# Bounded operators leaving dense subspace invariant

Let $A$ be a C$^*$-algebra. A pre-Hilbert $A$-module $H$ is a right $A$ module with a $A$-valued inner product (which is linear in the second variable and conjugate linear in the first variable) such that

1.$\langle \xi, \beta \cdot T \rangle_{A} = \langle \xi, \beta\rangle_{A}T$,

2.$\langle \xi, \xi\rangle_{A} \geq 0$, and $\langle \xi, \xi\rangle_{A}=0$ implies $\xi=0$,

where $\xi, \beta \in H$ and $T \in A$. If $H$ is complete with respect to the norm $\|\langle \xi, \xi\rangle_{A}\|^{1/2}$, then $H$ is called a Hilbert $A$-module.

If $H$ is a pre-Hilbert $A$-module, let $\overline{H}$ be the completion of $H$ with respect to the norm $\|\langle \xi, \xi\rangle_{A}\|^{1/2}$. Let $B(H)$ be the set of bounded adjointable mapping from $H$ to $H$. It is clear that each $T \in B(H)$ extends to a element in $B(\overline{H})$ (the bounded adjointable mapping from $\overline{H}$ to $\overline{H}$). Thus, we can identify $B(H)$ as a subalgebra in $B(\overline{H})$. My question is if $B(H)$ is dense in $B(\overline{H})$?

• Just to check that I understand the question: if $A$ is just the complex numbers, then does your question become: suppose $V_0$ is a dense subspace of a Hilbert space $V$; is $B(V_0)$ dense in $B(V)$? – Yemon Choi Jun 7 '17 at 18:44
• If so, then I know examples where this would fail – Yemon Choi Jun 7 '17 at 18:44
• Yes, you are absolutely right. Would you please let me know such an example? Thank you. – heller Jun 8 '17 at 1:44
• I think I worked this out for myself about 10 years ago -- see the comments in mathoverflow.net/questions/30353/… -- but it surely must have been observed before. I will try to find time later to write out the details here – Yemon Choi Jun 8 '17 at 2:26
• Thanks Yemon, I add details and post a example below. If you find any mistakes, please let me know. Cheers. – heller Jun 8 '17 at 10:11

The following is based on a copy of some old handwritten notes; I haven't had time to refresh my memory on all the details. I will try to fill in these details in the next day or so. $\newcommand{\Nat}{{\bf N}}\newcommand{\veps}{\varepsilon}$

Using the notation of my comment: take $V_0=\ell_1(\Nat)$ sitting inside $V=\ell_2(\Nat)$, and let $A_0$ denote the space of all bounded operators on $V$ that map $V_0$ to itself. We seek a bounded operator $S:\ell_2(\Nat)\to \ell_2(\Nat)$ whose distance from $A_0$ is strictly positive.

Fix a sequence of non-empty finite subsets of $\Nat$, denoted by $(F_j)_{j\geq 1}$, which satisfy $\max F_j < \min F_{j+1}$. (We will later want $|F_j|\to\infty$ at a certain rate.) Let $$u_j := |F_j|^{-1/2} {\bf 1}_{F_j}$$ and, denoting the standard basis of $\ell_2$ by $(e_j)_{j\geq 1}$, define $S:c_{00}(\Nat)\to \ell_2(\Nat)$ by $Se_j = u_j$. A quick calculation shows that $S$ extends to an isometry on $\ell_2(\Nat)$.

Now suppose $T\in {\mathcal B}(\ell_2(\Nat))$ satisfies $\Vert S-T\Vert = \veps <1$. The idea is that since $\Vert Te_n-u_n\Vert_2 \leq\veps$, $Te_n$ must be close in $\ell_2$-norm to something that will have large $\ell_1$-norm (provided we make $|F_j|\nearrow \infty$ at a suitable rate). Moreover, we can inductively choose a subsquence $(e_{n(j)})$ so that the vectors $Te_{n(j)}$ have approximately disjoint support in some sense. One then considers $$x:= \sum_{j=1}^\infty c^j e_{n(j)}$$ for appropriately chosen $c\in (0,1)$, to obtain $Tx = \sum_{j=1}^\infty c^j Te_{n(j)} \notin \ell_1(\Nat)$.

Thanks Yemon. I fill in the details of you example.

Let $H=l^1(N)$. Then $\overline{H}=l^2(N)$.

Let $\xi_j = 1/\sqrt{2^n} \sum_{i=0}^{2^n-1} e_{2^n+i}$ where $\{e_1, e_2, \ldots \}$ is the canonical orthonormal basis of $l^2(N)$.Define $V \in B(l^2(N))$ by $Ve_j = \xi_j$. Then it is clear that $V$ is a partial isometry.

Assume that $T \in B(H)$ and $\|T-V\| < 1/2$. Note that $\|Te_n - \xi_n\| < 1/2$. Thus $$\sum_{i=0}^{2^n-1} |\langle{e_{2^n+i},Te_n\rangle}| \geq \sqrt{2^{n}}-\sum_{i=0}^{2^n-1} |\langle{e_{2^n+i},Te_n\rangle} - 1/\sqrt{2^n}| \geq \sqrt{2^{n}}(1-\|Te_n - \xi_n\|)\geq \frac{\sqrt{2^{n}}}{2}.$$

We now choose inductively a two subsequence of non negative numbers $1=n_0 < n_1< n_2 < \cdots$ and $n(1) < n(2) < \cdots$ such that $n(k) \geq k^2$ and $$\sum_{i \notin [n_{l-1}, n_{l}-1]} |\langle{e_{i},Te_{n(l)}\rangle}| < 1, \quad l = 1, 2, \ldots$$

Let $k(1) =1$. Since $Te_1 \in l^1(N)$, we can choose $n_1 \in N$ satisfies the above condition. Assume that $\{n_1, \ldots, n_k\}$ and $\{n(1), \ldots, n(k)\}$ are chosen. Recall that $T^*$ also maps $l^1(N)$ into $l^1(N)$ and $\{T^*e_i\}_{i=1}^{n_k-1} \subset l^1(N)$. We may choose $n(k+1) > \max(n_k, n(k))$ such that $\sum_{i \in [1, n_{k}-1]} |\langle{e_{i},Te_{n(k+1)}\rangle}| < 1/2$. Now it is clear that we can choose $n_{k+1} > n_k$ such that $\sum_{i \notin [n_{k}, n_{k+1}-1]} |\langle{e_{i},Te_{n(k+1)}\rangle}| < 1$.

Let $\beta = \sum_{k=1}^{\infty} 1/k^2 e_{n(k)} \in l^1(N)$. Then $$\|T\beta\|_1 = \sum_{l=1}^{\infty} \sum_{i \in [n_{l-1}, n_l -1]} |\langle{e_i, \sum_{k=1}^{\infty} 1/k^2 Te_{n(k)}\rangle}|$$ $$\geq \sum_{l=1}^{\infty} \frac{\sqrt{2^{n(l)}}}{2l^2} - \sum_{l=1}^{\infty} \frac{1}{l^2} = +\infty.$$