# Analytic Vectors (Nelson’s Theorem)

Let \$A\$ be a symmetric operator in a Hilbert space \$mathcal H\$. For \$f in mathcal C^infty(A)\$ let \$mathcal L_f\$ be the
vector space spanned by \${A^k f | k=0, 1,… }\$ and \$mathcal H_f := overline{mathcal L_f} \$.

Nussbaum’s “Vector of Uniqueness” Lemma: If for non-real \$lambda\$, \$mathcal D(A)\$ contains a dense set of vectors \$u\$ such that \$mathcal R(A | mathcal L_u-lambda)\$ is dense in \$mathcal H_u\$ then \$mathcal R(A-lambda)\$ is dense in \$mathcal H\$ (\$A\$ is essentially self-adjoint).
Proof: There are vectors in \$mathcal R(A | mathcal L_u-lambda)\$ arbitrarily close to \$u\$ and \$mathcal R(A | mathcal L_u-lambda) subseteq mathcal R(A-lambda)\$. \$blacksquare\$

\$f\$ is called an analytic vector for \$A\$ if \$ sum_{n=0}^infty frac{Vert A^nf Vert s^n}{ n!} lt infty \$ for some \$sgt 0\$ (depending on \$f\$) \$(Leftarrow Vert A^nf Vert = O(n!) )\$.

Nelson’s Theorem: If \$mathcal D(A)\$ contains a dense set of analytic vectors then \$A\$ is essentially self-adjoint.
Proof:
We prove a stronger theorem of Nussbaum with the weaker condition “quasi-analytic” vector:
\$sum_{n=1}^infty frac{1}{Vert A^nf Vert ^{frac{1}{n}}} = infty\$
(\$LeftarrowVert A^nf Vert = O( n^n)\$). W.l.o.g. \$Vert f Vert=1\$.
On an orthonormal basis \${e_n}\$ constructed from \${A^nf }\$ by the Gram-Schmidt process, \$A | mathcal L_f\$ is a Jacobi matrix operator

\$\$ begin{bmatrix}
a_0 & b_0 & 0 & 0 & cdots \
b_0 & a_1 & b_1 & 0 & cdots\
0 & b_1 & a_2 & b_2 & cdots \
0 & 0 & b_2 & a_3 & cdots \
vdots & vdots & vdots & vdots & ddots \
end{bmatrix}.
\$\$
(\$Ae_m in text{span}(e_0,…e_{m+1})\$ so \$(e_n, Ae_m) = 0\$ for \$ngt m+1\$. Since \$A\$ is symmetric, the matrix must be tridiagonal. The off diagonal elements can be made positive real by choice of phase of the \$e_n\$
.)

Then \$b_0b_1…b_{n-1}=(e_n, A^n f)le Vert A^nf Vert\$. Using Carleman’s inequality:
\$\$infty = sum_{n=1}^inftyfrac{1}{Vert A^nf Vert ^{frac{1}{n}}} le sum_{n=1}^infty frac{1}{b_0 b_1…b_{n-1}}^frac{1}{n} le esum_{n=0}^infty frac{1}{b_n}\$\$
By Carleman’s test \$mathcal A | mathcal L_f\$ is essentially self-adjoint and by Nussbaum’s lemma so is \$A\$. \$blacksquare\$

Carleman’s test: A Jacobi matrix operator \$J\$ is essentially self-adjoint if \$sum_{n=0}^infty frac{1}{b_n} =infty\$.
Proof:
Suppose a vector \${p_k}\$ is orthogonal to the range of \$J-lambda\$ for non-real \$lambda\$. Then:
begin{align}
lambda p_0 = a_0 p_0 +b_0 p_1 \
lambda p_n = b_{n-1}p_{n-1} + a_np_n +b_n p_{n+1} && n=1, 2, …
end{align}
Multiplying this and its conjugate by \$bar p_n\$ and \$p_n\$ respectively and subtracting leads to:
begin{align}
frac{b_0(p_1 bar p_0 -bar p_1 p_0)} {lambda – bar lambda} =
|p_0|^2 \
frac{b_n(p_{n+1} bar p_n -bar p_{n+1} p_n )}{lambda – bar lambda} =
frac{b_{n-1}(p_n bar p_{n-1} -bar p_n p_{n-1})}{lambda – bar lambda}+ |p_n|^2 && n=1, 2, …
end{align}
Note that the LHSs are \$ ge |p_0|^2\$. Using the Cauchy-Schwarz inequality:
\$\$infty = |p_0|^2sum_{n=0}^infty frac{1}{b_n} le frac{1}{lambda – barlambda}sum_{n=0}^infty (p_{n+1} bar p_n -bar p_{n+1} p_n) le frac{1}{|text{Im}(lambda)|}sum_{n=0}^infty |p_n|^2\$\$ \$blacksquare\$

Analytic vectors are used to construct continuous unitary group representations on a dense subset, it extends to the entire space and the generator is the self-adjoint closure of the original operator, see p.200-202 in Reed and Simon, Vol. II.