Abstract smoothing does not imply a spatial derivative without a PDE realisation

The diagonal operator is self-adjoint and nonpositive. $D(A)$ contains the finitely supported vectors, hence is dense in $X$ by [L1]; for $x,y\in D(A)$ the series $⟨Ax,y⟩={\sum }_n(-n^2)x_n\overline{y_n}$ converges absolutely and equals $\overline{⟨Ay,x⟩}$ because the diagonal entries are real, so $A$ is symmetric. If $y\in D(A^{\ast })$, the adjoint identity tested against each $e_n\in D(A)$ gives $(A^{\ast }y{)}_n=-n^2{y}_n$ for every $n$; since $A^{\ast }y\in X={\mathrm{\ell }}^2({\mathbb{N}}_{\ge 1})$ and $(e_n)$ is an orthonormal basis by [L1], Parseval gives ${\sum }_n{n}^4\mathrm{\mid }{y}_n{\mathrm{\mid }}^2=\mathrm{\parallel }{A}^{\ast }y{\mathrm{\parallel }}^2<\mathrm{\infty }$, so $y\in D(A)$. Symmetry gives the reverse inclusion $D(A)\subseteq D(A^{\ast })$, hence $D(A^{\ast })=D(A)$ and $A$ is self-adjoint. Finally $⟨Ax,x⟩=-{\sum }_n{n}^2\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\le 0$ for every $x\in D(A)$.

The diagonal family is the semigroup generated by $A$. For $t\ge 0$ one has $\mathrm{\parallel }T(t)x{\mathrm{\parallel }}^2={\sum }_n{e}^{-2n^{2}t}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\le e^{-2t}\mathrm{\parallel }x{\mathrm{\parallel }}^2$, so $T(t)$ is a contraction for $t\ge 0$; the functional equation is coefficientwise and strong continuity at $0$ follows from $\mathrm{\parallel }T(t)x-x{\mathrm{\parallel }}^2={\sum }_n(1-e^{-n^{2}t}{)}^2\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\to 0$ by dominated convergence; for $x\in D(A)$ the difference quotients satisfy $\mathrm{\parallel }(T(t)x-x)\mathrm{/}t-Ax{\mathrm{\parallel }}^2={\sum }_n(\frac{1-e^{-n^{2}t}}{n^{2}t}-1{)}^2{n}^4\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\to 0$ by dominated convergence, since $\mid \frac{1-e^{-s}}{s}-1\mid \le 1$ for $s\ge 0$ and ${\sum }_n{n}^4\mathrm{\mid }{x}_n{\mathrm{\mid }}^2<\mathrm{\infty }$, so $A$ is contained in the generator; conversely, if $x$ lies in the domain $G$ of the generator then for each $n$ continuity of the $n$-th coordinate functional gives $G{x}_n={\lim }_{t\downarrow 0}(e^{-n^{2}t}-1)x_n\mathrm{/}t=-n^2{x}_n$, so ${\sum }_n{n}^4\mathrm{\mid }{x}_n{\mathrm{\mid }}^2=\mathrm{\parallel }Gx{\mathrm{\parallel }}^2<\mathrm{\infty }$ and $x\in D(A)$ with $Ax=Gx$; hence the generator of $T$ is exactly $A$.

The abstract semigroup is this diagonal semigroup. By [step 1.1] $A$ is self-adjoint and nonpositive, so [L2] makes $A$ sectorial of angle $\pi \mathrm{/}2$ and the generator of a bounded analytic semigroup, while [step 1.2] exhibits $T$ as an exponentially bounded strongly continuous semigroup with generator $A$; by the uniqueness in [L3] these semigroups coincide, so the diagonal family $T(t)x=(e^{-n^{2}t}{x}_n)$ is the semigroup generated by $A$, which is the assertion of the statement.

The iterated domains and the smoothing identities. By induction from [step 1.2] the graph domain is $D(A^m)=\{x\in X:{\sum }_n{n}^{4m}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2<\mathrm{\infty }\}$ with $(A^{m}x{)}_n=(-n^2{)}^m{x}_n$: the case $m=1$ is the definition of $D(A)$, and if the description holds for $m$ then $A^{m}x\in D(A)$ exactly when ${\sum }_n{n}^4\mathrm{\mid }(A^{m}x{)}_n{\mathrm{\mid }}^2={\sum }_n{n}^{4m+4}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2<\mathrm{\infty }$; consequently for $t>0$ the vector $T(t)x$ has $A^{m}T(t)x=((-n^2{)}^m{e}^{-n^{2}t}{x}_n{)}_n$ and $\mathrm{\parallel }{A}^{m}T(t)x{\mathrm{\parallel }}^2={\sum }_n{n}^{4m}{e}^{-2n^{2}t}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\le {\sup }_n(n^{2m}{e}^{-n^{2}t}{)}^2\mathrm{\parallel }x{\mathrm{\parallel }}^2<\mathrm{\infty }$, so $T(t)x\in D(A^m)$ and $\mathrm{\parallel }{A}^{m}T(t)\mathrm{\parallel }\le {\sup }_{n\ge 1}{n}^{2m}{e}^{-n^{2}t}\le {\sup }_{s\ge 0}{s}^m{e}^{-st}=(m\mathrm{/}(et){)}^m$; this reproduces the abstract membership $T(t)X\subseteq D(A^m)$ of [L3] with an explicit constant.

The witness is not in $D(A)$ but is smoothed. For $x=(1\mathrm{/}n{)}_{n\ge 1}$ one has ${\sum }_n\mathrm{\mid }{x}_n{\mathrm{\mid }}^2={\sum }_n{n}^{-2}<\mathrm{\infty }$, so $x\in X$, while ${\sum }_n{n}^4\mathrm{\mid }{x}_n{\mathrm{\mid }}^2={\sum }_n{n}^2=+\mathrm{\infty }$, so $x\mathrm{\notin }D(A)$ by [step 3.1]; for every $t>0$ and every $m\ge 1$, however, ${\sum }_n{n}^{4m}{e}^{-2n^{2}t}{n}^{-2}<\mathrm{\infty }$ because the exponential decay dominates every polynomial, so $T(t)x\in D(A^m)$ with the series of [step 3.1], and the smoothing thus raises the abstract regularity of a vector that is not even in the domain of $A$.

No spatial derivative is produced. The statements of steps 3.1 and 4.1 are identities between sequences: $A^m$ acts by the multiplier $(-n^2{)}^m$ and the index $n$ carries no spatial or differential meaning, so the inclusion $T(t){\mathrm{\ell }}^2\subseteq {\bigcap }_{m}D(A^m)$ is purely operator-theoretic; by [L4] the graph domain $D(A^m)$ acquires the interpretation of Sobolev derivatives only after an elliptic-regularity theorem identifies $A$ with a differential operator, and no such identification is present for this diagonal sequence operator, which is why the example is the companion witness to that remark.