# log|x| is a tempered distribution

How do I prove that $$\log|x|$$ is a tempered distribution on $$\mathcal{S}(\mathbb{R}^n)$$, ie., I need to prove that the linear functional $$\phi \in \mathcal{S}(\mathbb{R}^n) \mapsto \int_{\mathbb{R}^n} \phi(x)\log|x|dx$$ is continuous. It's suficcient to prove that $$\phi_k \rightarrow 0$$ in $$\mathcal{S}(\mathbb{R}^n)$$ implies that $$\int_{\mathbb{R}^n} \phi_k(x)\log|x| \rightarrow 0$$ as $$k \rightarrow \infty$$, but I don't know estimate $$\phi_k(x)\log|x|$$ by an integrable function, so I could use dominated convergence theorem, or estimate it by a sum seminorms $$\|\phi_k\|_{\alpha,\beta}$$. (this is the Example 2.3.5 (7) from Grafakos's Book - Classical Fourier Analysis - third edition)

Suppose $$\phi_k \to 0$$ in $$\mathcal S.$$ Then

$$\sup_x (1+|x|)^{n+1}|\phi_k(x)| =M_k \to 0$$

as $$k\to \infty.$$ Thus the sequence $$M_k$$ is bounded by some $$M.$$ We then have

$$|\phi_k(x)||\ln |x|| \le M\frac{|\ln |x||}{(1+|x|)^{n+1}}$$

for all $$k$$ and $$x.$$ The function on the right is in $$L^1(\mathbb R^n)$$ (verify by going to polar coordinates in $$\mathbb R^n$$). Since $$\phi_k\to 0$$ pointwise, the DCT gives the desired result.

• Hi Z-man! I hope that you're staying safe and healthy. Just curious … are you restricting your analysis to $\mathbb{R}^1$ only? For example, in $\mathbb{R}^3$, $$\int_0^{2\pi}\int_0^\pi \int_0^\infty \frac{\log(r) }{(1+r)^2}\,\sin(\theta)\,r^2 \sin(\theta)\,dr,,d\theta\,\,d\phi$$clearly diverges. So, perhaps instead, just start with $(1+|x|^n)^2$. That should suffice. – Mark Viola May 8 '20 at 3:55
• @MarkViola Hey MV, staying safe, healthy and bored. You are right, I was in $\mathbb R^1$ while the problem is posed for $\mathbb R^n.$ Will edit later. Thank you! – zhw. May 8 '20 at 15:27
• Hey! Pleased to hear you're doing well; unhappy to hear that you're bored. And you're welcome for the catch. – Mark Viola May 8 '20 at 16:46
• @MarkViola Fixed! – zhw. May 8 '20 at 17:31
• It's not trivial if you haven't dealt with radial functions too much. Here's what will do it: If $f\in L^1$ and $f$ is radial, i.e., $f(x)=g(|x|)$ for some $g$ on $[0,\infty),$ then $$\int_{\mathbb R^n} f(x)\,dx = C\int_0^\infty g(r)r^{n-1}\,dr.$$ Here $C$ is a constant independent of $f.$ – zhw. May 8 '20 at 20:03