I am stuck with the following problem : Every normal element in a $C^*$ algebra can be written as a linear combination of four commuting positive elements.
I had tried along the following lines. Decompose the normal element $a$ as $a=x+iy$. We know that x and y commute. As both x and y are self-adjoint, let us decompose them further as $x=x_{+} - x_{-}$ and $y=y_{+}-y_{-}$ where the subscripts + and - are (loosely speaking) the unique positive and negative parts of x and y. Hence I have four positive elements now whose linear combination is $a$. I also have $x_{+}x_{-}=x_{-}x_{+}$ and $y_{+}y_{-}=y_{-}y_{+}$. But I cannot show the other commuting relations.
So my question is, am I progressing with the correct decomposition of $a$? If yes, how to show the other commuting relations?
If you think about your argument, you haven't used that $a $ is normal. That will give you $xy=yx $.
From $xy=yx$, you get $x^ny=yx^n$, and so $p(x)y=yp(x)$ for any polynomial $p$, and from here $f(x)y=yf(x)$ for any continuous function $f$.