Is $\ln |x| \in L^1_{loc}$?

Question

I want to prove it to define an distribution in $S'(\mathbb{R})$, but I don't know if $\ln|x|$ is in this space.

Answer 1

Yeah, it is.

You can start using the following inequality:

$|\ln t| \leq t$

For the unidimensional case. Then, applying it to $|t|^\epsilon$, we have:

$|\ln|t||\leq \epsilon^{-1}\max\{|t|^\epsilon,|t|^{-\epsilon}\}$

Now, you can use this inequality in:

$\int_{\mathbb{R}^n} |\ln |x|| dx \leq \int_{\mathbb{R}^n}\epsilon^{-1}\max\{|t|^\epsilon,|t|^{-\epsilon}\}dx < \infty $

Concluding that $\ln |x| \in L^1_{loc}(\mathbb{R}^n)$

In fact, using the hint of Rocco Maggi, you can choose $m=n+\epsilon$ and then $\ln |x| \in S'(\mathbb{R}^n)$

Answer 2

Since $\log$ only has a singularity at zero, it suffices to check whether $$ \int_{-a}^a |\log(|x|)\vert dx $$ exists for all $a>0$, which is indeed the case.

Answer 3

As an addition to the answer by daw, maybe it is important to point out also that local integrability is not a sufficient condition for a distribution to be tempered: even if $f\in\mathrm{L}^1_{\mathrm{loc}}$, you can have that $T_f$, the regular distribution in $\mathcal{D}'$ associated to $f$ by $$\langle T_f,\phi\rangle=\int f(x)\phi(x)\,\mathrm{d}x$$ for every $\phi\in\mathrm{C}^\infty_\mathrm{c}=\mathcal{D}$, is not $\mathcal{S}'$; eg: $x\mapsto\mathrm{e}^x$, which is $\mathrm{L}^1_{\mathrm{loc}}(\mathbb{R})$ but not $\mathcal{S}'(\mathbb{R})$.

A sufficient condition, for $f\in\mathrm{L}^1_{\mathrm{loc}}$, to be in $\mathcal{S}'$, is to have at most polynomial growth, which is to say that there exist $C>0$ and $m\in\mathbb{N}$ such that $\vert f(x)\vert\le C(1+\vert x \vert^m)$ — this is not a necessary condition!, eg: $x\mapsto\mathrm{e}^x \mathrm{e^{\mathrm{i}\mathrm{e}^x}}$, which has exponential growth, but still is $\mathcal{S}'(\mathbb{R})$ (because linear on the derivative of $x \mapsto \mathrm{e^{\mathrm{i}\mathrm{e}^x}}$, which is $\mathrm{L}^\infty(\mathbb{R})\subset\mathcal{S}'(\mathbb{R})$: $\mathcal{S}'$ is "stable" under derivative).

In your case, $x\mapsto\log\vert x\vert$ is $\mathrm{L}^1_{\mathrm{loc}}(\mathbb{R})$ and has less than polynomial growth at $\pm\infty$, but then you have to mind the singularity in $0$… So: why don't you combine all the hints by looking for a distribution (a) which is $\mathcal{S}'(\mathbb{R})$, and (b) whose derivative is the function you're after?