Let $\mathcal{H}$ be a separable Hilbert space and $A_m:\mathcal{H}\to \mathbb{R}^m$ linear with $\sup_m \|A_m\|<\infty$. Assume that for all $x\in\mathcal{H}\setminus \{0\}$ there is a $\varepsilon>0$ such that $\liminf_{m\to\infty}\|A_m x \| \ge \varepsilon$. What can one say then about the (weak) convergence of the sequence $x_m:=P_{\mathcal{N}(A_m)}x$ (orthogonal projection onto the kernel) for fixed $x\in\mathcal{H}$? I first thought, that the weak limit should be $0$, but I wasn't succesfull in proving this, at least without any additional (kind) of monotonicity assumption…
Non-monotone sequence of orthogonal projections
functional-analysishilbert-spacesprojection
Related Solutions
Let $x\in X$. We have $P_n(x)=\sum_{j=1}^n\langle x,e_j\rangle e_j$ so , by Bessel-Parseval equality $$\lVert P_nx-x\rVert^2=\sum_{j\geq n+1}|\langle x,e_j\rangle|^2.$$ As the latest series is convergent we have the result.
Now, let $K\subset H$ compact. Fix $\varepsilon >0$. Then we can find an integer $N$ and $x_1,\dots,x_N\in K$ such that for each $x\in K$, we can find $1\leq k \leq N$ such that $\lVert x-x_k\rVert\leq\varepsilon$. Fix $x\in K$. Then $$\lVert P_nx-x\rVert^2=\sum_{j\geq n+1}|\langle x-x_k+x_k,e_j\rangle|^2\leq \varepsilon^2+\max_{1\leq k\leq N}\sum_{j\geq n+1}|\langle x_k,e_j\rangle|^2.$$ As the RHS doesn't depend on $x$, we have $$\sup_{x\in K}\lVert P_nx-x\rVert^2\leq \varepsilon^2+\max_{1\leq k\leq N}\sum_{j\geq n+1}|\langle x_k,e_j\rangle|^2.$$ Now take the $\limsup_{n\to+\infty}$ to get the result.
- If $T$ is compact, then $K:=\overline{T(B(0,1))}$ is compact, so we apply the previous result to this $K$.
Note that the property of approximation of a compact operator by a finite rank operator is true in any Hilbert space, not only in separable ones. To see that, fix $\varepsilon>0$; then take $v_1,\dots,v_N$ such that $T(B(0,1))\subset \bigcup_{j=1}^NB(y_j,\varepsilon)$. Let $P$ the projection over the vector space generated by $\{y_1,\dots,y_N\}$ (it's a closed subspace). Consider $PT$: it's a finite ranked operator. Now take $x\in B(0,1)$. Then pick $j$ such that $\lVert Tx-y_j\rVert\leq\varepsilon$. We also have, as $\lVert P\rVert\leq 1$, that $\lVert PTx-Py_j\rVert\leq \varepsilon$. As $Py_j=y_j$, we get $\lVert PTx-Tx\rVert\leq 2\varepsilon$.
Suppose that $\{F_n\}$ is a sequence of finite-rank operators such that $F_n\to I$ strongly. Note that by the uniform boundedness principle the sequence is bounded, i.e. there exists $k>0$ with $\|F_n\|<k$ for all $n$ (thanks julien for reminding me of this). I will assume that all $F_n$ are selfadjoint (I need for my estimates, but didn't think if there is a counterexample or not).
We have $$ \|(F_n-I)A\|_{HS}^2=\sum_j\|(F_n-I)Ae_j\|^2=\sum_{j=1}^n\|(F_n-I)Ae_j\|^2+\sum_{j=n+1}^\infty\|(F_n-I)Ae_j\|^2\\ \leq\sum_{j=1}^n\|(F_n-I)Ae_j\|^2+(2k^2+2)\sum_{j=n+1}^\infty\|Ae_j\|^2 $$ and now we can reason as in Nate's proof to say that this goes to zero (i.e. use that $A$ is HS for the tail, and the pointwise convergence of $F_n-I$ to zero in the first finite sum).
This also implies that $A(F_n-I)\to0$ in HS norm. Indeed, $$ \|A(F_n-I)\|_{HS}^2=\mbox{Tr}((F_n-I)A^*A(F_n-I))=\mbox{Tr}(A(F_n-I)(F_n-I)A^*)\\ =\|(F_n-I)A^*\|_{HS}^2 $$ (as $A$ is HS if and only if $A^*$ is, the above works).
Now $$ \|F_nAF_n-A\|_{HS}^2=\|F_nAF_n-F_nA-(I-F_n)A\|_{HS}^2=\|F_nA(F_n-I)+(F_n-I)A\|_{HS}^2\\ \leq2\|F_nA(F_n-I)\|_{HS}^2+2\|(F_n-I)A\|_{HS}^2\leq2k^2\|A(F_n-I)\|_{HS}^2+2\|(F_n-I)A\|_{HS}^2\to0. $$
Best Answer
Let $A_m(x_1, x_2, \ldots) = \left(\frac{x_1 - x_{m + 1}}{\sqrt 2}, x_2, \ldots, x_m\right)$ (essentially projection to subspace $\langle e_1 - e_{m+1}, e_2, \ldots, e_m\rangle$).
Take $x \in \mathcal{H}\setminus \{0\}$.
If $x_n \neq 0$ for some $n > 1$, we can take $\epsilon = |x_n|$ and get $\|A_m x\| \geq \epsilon$ for $m > n$.
If $x_n = 0$ for all $n > 1$ (so $x = x_1 e_1$, as $x \neq 0$), we have $A_m x = \frac{1}{2} e_1 - \frac{1}{2}e_{m + 1}$ and can take $\epsilon = \frac{|x_1|}{2}$. So $\liminf\limits_{m\to\infty}\|A_mx\|>0$.
Now, $\{\frac{e_1 + e_{m + 1}}{\sqrt 2}, e_{m + 2}, e_{m + 3}, \ldots\}$ is orthonormal basis of $\mathcal{N}(A_m)$. So $P_{\mathcal{N}(A_m)}(e_1) = \frac{e_1 + e_{m + 1}}{2}$ which doesn't converge weakly to $0$.