Let $X$ separable Hilbert space with basis $\{e_n\}$, $U_N=\{e_1,…e_N \} \subset X$, $P_N$ the projection operator over $U_N$. Define $Q_N=I-P_N$ the orthogonal complement ($I$ identity) and $B \colon X \to X$ compact linear operator.
I want to prove that $$\lim_{N \to \infty}||BQ_N||= 0$$
I would proceed in the following way:
For every $x \in X$ we have $P_N x \to x$ as $N \to \infty$. Then $Q_N x \to 0$ so that $BQ_N x \to 0$. Now since $B$ is compact we have that the set $$\overline{B Q_N \{ x \in X: |x|=1\}}=B Q_N \{ x\in X: |x|=1\} $$ is compact. This should imply that the convergence $BQ_N x \to 0$ is uniform in $\{x \in X, |x|=1 \}$, i.e. $$\sup_{|x|=1} |B Q_Nx| \to 0$$
which implies my claim. How do I justify the uniform convergence in $\{x \in X, |x|=1 \}$ ?
Best Answer
Not sure if having infinitely many compact sets leads to a nice argument. Below I can show you two ways of going about this.
It is easier to prove that $Q_NB\to0$. This is enough, because we can then get $Q_NB^*\to0$ and take adjoints.
Fix $\varepsilon>0$. The compactness of $\overline{\{ Bx : |x|=1\}}$ (you cannot remove the closure, in general the range of a bounded operator is not closed) allows you to choose $\{Bx_1,\ldots,Bx_m\}$, with $|x_k|=1$ for all $k$, such that for all $x$ with $|x|=1$ you have $k\in\{1,\ldots,m\}$ with $$\|Bx-Bx_k\|<\varepsilon/2.$$ Since $Q_NBx_k\to0$ for $k=1,\ldots,m$, there exists $N_0$ such that $\|Q_NBx_k\|<\varepsilon/2$ for all $N\geq N_0$. Then $$ \|Q_NBx\|\leq\|Q_NBx-Q_NBx_k\|+\|Q_NBx_k\|\leq\|Bx-Bx_k\|+\frac\varepsilon2<\varepsilon. $$ So $\|Q_NB\|<\varepsilon$ for all $N\geq N_0$.
The second way is to use that $B=\lim B_n$, with $B_n$ finite rank. Since $B_nQ_N$ is finite rank, convergence in a ball is uniform due to being in a finite-dimensional space. So $B_nQ_N\xrightarrow{N\to\infty}0$. Then $$ \|BQ_N\|\leq\|(B-B_n)Q_N\|+\|B_nQ_N\|\leq\|B-B_n\|+\|B_nQ_N\|. $$ Thus $$ \limsup_N\|BQ_N\|\leq\|B-B_n\|, $$ and as $n$ is arbitrary we get that $\lim_N\|BQ_N\|$ exists and is zero.