The Euclidean Dirac operator has nonclosed range at zero

Under the unitary Fourier transform [F2], $D$ is multiplication by $2\pi ic(\xi )$. By [F1], its maximal multiplier domain is exactly $\{u\in L^2:\mathrm{\mid }\xi \mathrm{\mid }\widehat{u}\in L^2\}$. If $u\in H^1$, [F3] and the distributional derivative identity in [F2] put $\mathrm{\mid }\xi \mathrm{\mid }\widehat{u}$ in $L^2$. Conversely, if $\mathrm{\mid }\xi \mathrm{\mid }\widehat{u}\in L^2$, then each $2\pi i{\xi }_j\widehat{u}$ is in $L^2$; its inverse Plancherel transform is the distributional derivative ${\mathrm{\partial }}_{j}u$, by the Fourier differentiation identity and injectivity of the Fourier transform on tempered distributions. Hence $u\in H^1$ by [F3]. A multiplication operator by a measurable finite matrix is closed, since convergence of $u_k$ and $c(\xi )u_k$ in $L^2$ has an almost-everywhere convergent subsequence and the pointwise multiplication identity passes to the limit. Schwartz sections are dense in $L^2$, so $D$ is densely defined and closed.

If $h$ is orthogonal to the range, then distributionally $D^{\ast }h=0$. Fourier transformation and [F2] give $c(\xi )\widehat{h}(\xi )=0$ almost everywhere. By [F1], $c(\xi )$ is invertible for every $\xi \mathrm{\ne }0$, and the singleton $\{0\}$ has Lebesgue measure zero; hence $\widehat{h}=0$ almost everywhere and $h=0$. Thus the range is dense.

Fix a unit vector $v\in S_0$ and define $\widehat{g}(\xi )={\mathbf{1}}_{\{0<\mathrm{\mid }\xi \mathrm{\mid }<1\}}\mathrm{\mid }\xi {\mathrm{\mid }}^{-(n\mathrm{/}2-1)}v$. Polar coordinates give $\mathrm{\parallel }\widehat{g}{\mathrm{\parallel }}_2^2=C_n{\int }_0^1{r}^{n-1-2(n\mathrm{/}2-1)}dr=C_n{\int }_0^{1}rdr<\mathrm{\infty }$, so [F2] gives $g\in L^2$. If $g=Du$ for some $u\in H^1$, then for almost every $\xi \mathrm{\ne }0$, [F1] forces $\mathrm{\mid }\widehat{u}(\xi )\mathrm{\mid }=(2\pi {)}^{-1}\mathrm{\mid }\xi {\mathrm{\mid }}^{-1}\mathrm{\mid }\widehat{g}(\xi )\mathrm{\mid }$. Consequently its squared norm near zero is a positive constant times ${\int }_0^1{r}^{n-1-2(n\mathrm{/}2)}dr={\int }_0^1{r}^{-1}dr=\mathrm{\infty }$, contradicting $u\in L^2$. Thus $g$ is not in the range, so the dense range is proper and not closed. A Fredholm operator must have closed range, hence $D$ is not Fredholm.

The construction works also for $n=1$, where $\widehat{g}(\xi )={\mathbf{1}}_{\{0<\mathrm{\mid }\xi \mathrm{\mid }<1\}}\mathrm{\mid }\xi {\mathrm{\mid }}^{1\mathrm{/}2}v$ and the displayed $L^2$ integral is still finite; the logarithmic divergence after division by $\mathrm{\mid }\xi \mathrm{\mid }$ is unchanged. The zero-frequency set has measure zero, but arbitrarily small nonzero frequencies cause the obstruction. Countable choice is used through the Fourier and Sobolev interfaces [F2]–[F3] and the almost-everywhere subsequence argument in step 1.1.