Decompose selfadjoint operator into difference of selfadjoint non-negative operators

Question

While preparing for an functional analysis exam I ran into the following exercise:

Let $H$ be a (complex) Hilbert space and $A \in L(H)$ (bounded) selfadjoint operator. Show there exists two non-negative (as in $\langle B x, x \rangle \ge 0$) selfadjoint operators $A_{\pm}$ such that

1. $A = A_+ - A_-$, $A \pm A_{\mp}$ are non-negative and $A_+ A_- = A_- A_+ = 0$
2. If $A$ commutes with $B \in L(H)$, so do $A_+$ and $A_-$.
3. If $A$ is compact, so are $A_{\pm}$.

I am looking for hints only!

I tried two constructions for $A_{\pm}$ inspired from this question.

1. $A_+ := A + \lambda I$, $A_- := \lambda I$
2. $A_+ := P^* D_1 P$, $A_- := P^* D_2 P$, where $D = D_1 - D_2$ in the finite case contains all the eigenvalues (not sure how to generalise this to the infinite-dimensional setting with the spectral theorem) and $P$ is unitary.

In both cases, the first two properties are easily verifiable and also $A_+ A_- = A_- A_+$, but $A_+ A_- = 0$ always leads to a contradiction, as demonstrated here for approach 1: $$ A_+ A_- = \lambda A + \lambda^2 I \overset{!}{=} 0 \implies \lambda = 0 \text{ or } A = - \lambda I. $$

Is there a mistake in the reasoning or the construction? If the constructions won't fulfil the desired properties, can somebody please give me a hint on how to construct $A_{\pm}$?

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With the hints of @flan this is how I have gotten so far:

On $\sigma(A)$ define the continuous functions $f_+(x) := \max(x,0) = \frac{x + | x |}{2}$ and $f_{-}(x) := \max(-x,0)$. We show that $A_{\pm} := f_{\pm}(A)$ fulfil the desired properties.

1. We have \begin{equation*} A_+ - A_- = \frac{A + | A |}{2} - \frac{|A | - A}{2} = A, \quad A + A_- = A + \frac{| A | - A}{2} = A_+, \end{equation*} and similarly $A - A_+ = A_-$. It therefore suffices to show $A_{\pm} \ge 0$ which follows from the properties of the continuous functional calculus for self-adjoint operators and that $f_+, f_- \ge 0$. Furthermore we have \begin{equation*} A_+ \cdot A_- = f_+(A) \cdot f_-(A) = (f_+ \cdot f_-)(A) = 0, \end{equation*} as $4(f_+ \cdot f_-)(x) = x |x| - x^2 + | x |^2 - | x | x = 0$. Analogously, $A_- A_+ = 0$ follows.

2. If $A B = B A$, then also $f(A) B = B f(A)$ for all polynomials $f$ on $\sigma(A)$. By continuity of the functional calculus we also have $A_{\pm} B = B A_{\pm}$.

3. ?

Answer 1

As @flan described in the comments: The idea that is missing from your argument for (2) is that since $f_{\pm}$ are continuous and $\sigma(A)$ is compact, then Stone-Weierstrass' theorem says that there is a sequence of polynomials $f_n$ converging uniformly on $\sigma(A)$ to $f_-$. Then $f_n(A)$ converges to $f_−(A)$ in the operator norm. Since $A B = BA$, then $f_n(A) B = B f_n(A)$. Taking limits we get $f_-(A) B = B f_{-}(A)$.

The argument for (3) is similar using the same uniformly convergent sequences of polynomials: If $A$ is compact, then so is $f_n(A)$. The argument finishes by noting that limit of compact operators is compact.