Describe the Kernel of this Linear Transformation I was trying to solve this:
Prove that $H_g$ is a linear transformation and find its kernel.
Specifically, describe the kernel if $g(x)=x$. The definition of $H_g$ is:
$$H_g(f)=(gf)'$$
($g$ is continuous in $[0,1]$ and $H_g$ is defined in $C[0,1]\rightarrow C[0,1]$)

I proved without problems that it's a linear transformation. To find its kernel I said that $(gf)'=g'f+gf'$. And then, $g'(x)f(x)+g(x)f'(x)$ must be equal to 0 for every $x$ in order to find the kernel. But I don't know how to continue it (Maybe with some integrals? I'm not sure)
 A: Let's say $f$ is in the kernel of $H_g$. This means that $H_g(f)$ is the neutral element of the codomain, but the neutral of the codomain is the zero function, so $f$ being in the kernel is equivalent to $H_g(f)=0$. That is equivalent to $(gf)'=0$ which is further equivalent to $gf=\text{const}$. I was not using $x$ in these expressions, but we can easily include it, then the previous statements would be written as: $\frac{d}{dx}g(x)f(x)=0,\forall x\in[0,1]$ and $g(x)f(x)=c,\forall x\in[0,1]$ for some $c\in\mathbb{R}$. As you can see, the notation without $x$ is shorter, but not much less clear. Anyway, we have derived enough to conclude that $f(x)=\frac{c}{g(x)},\forall x\in[0,1]$. But this only makes sense if $g(x)\neq0,\forall x\in[0,1]$. And in that case this is precisely the kernel ($f$ is parametrized by $c$). If $g(x)=0$ for some $x$ then we can clearly see that the equation $g(x)f(x)=c$ must yield $c=0$. Then we can further conclude that $f(x)=0$ whenever $g(x)\neq0$. You stated that $f$ is an element of $C[0,1]$, i.e. it is continuous, and we know that the preimage of a point of a continuous function is a closed set. We have that $f^{-1}(0)$ is closed in $[0,1]$ and we have that $f^{-1}(0)$ contains all points $x$ for which $g(x)\neq0$. This means that $f^{-1}(0)$ must contain the closure (the minimal closed superset) of $\{x\in[0,1]|g(x)\neq0\}$ which is exactly the support of $g$ by definition. So the answer in case $g^{-1}(0)\neq\emptyset$ is the set of all functions $f$ which vanish at the support of $g$.
A: Hint:
The condition $H_g(f)=f+xf'=0$, with $g(x)=x$, serves to determine its kernel.
So, $f'(x)=\dfrac{f(x)}{x}$, that is a differential equation which characterizes, for $f$ differentiable.
