Every contractive self-adjoint operator is the weak limit of projectors I try to show that every self-adjoint operator $A$ on Hilbert space $H$ with $\|A\| \leq 1$ is the weak limit of the projectors. My teacher told me that the most important part is to show that for one-dimensional operator and I have done that. But how can I obtain the general case form one-dimensional? I have no idea.
Any help is appreciated. Thanks in advance.
 A: First note that this is not true in finite-dimension, as a weak limit is a norm limit, and a norm-limit of projections is a projection. So we may assume that $\dim H=\infty$.
Second, it's only true if $A$ is positive. A weak limit of projections is positive, so you cannot have a non-positive selfadjoint operator as a limit of projections.
They key point is that you only need to find, given $\varepsilon>0$ and $x_1,\ldots,x_m,y_1,\ldots,y_m\in H$, a projection $P$ such that
$$
|\langle(P-A)x_k,y_k\rangle|<\varepsilon,\qquad\qquad k=1,\ldots,m.
$$
It doesn't really matter what $P$ does to elements not in the span of $x_1,\ldots,x_m,y_1,\ldots,y_m$. So let $Q_1$ be the orhtogonal projection onto $\operatorname{span}\{x_1,\ldots,x_m,y_1,\ldots,y_m\}$, and let $Q_2$ be a projection with equal rank than $Q_1$ and orthogonal to it. Put
$$
H_0=\operatorname{span}\big\{Q_1H\cup Q_2H\big\}.
$$
With respect to the decomposition $H_0=Q_1H\oplus Q_2H$, using an isometry between $Q_1H$ and $Q_2H$ we may think  of $B(H_0)$ and $M_2(B(Q_1H))$. In this view, define
$$
P=\begin{bmatrix} A_0& (A_0-A_0^2)^{1/2} \\ (A_0-A_0^2)^{1/2} & I-A_0\end{bmatrix},
$$
where $A_0=Q_1AQ_1$.
Then $P$ is a projection, that we may extend as $0$ to $H_0^\perp$. We have
\begin{align}
\langle Px_k,y_k\rangle&=\langle PQ_1x_k,Q_1y_k\rangle=\langle Q_1PQ_1x_k,y_k\rangle=\langle A_0x_k,y_k\rangle\\[0.3cm]
&
=\langle Q_1AQ_1x_k,y_k\rangle=\langle Ax_k,y_k\rangle. 
\end{align}
That is,
$$
\langle (P-A)x_k,y_k\rangle=0,\qquad\qquad k=1,\ldots,m.
$$
