-1/x^2} & \text{for $0<x\le 1$}
\end{cases}
\qquad
g(x)=\begin{cases}
e^{-1/x^2} & \text{for $-1\le x<0$}\\
0 & \text{for $0\le x\le 1$}
\end{cases}
$$
The functions are actually of class $C^{\infty}$.
The assertion is true for the analytic functions.

Analytic functions on a closed real interval are actually, by definition, analytic in an larger open interval. Such functions can be thought as complex analytic functions on an open connected set of the complex plane. These functions have a very pleasant property: if the set of zeros of an analytic function has an accumulation point, then the function is the constant zero.
If $f$ and $g$ are analytic functions on the region (open connected set) $\Omega$, and $fg=0$, then $Z(f)\cup Z(g)=\Omega$. Thus at least one of the two sets (where $Z(f)=\{z\in\Omega:f(z)=0\}$) has an accumulation point, or their union wouldn't be open.
A: Let $f(x)=(x-\dfrac{a+b}{2})^2$ if $a\leq x\leq \dfrac{a+b}{2}$ and $f(x)=0$ if $\dfrac{a+b}{2}\leq x\leq b$.
Let $g(x)=0$ if $a\leq x\leq \dfrac{a+b}{2}$ and $g(x)=(\dfrac{a+b}{2}-x)^2$ if $\dfrac{a+b}{2}\leq x\leq b$.
Note that $f$ and $g$ meet all your requirements but not identically $0$.
A: Answer:
Consider $$f:[0,1] \longrightarrow \mathbb{R}$$as$$f(x)=\begin{cases} 
      0 & 0\leq x \leq \frac{1}{2} \\
      (\frac{1}{2}-x)^2 & \frac{1}{2}\leq x\leq 1 
   \end{cases}$$
And$$g:[0,1] \longrightarrow \mathbb{R}$$as$$g(x)=\begin{cases} 
    (\frac{1}{2}-x)^2 & 0\leq x \leq \frac{1}{2} \\
       0& \frac{1}{2}\leq x\leq 1 
   \end{cases}$$ then $$f.g=0$$ but neither $f=0$ nor $g=0$
