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This is supposedly quite a simple fact, yet I haven't been able to show it. If we suppose that $T$ is a bounded self-adjoint operator, then $$\langle Tv,v \rangle = \langle v,Tv \rangle$$ for all $v \in \mathcal{H}$. I don't see how we are allowed to then decompose it into the form $A - B$ for $A$ and $B$ both positive operators.

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    One way to do this is to use functional calculus to take the positive and negative parts of $T$ respectively. You could also do $A = \frac{1}{2}(T+|T|)$ and $B = \frac{1}{2}(|T|-T)$ which works I think. – 2017-02-19
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    @Shalop Can you prove that $A$ and $B$ are positive? How is $\left| T \right|$ defined? – 2017-02-25
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    @user3359 For a general operator $T$, we define $|T|:= \sqrt{T^*T}$ where $\sqrt{\cdot}$ denotes the square root (which exists for positive operators). The polarization identity basically gives a relation that $|\langle Tx,x \rangle| \leq \langle |T|x,x\rangle$, from which it follows that $\langle (|T|-T)x,x \rangle \geq 0$ and similarly for $T+|T|$. – 2017-02-25
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    @Shalop Thankyou very much – 2017-02-25
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    @Shalop Could you explain how to obtain the result by polarization identity? i.e. $|\langle Tx,x \rangle| \leq \langle |T|x,x\rangle$ – 2017-11-16
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    @Shalop@JiaqiLi I asked this here: https://math.stackexchange.com/questions/2749460 – 2018-04-22

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Since $T$ is self-adjoint, $C^*(T)$ (the $C^*$-algebra generated by $1$ and $T$) is commutative, and therefore the Gelfand transform $\phi$ is an isometric isomorphism of $C^*(T)$ onto $C(X)$ for some compact Hausdorff space $X$. Furthermore, $\phi(T)(X)=\sigma(T)\subset\mathbb{R}$ so $\phi(T)^+,\phi(T)^-\in C(X)$ (the positive and negative parts of $\phi(T)$, respectively), and thus there are $A,B\in C^*(T)$ such that $\phi(A)=\phi(T)^+$ and $\phi(B)=\phi(T)^-$. From this, it is clear that $A$ and $B$ are positive and that $T=A-B$.

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    Is it possible to do this without $C^{\ast}$-algebras and the Gelfand transform. I haven't seen this material yet. – 2017-02-21
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    @user412674 I will look around, but at the moment I'm not aware of a more simple solution. If you would like some references, see Conway's *A Course in Functional Analysis* (the chapter on $C^*$-algebras is relatively short, and the results at play are contained in the first few sections) or Murphy's *$C^*$-algebras and Operator Theory* (The results necessary are contained within the first 50 pages of the book). – 2017-02-21