Let $\mathcal{H}$ be a Hilbert space and $r$ be a fixed natural number. Suppose $\{P_n\}_{n=1}^\infty$ be a sequence of projections on $\mathcal{H}$ with $\text{dim ran }P_n<r$ for all $n$. Let $P$ be another projection on $\mathcal{H}$ such that $P_n\rightarrow P$ in SOT. Then show that $\text{dim ran }P<r$.
Comments: My attempts are mainly in the following two ways so far:
- Suppose $\mathcal{H}$ be separable Hilbert space with orthonormal basis $\{e_n:n\geq 1\}$. Let $f\in\text{ran}P$. Now $f=\sum\limits_{n\geq 1}\langle f,e_n\rangle e_n$. Using the SOT convergence, I was trying show $\langle f,e_n\rangle=0$ for all but finitely many $n$ and those finitely many could be atmost $r$. But I could not be successful by this approach.
- I assumed that $\text{dim ran}P>r$ and trying to get a contradiction using the SOT limit of $P_n$ but could not be successful here also.
Any comment is highly appreciated. Thanks in advance.
Arguing by contradiction, if $\text{dim ran } P\geq r$, there exists an orthonormal set $$ \{e_i:1\leq i\leq s\}\subseteq \text{ran } P. $$ Observing that $$ \lim_{n\to \infty }P_n(e_i) = P(e_i) = e_i, $$ we see that the coeficients of the matrix $$ A_n=\Big (\langle P_n(e_i), P_n(e_j) \rangle \Big)_{i, j=1}^r $$ can be made as close as desired to the coefficients of $\big (\langle e_i, e_j \rangle \big )_{i, j=1}^r$, a.k.a. the identity $r\times r$ matrix.
Since the invertible matrices form an open set, we may choose $n$ such that $A_n$ is invertible, and this implies that $\{P_n(e_i):1\leq i\leq r\}$ is a linearly independent set, contradicting the assumption that $\text{dim ran } P_n<r$.