# Adjoint of an Operator in $l^2$

Let $l^2$ be the Hilbert space of all complex sequences $\phi =(\phi_j)_{j=0}^{\infty}$ such that $\sum_{j=0}^{\infty} |\phi_j |^2 < \infty$. Set

$D= \{ \phi \in l^2 : \sum_{j=0}^{\infty} j |\phi_j |^2 < \infty \}$,

and consider the operator $X$ on $D$ which associates to each $\phi \in D$ the vector $X \phi$, whose j-th component (j=0,1,2,...) is

$(X \phi)_j = \sqrt{j+1} \psi_{j+1} + \sqrt{j} \psi_{j-1}$,

(we set $\psi_{-1}=0$). We have for every $\phi \in l^2 , \psi \in D$ $$\sum_{j=0}^{\infty} \bar{\phi_j} [\sqrt{j+1} \psi_{j+1} + \sqrt{j} \psi_{j-1}] = \sum_{j=0}^{\infty} \psi_j \overline{ [\sqrt{j+1} \phi_{j+1} + \sqrt{j} \phi_{j-1}]},$$ where again we set $\phi_{-1}=0$. So in particular, $X$ is a symmetric operator. But it is not self-adjoint. To see this, consider the vector $\phi$ whose j-th component is $\phi_j = (-1)^{\lfloor j/2 \rfloor} j^{-\beta}$, where $1/2 < \beta <1$. Is is easy to see that $\phi \in l^2 \backslash D$, and that the vector whose j-th component is $\sqrt{j+1} \phi_{j+1} + \sqrt{j} \phi_{j-1}$ is in $l^2$, so $\phi$ belongs to the domain of the adjoint. I conjectured that the domain of the adjoint $X^{*}$ is exactly the set of all vectors $\phi \in l^2$ such that $\sum_{j=0}^{\infty} |\sqrt{j+1} \phi_{j+1} + \sqrt{j} \phi_{j-1} |^2 < \infty$. Anyhow, I could not prove it, and now I am starting to think that it is false. See also my related post "Null functional on $l^2$".

Your proposed domain for the adjoint $X^\star$ appears to me to be correct.

As defined, $$(Xf,g) = \sum_{j=0}^{\infty}(\sqrt{j+1}f_{j+1}+\sqrt{j}f_{j-1})\overline{g_j}.$$ By definition of adjoint, $g\in\mathcal{D}(X^{\star})$ iff there exists $h \in \ell^2$ such that the following holds for all $f \in \mathcal{D}(X)$: $$\sum_{j=0}^{\infty}(\sqrt{j+1}f_{j+1}+\sqrt{j}f_{j-1})\overline{g_j}=(Xf,g) = (f,h) = \sum_{j=0}^{\infty}f_j\overline{h_j}$$ In particular, it must hold for $f=(0,1,0,0,\cdots)$, which leads to $$\overline{g_0}+\sqrt{2}\overline{g_2}=\overline{h_1} \implies h_1=g_0+\sqrt{2}g_2.$$ And it must hold for $f=(0,0,1,0,\cdots)$, which gives $$\sqrt{2}\overline{g_1}+\sqrt{3}\overline{g_3}=\overline{h_2} \implies h_2= \sqrt{2}g_1+\sqrt{3}g_3$$ So it is necessary that $h_j=\sqrt{j}g_{j-1}+\sqrt{j+1}g_{j+1}$. And it is necessary that $\sum_{j}|h_j|^2 < \infty$. So $g$ is in your proposed domain $\mathscr{D}$. Conversely, suppose $g\in\mathscr{D}$, and suppose $f \in \mathcal{D}(X)$. Then $$\sum_{j=0}^{\infty}(\sqrt{j+1}f_{j+1}+\sqrt{j}f_{j-1})\overline{g_j} -\sum_{j=0}^{\infty}f_j(\sqrt{j+1}\overline{g_j}+\sqrt{j}\overline{g_{j-1}}) = 0,$$ because you can rearrange the terms in the first sum without affecting convergence in order to match the terms in the second sum. (This is because $f\in\mathcal{D}(X)$ implies the absolute convergence of $\sum_{j=0}^{\infty}\sqrt{j+1}f_{j+1}\overline{g_{j}}$ and of $\sum_{j=0}^{\infty}\sqrt{j}f_{j-1}\overline{g_{j}}$.)

• For sure, your argument is correct, and I thank you very much for having clearly answered my question. Still I have a big doubt about this operator. Is the closure of $X$ equal to the adjoint $X^{*}$? Clearly, the domain of $\bar{X}$ is contained in the domain $D$ of $X^{*}$, but I can't show the reverse inclusion. Any help is welcome. Jan 27, 2016 at 15:39
• @MauryBarbato : Have you tried to solve $X^\star f= \pm i f$? If either has a non-trivial solution with $f \in \ell^2$, then $X$ cannot be essentially selfadjoint, meaning that the closure of $X$ is not $X^{\star}$. That actually resolves the problem one way or the other. I don't have much time at the moment, or I would try. Jan 27, 2016 at 18:30
• I checked that the equation $X^{*}f= \pm i f$ has only the trivial solution, but I didn't understand why this should imply that the closure of $X$ is strictly contained in the adjoint $X^{*}$ of $X$. Could you give me some hint or some reference where I could learn this fact, please? Thank you very much in advance. Jan 27, 2016 at 20:18
• I have found the result you were referring to in Schmudgen, Unbounded Self-Adjoint Operators on Hilbert Space, Proposition 3.8. Thank you very much for your invaluable help. I wouldn't be able to answer these questions by myself! Jan 27, 2016 at 20:36
• @MauryBarbato : You're very welcome. I'm glad I could help. Knowing that $\mathcal{N}(X^{\star}\pm iI)=\{0\}$ means that $X^{\star}$ must be the closure of $X$ because $\mathcal{D}(X^{\star})=\mathcal{D}(X^c)\oplus\mathcal{N}(X^{\star}-iI)\oplus \mathcal{N}(X^{\star}+iI)$, where $X^c$ is the closure of a densely-defined symmetric operator. These null spaces are the deficiency spaces. Jan 27, 2016 at 21:13

It occurred to me that your problem has to do with a creation and annihilation operator, according to \begin{eqnarray*} X &=&U+V \\ a^{\ast } &=&V,\;a=U \end{eqnarray*} see below.

Let $\mathcal{H}=l^{2}$ \ with elements $u=u_{1},u_{2},\cdots$ and let $K$ be defined by \begin{eqnarray*} \mathcal{D}(K) &=&\mathcal{D},\;\mathcal{D}=\{u\in l^{2}|\sum_{j=0}^{\infty }j|u_{j}|^{2}<\infty \} \\ (Ku)_{j} &=&\sqrt{j}u_{j},\;u\in \mathcal{D} \end{eqnarray*} $K$ is symmetric, non-negative on $\mathcal{D}$ and its null space consists of the elements $(u_{0},0,0,\cdots )$. In fact it is self-adjoint on $% \mathcal{D}$ according to \begin{eqnarray*} (Ku,v) &=&(u,f) \\ \sum_{j=0}^{\infty }\sqrt{j}u_{j}\bar{v}_{j} &=&\sum_{j=1}^{\infty }\sqrt{j}% u_{j}\bar{v}_{j}=\sum_{j=0}^{\infty }u_{j}\bar{f}_{j}\Rightarrow f_{j}=\sqrt{% j}\bar{v}_{j},\;j\neq 0 \\ j &=&0\Rightarrow 0=(u_{0},f_{0})\Rightarrow f_{0}=0 \end{eqnarray*} We introduce the scale of spaces \begin{equation*} \mathcal{H}_{k}=[K+i]^{-k}\mathcal{H} \end{equation*} As a set $\mathcal{H}_{k}$ is dense in $\mathcal{H}$ and is itself a Hilbert space under the norm (or an equivalent one) \begin{equation*} \parallel f\parallel _{k}=\parallel Kf\parallel _{\mathcal{H}},\;f\in \mathcal{H}_{k}. \end{equation*} Thus $\mathcal{H}=\mathcal{H}_{0}$, $\mathcal{D=H}_{1}$. For $u\in \mathcal{D% }$ the operator $X$ is given by \begin{eqnarray*} (Xu)_{j} &=&\sqrt{j+1}u_{j+1}+\sqrt{j}u_{j-1}=(Uu)_{j}+(Vu)_{j},\;j>0, \;(Xu)_{j}=u_{1} \\ (Uu)_{j} &=&\sqrt{j+1}u_{j+1},\;(Vu)_{j}=\sqrt{j}u_{j-1},\;j>0, \;(Uu)_{0}=u_{1},\;(Vu)_{0}=0 \end{eqnarray*} $X$, $U$ and $V$ are bounded operators from $\mathcal{H}_{1}$ onto $\mathcal{ H}$. Next we note that $V=U^{\ast }.$ We note that for $u,v\in \mathcal{D}$ \begin{equation*} (Uu,v)=\sum_{j=0}^{\infty }\sqrt{j+1}u_{j+1}\bar{v}_{j}=\sum_{j=1}^{\infty }u_{j}\sqrt{j}\bar{v}_{j-1}=(u,Vv) \end{equation*} so $U^{\ast }\subset V$. Let now \begin{eqnarray*} (Uu,f) &=&(u,g) \\ \sum_{j=0}^{\infty }\sqrt{j+1}u_{j+1}\bar{f}_{j} &=&\sum_{j=0}^{\infty }u_{j} \bar{g}_{j} \\ \sum_{j=1}^{\infty }u_{j}\sqrt{j}\bar{f}_{j-1} &=&\sum_{j=0}^{\infty }u_{j} \bar{g}_{j} \end{eqnarray*} Choosing a specific $u$ we find \begin{eqnarray*} g_{0} &=&0 \\ g_{j} &=&\sqrt{j}\bar{f}_{j-1},\;j>0 \end{eqnarray*} so $U^{\ast }=V$. For $u\in \mathcal{H}_{2}=[K+i]^{-2}\mathcal{H}$, denoting $u=[K+i]^{-2}f$ \begin{eqnarray*} (U^{\ast }Uu)_{j} &=&(U^{\ast }U[K+i]^{-2}f)_{j}=\sqrt{j} (U[K+i]^{-2}f)_{j-1}=j([K+i]^{-2}f)_{j}=(K^{2}[K+i]^{-2}f)_{j},\;j\neq 0 \\ (U^{\ast }Uu)_{0} &=&0=(K^{2}[K+i]^{-2}f)_{0} \\ (UU^{\ast }u)_{j} &=&(UU^{\ast }[K+i]^{-2}f)_{j}=\sqrt{j+1}(U^{\ast }[K+i]^{-2}f)_{j+1}=(j+1)([K+i]^{-2}f)_{j}=((K^{2}+1)[K+i]^{-2}f)_{j} \\ (UU^{\ast }\varphi )_{0} &=&(UU^{\ast }[K+i]^{-2}f)_{0}=(U^{\ast }[K+i]^{-2}f)_{1}=([K+i]^{-2}f)_{0}\neq 0 \\ \{UU^{\ast }-U^{\ast }U\}\varphi &=&\{UU^{\ast }-U^{\ast }U\}[K+i]^{-2}f=[K+i]^{-2}f \\ UU^{\ast }-U^{\ast }U &=&[U,U^{\ast }]==1\;\mathrm{on}\;\mathcal{H}_{2}\; \mathrm{extends\;to\;}\mathcal{H} \end{eqnarray*} Recall that creation and annihilation operators satisfy \begin{equation*} \lbrack a,a^{\ast }]=1 \end{equation*} so we can identify \begin{equation*} a=U,\;a^{\ast }=U^{\ast }=V \end{equation*} Then $K^{2}=U^{\ast }U$ is the number operator.