Im working on a corolary in this article: http://www.people.virginia.edu/~jlr5m/Papers/p46.pdf corolary 2, page 5.
I understand the argument shown, more precisely how to prove that
$$ f'_+=Xf_++(I-Y)f_-, $$
$$ f'_-=(I-X)f_++Yf_- $$
But im stuck getting a costant $m$ to show that $m|| f ||_{|\mathfrak{H}|} \le || f ||_{|\mathfrak{H}|'}$, i know those relations will make it but i dont really know how to approach. Any help is welcome.
The theorem at the top of page 5 shows that $X$ and $Y$ are bounded. Then \begin{align} \|f\|_{|\mathfrak H|'}^2&=\|f_+'\|_{|\mathfrak H|'}^2+\|f_-'\|_{|\mathfrak H|'}^2\\ \ \\ &=\|Xf_++(I-Y)f_-\|_{|\mathfrak H|'}^2+\|(I-X)f_++Yf_-\|_{|\mathfrak H|'}^2\\ \ \\ &\leq (\|X\|\,\|f_+\|_{|\mathfrak H|}+\|I-Y\|\,\|f_-\|_{|\mathfrak H|})^2 +(\|I-X\|\,\|f_+\|_{|\mathfrak H|}+\|Y\|\,\|f_-\|_{|\mathfrak H|})^2\\ \ \\ &\leq M(\|f_+\|_{|\mathfrak H|}^2+\|f_-\|_{|\mathfrak H|}^2)\\ \ \\ &=M\,\|f\|_{|\mathfrak H|}^2, \end{align} where $M$ is obtained by collecting terms after expanding the squares and using the easy inequality $|2ab|\leq a^2+b^2$).
The other inequality is obtained in the same way after solving your two equation linear system in terms of $f_+$ and $f_-$.