I'm self studying generalized functions, and it's been the third time or so that I come across the following statement :
If $\phi(x)$ is differentiable at $0$, then there exists a function $\psi(x)$ continuous at $0$ such that $\phi(x) = \phi(0) + x\psi(x)$.
Which theorem or result is used here? Is there a higher order version (i.e. in the case where $\phi$ is twice differentiable or more).
At first I used to tell myself it's kind of like a taylor expansion, say :
$$\phi(x) = \phi(0)+x\phi'(0) + x^2\epsilon(x)$$
but then $\phi'(0)+x\epsilon(x)$ is not guaranteed to be defined at $0$; well, if it were, then we'd have the continuity at $0$ due to $\lim_{x \to 0} \epsilon(x) = 0$.
Hint
$$\phi(x) = \phi(0) + x\psi(x) \Leftrightarrow \\ \psi(x)= \frac{\phi(x)-\phi(0)}{x-0} \forall x \neq 0$$
$\phi$ is differentiable at $x=0$ if and only if $$\lim_{x \to 0} \psi(x)= \phi'(0)$$
In order to make $\psi$ continuous at $x=0$ you need to define $\psi(0)$ such that $$\lim_{x \to 0} \psi(x)= \psi(0)$$
SOOO You know what $\psi(x)$ should be for $x \neq 0$ and what $\psi(0)$ should be.