A Dirac mass has precisely sufficiently negative Sobolev order

The Fourier transform of the Dirac mass. By [F1], $\mathcal{F}{\delta }_0=u_1$ is the regular distribution of the constant function $1$.

The polar reduction. Apply [F6] to $f(\xi )=⟨\xi {⟩}^{2s}$; since $f(r\omega )=(1+r^2{)}^s$ is radial, the total mass of $\sigma$ factors out: ${\int }_{{\mathbb{R}}^n}⟨\xi {⟩}^{2s}d\xi =\sigma (S^{n-1}){\int }_0^{\mathrm{\infty }}\phi (r)dr,\phi (r):=(1+r^2{)}^s{r}^{n-1},$ with $\sigma (S^{n-1})\in (0,\mathrm{\infty })$. The function $\phi$ extends continuously to $r=0$, with value $1$ if $n=1$ and $0$ if $n>1$. By [F14], its Lebesgue integral on each compact interval equals its Riemann integral, and monotone convergence of $\phi {\mathbf{1}}_{[0,N]}$ identifies the full nonnegative Lebesgue integral with the supremum of these truncations, including when infinite. On $0<r\le 1$ one has $(1+r^2{)}^s\le \max (1,2^s)$ and $r^{n-1}\le 1$ by [F11], so ${\int }_0^1\phi \le \max (1,2^s)<\mathrm{\infty }$ by [F9]. Since ${\int }_0^R\phi ={\int }_0^1\phi +{\int }_1^R\phi$ for $R>1$ by additivity [F9], the integral over $(0,\mathrm{\infty })$ is finite if and only if the tail truncations ${\int }_1^R\phi$ are bounded, that is, if and only if the tail ${\int }_1^{\mathrm{\infty }}\phi$ is finite in the sense of [F7].

Block bounds. Put $q:=2s+n-1$ and write, by [F11], $\phi (r)=r^q(1+r^{-2}{)}^s$. For an integer $k\ge 1$ and $r\in [k,k+1]$ one has $1+r^{-2}\in [1,2]$, so $(1+r^{-2}{)}^s\in [c_1,c_2]$ with $c_1=\min (1,2^s)>0$ and $c_2=\max (1,2^s)$; moreover $k+1\le 2k$ for $k\ge 1$, so $2^{\min (q,0)}\le r^q\mathrm{/}{k}^q\le 2^{\max (q,0)}$, and with $c_3=c_1{2}^{\min (q,0)}$, $c_4=c_2{2}^{\max (q,0)}$, $c_3{k}^q\le \phi (r)\le c_4{k}^q,r\in [k,k+1],k\ge 1.$

The membership criterion. By [F2], ${\delta }_0\in H^s$ if and only if there is $g\in L^2$ with $⟨\xi {⟩}^s\mathcal{F}{\delta }_0=u_g$. By step 1.1 and [F3] applied to the smooth symbol $a=⟨\xi {⟩}^s$ and the locally integrable $h=1$, this product is $⟨\xi {⟩}^s{u}_1=u_{⟨\xi {⟩}^s}$. The condition is therefore $u_{⟨\xi {⟩}^s}=u_g$ for some $g\in L^2$. Both $⟨\xi {⟩}^s$ and $g$ are locally integrable by continuity and [F5], so [F4] forces $⟨\xi {⟩}^s=g$ almost everywhere; hence a qualifying $g$ exists exactly when $⟨\xi {⟩}^s\in L^2({\mathbb{R}}^n)$, that is, by [F13], exactly when ${\int }_{{\mathbb{R}}^n}⟨\xi {⟩}^{2s}d\xi <\mathrm{\infty }$, and then $\mathrm{\parallel }{\delta }_0{\mathrm{\parallel }}_{H^s}=\mathrm{\parallel }g{\mathrm{\parallel }}_2=\mathrm{\parallel }⟨\xi {⟩}^s{\mathrm{\parallel }}_2$.

The block comparison. For every integer $N\ge 1$, additivity and monotonicity of the integral [F9] applied to step 1.3 give $c_3{\sum }_{k=1}^N{k}^q\le {\int }_1^{N+1}\phi \le c_4{\sum }_{k=1}^N{k}^q\mathrm{.}$ If ${\sum }_{k\ge 1}{k}^q$ converges with value $S$, then for every $R>1$ [F12] supplies an integer $N\ge R$, and monotonicity [F9] together with step 1.3 gives ${\int }_1^R\phi \le {\int }_1^{N+1}\phi \le c_{4}S$, so the truncations are bounded and ${\int }_1^{\mathrm{\infty }}\phi \le c_{4}S<\mathrm{\infty }$ by [F7]. Conversely, if ${\int }_1^{\mathrm{\infty }}\phi <\mathrm{\infty }$, then ${\sum }_{k=1}^N{k}^q\le c_3^{-1}{\int }_1^{N+1}\phi \le c_3^{-1}{\int }_1^{\mathrm{\infty }}\phi$ for every $N$, so the partial sums are bounded and ${\sum }_{k\ge 1}{k}^q$ converges by [F8].

The threshold. By step 2.2, ${\int }_1^{\mathrm{\infty }}\phi <\mathrm{\infty }$ if and only if ${\sum }_{k\ge 1}{k}^q={\sum }_{k\ge 1}1\mathrm{/}{k}^{-q}$ converges, which by [F8] happens exactly when $-q>1$, that is, exactly when $2s+n-1<-1$, i.e. $s<-n\mathrm{/}2$. Combining with steps 2.1 and 1.2, this gives ${\delta }_0\in H^s⟺s<-n\mathrm{/}2$.

The strict endpoint. Let $s=-n\mathrm{/}2$, so $q=-1$ and $\phi (r)=r^{-1}(1+r^{-2}{)}^s\ge c_1{r}^{-1}$ for $r\ge 1$ by [F11]. Hence for every $R>1$, ${\int }_1^R\phi \ge c_1{\int }_1^R\frac{dr}{r}=c_1\log R⟶+\mathrm{\infty }(R\to \mathrm{\infty })$ by [F9] and [F10]; the truncations are unbounded, so ${\int }_1^{\mathrm{\infty }}\phi =\mathrm{\infty }$ by [F7], and by steps 2.1 and 1.2, ${\delta }_0\mathrm{\notin }{H}^{-n\mathrm{/}2}$. The divergence is logarithmic: the truncated integral grows like $\log R$ because the radial integrand behaves like $1\mathrm{/}r$.

Conclusion. Step 3.1 proves the equivalence ${\delta }_0\in H^s⟺s<-n\mathrm{/}2$ and exhibits the strict threshold $2s+n<0$, while step 3.2 proves that the borderline case $s=-n\mathrm{/}2$ fails by logarithmic divergence; step 2.1 identifies the exact norm with the $L^2$ norm of the unique weighted Fourier class, and no pointwise-function assumption is used anywhere. Countable Choice is used exactly through the cited characterization, regular-distribution and polar-coordinate interfaces, which carry it as their hypothesis.