# Self-adjoint operator as difference of two positive operators

The problem comes from my functional analysis homework.

Let $H$ be a complex Hilbert space and $A:H \to H$ be a bounded, self-adjoint linear operator. Prove that there exist positive operators $P$ and $N$ such that $A=P-N$ and $PN=0$. (An operator $T$ is positive if $\langle Tx,x \rangle \ge 0$ for all $x \in H$.)

I found a question similar to this one: Bounded self adjoint operator can be written as difference of positive operators. An answer using $C^*$-algebra was provided there. However, we didn't learn anything on $C^*$-algebra in this class (and I know nothing about it), so the problem is supposed to be proven in an "elementary" way. Here is what I have done so far:

Define $B=(A^2)^{1/2}$, and let $$P=(A+B)/2$$ $$N=(B-A)/2$$ Then $P$ and $N$ are bounded, self-adjoint linear operators. It is easily verified that $$A=P-N$$ and $$PN=0$$ My question: how can we prove that $P$ and $N$ are positive?

By direct calculation, this is equivalent to $|\langle Ax,x \rangle| \le \langle Bx,x \rangle$. @Shalop said in the comments that this inequality can be proven with polarization identity, but I don't see how to do that. Any ideas?

• Using that $|\langle Tx,x \rangle| \leq \langle |T|x,x \rangle$, you have it directly. Is proving this inequality what your question is? – AlexanderJ93 Nov 19 '17 at 3:59
• @AlexanderJ93 Yes, I'd like to know how to prove this inequality. – Jiaqi Li Nov 19 '17 at 4:10
• How does your course define the operator square root? (Eg., using the spectral theorem?) – Keith McClary Nov 20 '17 at 5:14
• @KeithMcClary We have a theorem stating that every positive bounded self-adjoint linear operator has a unique positive square root. The proof was before and independent of the spectral theorem. – Jiaqi Li Nov 20 '17 at 5:17
• I see one here that I don't remember learning. I'm just trying to figure out what is considered "elementary". – Keith McClary Nov 20 '17 at 5:37

If $A,B$ are positive, commuting operators, then $AB$ is positive. This is because the unique positive $\sqrt{A}$ must also commute with $B$ and, hence, $$\langle ABx,x\rangle = \langle \sqrt{A}Bx,\sqrt{A}x\rangle = \langle B\sqrt{A}x,\sqrt{A}x\rangle \ge 0.$$ This result is useful in what follows.

Suppose $A$ is selfadjoint. Let $P=\frac{1}{2}(|A|+A)$ and $N=\frac{1}{2}(|A|-A)$, where $|A|$ is the unique positive square root of $A^2$. Then $PN=NP=0$ and $A=P-N$. This is the desired decomposition of $A$, and the trick is to show that $P,N$ are positive operators.

Let $E$ be the orthogonal projection onto $\mathcal{N}(|A|+A)$. Then $(|A|+A)E=0$ gives $E(|A|+A)=0$ by taking adjoints. And $(|A|+A)(|A|-A)=0$ gives $E(|A|-A)=|A|-A$. Hence, $$2EA=E(|A|+A)-E(|A|-A) = A-|A| \\ |A| = (I-2E)A \\ 2E|A| = E(|A|+A)+E(|A|-A)=|A|-A \\ A = (I-2E)|A|.$$ These two equations are consistent because $(I-2E)^2=I-4E+4E=I$ establishes $I-2E$ as its own inverse. Taking adjoints of the above equations shows that $E$ commutes with $A$ and with $|A|$, which is useful in what follows. Now the operators $P$ and $N$ may be written as $$P=\frac{1}{2}(|A|+A)=\frac{1}{2}(|A|+(I-2E)|A|)=(I-E)|A|, \\ N=\frac{1}{2}(|A|-A)=\frac{1}{2}(|A|-(I-2E)|A|)=E|A|$$ Because $E$ commutes with $A$, then $E$ must also commute with $A^2$ and, hence, also with $|A|=(A^2)^{1/2}$. By the result of the first paragraph, $P=(I-E)|A|$ and $N=E|A|$ are positive.

• +1 Your answers are always a pleasure to read. Excellent work. – AmorFati Nov 23 '17 at 7:47
• @KyleBroder : Thank you. – DisintegratingByParts Nov 23 '17 at 8:11

If $A=\int_{-\infty} ^\infty x dE_x$, take $P= \int_0 ^\infty x dE_x$ and $N=\int_{-\infty} ^0 \vert x\vert dE_x =-\int_{-\infty} ^0 x dE_x$ .

• downvoters should explain the reason for downvotes. This answer is actually correct. – Diesirae92 Nov 23 '17 at 16:30
• I upvoted DisintegratingByParts' nice proof which uses more elementary methods. Using the spectral theorem is a "sledgehammer" proof, but probably what is expected for an exercise in that chapter. – Keith McClary Nov 23 '17 at 18:22
• Thank you for providing the answer! However, I guess what we learn for the spectral theorem of self-adjoint operator is more elementary... (I didn't learn the Riemann-Stieltjes integral representation of the operator) – Jiaqi Li Nov 23 '17 at 19:29