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.

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