Sequence of isometries with mutually orthogonal ranges

Question

Let $(W_n)_{n\in \Bbb {N}} $ be a sequence of isometries in $B (H ) $ with pairwise orthogonal ranges, i.e. $W_n^*W_m=0$ for all $n\neq m $. I want to show that if $F $ is a finite rank operator then there exists $n_0$ sufficiently large so that for all $n\geq n_0$ we have $FW_n=0$.
I think that this should be equivalent to the claim that $W_n $ converges SOT to zero. However, I don't know how to show it.

Answer 1

If $(W_n)_{\mathbb{N}}$ is a sequence of isometries, you can never have $W_n x \rightarrow 0$ for $x\neq 0$, since $\lVert W_nx \rVert = \lVert x \rVert$. That is, you can never have $W_n\rightarrow 0$ in the strong operator topology. But with your assumptions, you do have $W_n x \rightarrow 0$ weakly, i.e. $\langle y,W_n x\rangle\rightarrow 0$ for all $y\in H$.

Let $P_n$ denote the orthogonal projection onto $\mathrm{Ran}(W_n)$. Then we have $$ \lvert \langle y, W_n x\rangle \rvert = \lvert \langle P_n y, W_n x\rangle \rvert \leqslant \lVert x\rVert \lVert P_n y \rVert.\tag{1} $$ Since $\sum_{n=1}^\infty \lVert P_n y \rVert ^2 \leqslant \lVert y \rVert<\infty$, we conclude that $\lVert P_n y \rVert\rightarrow 0$ as $n\rightarrow \infty$, and therefore $W_n\rightarrow 0$ in the weak operator topology.

Now, if $F$ has finite rank, pick an orthonormal basis $(f_1,\ldots,f_k)$ of $\mathrm{Ran}(F)$. Then we find from (1) that $$ \lVert FW_n x\rVert^2=\sum_{j=1}^k \lvert \langle f_j , W_n x\rangle \lvert^2 \leqslant \lVert x \rVert ^2 \sum_{j=1}^k \lVert P_n f_j \rVert^2. $$ Since $\sum_{j=1}^k \lVert P_n f_j \rVert^2 \rightarrow 0$ as $n\rightarrow \infty$, it follows that $FW_n\rightarrow 0$ in norm.

So far so good. As for your actual question, here is a counter example: Let $\phi:\mathbb{N}\rightarrow \mathbb{N}\times\mathbb{N}$ be a bijection. For $n\in\mathbb{N}$, let $\mathcal{H}_n=\ell^2(\mathbb{N})$ with standard orthonormal basis $$ e_{n,j}(k)=\begin{cases} 1, &j=k, \\ 0, & j\neq k. \end{cases} $$ Then we may consider the separable Hilbert space $\mathcal{H}=\bigoplus_{n=1}^\infty \mathcal{H}_n$, with orthonormal basis $(e_{n,j})_{(n,j)\in \mathbb{N}^2}$. Define a sequence of isometries $W_k:\mathcal{H}\rightarrow \mathcal{H}$ $$ W_k(\sum_{n,j\in\mathbb{N}}x_{n,j}e_{n,j})=\sum_{l=1}^\infty x_{\phi(l)}e_{k,l}. $$ Then $\mathrm{Ran}(W_n)=\mathcal{H}_n$. Finally, define $y=\sum_{n,j\in \mathbb{N}}y_{n,j} e_{n,j}=\sum_{n,j\in \mathbb{N}}2^{-\frac{n+j}{2}} e_{n,j}$ and let $F(x)=\langle y,x\rangle y$ be the orthogonal projection onto the subspace spanned by $y$. Then $$ \langle y, FW_k y\rangle = \langle y,W_k y\rangle = \sum_{l=1}^\infty y_{k,l}y_{\phi(l)}>0, $$ which rules out $FW_k=0$.

Answer 2

The subspaces $$(\text{ran}\,W_n)\cap (\ker F)^\perp=(\text{ran}\,W_n)\cap \overline{\text{Im}\,F^*}=(\text{ran}\,W_n)\cap {\text{Im}\,F^*}$$ are pairwise orthogonal subspaces of $\text{Im}\,F^*$, which is finite-dimensional. Thus, only finitely many are nonzero.

In consequence, since $$ (\text {ran}\,W_n)\cap (\ker F)^\perp\ne\{0\} \iff FW_n\ne0, $$ we have that $FW_n\ne0$ for only finitely many $n$.