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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)

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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.

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  • $\begingroup$ 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. $\endgroup$ – Mark Viola May 8 '20 at 3:55
  • $\begingroup$ @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! $\endgroup$ – zhw. May 8 '20 at 15:27
  • $\begingroup$ Hey! Pleased to hear you're doing well; unhappy to hear that you're bored. And you're welcome for the catch. $\endgroup$ – Mark Viola May 8 '20 at 16:46
  • $\begingroup$ @MarkViola Fixed! $\endgroup$ – zhw. May 8 '20 at 17:31
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    $\begingroup$ 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 $f.$ $\endgroup$ – zhw. May 8 '20 at 20:03

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