I want to calculate $$ \lim_{x \rightarrow 0} \frac{\phi'(x)}{2x} $$ where $\phi$ is a Schwartz function and $\phi(0) =0$ if that matters. Wondering if this is valid:
$$ \lim_{x \rightarrow 0} \frac{\phi'(x)}{2x} = \lim_{x \rightarrow \infty} \frac{x\phi'(1/x)}{2}$$
Chain rule:
$$\phi'(1/x) = -\frac{1}{x^2}\phi'(x)$$
Gives:
$$\lim_{x \rightarrow \infty} \frac{x\phi'(1/x)}{2} = \lim_{x \rightarrow \infty} - \frac{\phi'(x)}{2x} = 0 $$
since $\phi'$ is Schwartz. Can I use the chain rule like this?
No, you can't.
$\phi'(1/x)$ is the derivative of $\phi$ at $1/x$ (just like $\phi'(4)$ is the derivative of $\phi$ at $4$), that is, you differentiate and after this, substitute.
If you define $g(x)=\phi(1/x)$ and you want to differentiate $g$ then you can apply the chain rule, but this is not what you want here.
To solve this problem, you must know $\phi'(0)$. Without further information about $\phi$ you can't find it out. Take for example $\phi(x)=e^{-(x+K)^2}$. These are clearly Schwartz functions for any $K\in\Bbb R$ and $\phi'(0)=-2Ke^{-K^2}$.