$\operatorname{Ker}(T)$, $T(f) = f'g + fg'$

Question

Let $g \in C^{1}[0,1]$

Let $T:C^{1}[0,1] \to C[0,1]$ linear transformation such that $T(f) = (fg)'$

I have to calculate $\operatorname{Ker}(T)$

I know if $f \in \operatorname{Ker}(T)$ then $(fg)' = 0$ so $fg = c$ in $[0,1]$

If $c \not = 0$, then $g(x) \not = 0$ $\forall x \in [0,1]$

Thus $f = \frac{c}{g}$

But if there exists $x \in [0,1]$ such that $g(x) = 0$ I don't know how to proceed

Can anybody help me?

Answer 1

If $c=0$ the answer $ker (T)=\{f:f(x)=0$ on the open set $\{y:g(y) \neq 0\} \}$.

Answer 2

Noting as you did that
$$ fg=c $$ for some constant $c$, we can deduce that if $g$ vanishes anywhere, $c=0$, and $f$ then is forced to vanish whenever $g$ doesn't, indeed $$ \ker(T)=\{ f\in C^1[0,1]:\;f(x)=0 \;\forall x\;\text{s.t}\; g(x)\ne 0\} $$ If $g$ does not vanish, we have the simpler expression $f=\frac{c}{g}$ for a constant $c$.