Example for $f,g\neq0$ but $f\cdot g=0$ 
Let $I=[0,1]\subset\mathbb R$. Find two infinitely differentiable functions $f$ and $g$ from $I$ to $\mathbb R$, so that neither $f$ nor $g$ is identically equal to $0$, but the product $f\cdot g$ is (identically zero).

There is an example using characteristic function for a not connected set, but here I don't know any example
 A: For example chose $f(x) = \exp(\frac1{x-\frac12}) \cdot \chi_{[0,\frac12]}(x)$ and $g(x) = \exp(\frac1{\frac12 - x}) \cdot \chi_{[\frac12,1]}(x)$. You can easily check both, $f,g\in C^\infty [0,1]$ and $fg = 0$.
A: You need $f(x)g(x)=0$ for every value of $x\in[0,1]$.  The only way you can multiply two numbers and get $0$ is that one of them is $0$.  So for every $x$, either $f(x)=0$ or $g(x)=0$.
To say that $f$ is not identically $0$ means for some $x\in[0,1]$ we have $f(x)\ne 0$.
To say that $g$ is not identically $0$ means for some $x\in[0,1]$ we have $g(x)\ne 0$.
So what you need is that for some values of $x$, $f(x)$ is $0$ and $g(x)$ is not $0$, and for all other values of $x$, $g(x)$ is $0$ and $g(x)$ is not $0$.
If $f(x)=0$ whenever $x<1/2$ and $g(x)=0$ whenever $x>1/2$, that will work.  In fact, we can just take $g(x) = f(1 - x)$.
Now the hard part: If a function is $0$ on $[0,1/2]$ and nonzero on $(1/2,1]$, how can we assure that it is infinitely differentiable?  A standard example is
$$
f(x) = \begin{cases} 0 & \text{if }x\le 1/2, \\  e^{-1/(x-1/2)^2} & \text{if }x>0.  \end{cases}
$$
Continuity at $x=1/2$ is easy.  Differentiability will require a proof that $f^{(n)}(1/2)=0$ for all $n$.  And here I find I must recapitulate the details for myself before posting anything further here.  Possibly you will have that done before I do.
A: Pick two bump functions with disjoint supports. 
