If $\phi \in C_c^{\infty}(\mathbb{R})$ then$ \frac{\phi}{x}\in L^1(\mathbb{R})$?

Question

If $\phi \in C_c^{\infty}(\mathbb{R})$ then$ \frac{\phi}{x}\in L^1(\mathbb{R})$ ?

Let $\phi$ be a smooth, compactly supported function and $\frac{1}{x}$ be an odd function defined on $\mathbb{R}$.

I want to show that $ \frac{\phi}{x}\in L^1(\mathbb{R})$. I need this result to prove that distributional derivative of $log(|x|)$ is $ 1/x$. Any hints or reference are appreciated. Thank you.

Answer 1

This is false. If $\phi =1$ in a neighborhood of $0$ and $x(t)=t$ then $\frac \phi x $ is not in $L^{1}$.

Answer 2

If $\phi \in C_c^\infty(\mathbb R)$ then, in general, $\frac{1}{x} \phi \not\in L^1(\mathbb R).$

But if $\phi(0) = 0$ then $\frac{1}{x} \phi \in C_c^\infty(\mathbb R)$ if we set $\left(\frac{1}{x}\phi\right)(0) = \phi'(0).$

As I see that you want to show that $\frac{1}{x} = \left(\ln |x|\right)'$ as a distribution, then first note that $\ln |x| \in L^1(\mathbb R).$ What you then should show is that $x \, (\ln |x|)' = 1$ as distributions.

Make a try, and come back if you don't succeed.