In dimension $4$, it is known that the curvature operator $\operatorname{Rm} : \Lambda^2(M) \to \Lambda^2(M)$ admits a block decomposition of the form $$\operatorname{Rm} = \begin{pmatrix} A & B \\ B^T & C\end{pmatrix}$$

Choosing convenient orthonormal basis, the matrices above can be written in diagonal form (i.e $A_{ij} = a_i \delta_{ij}$, $B_{ij} = b_i \delta_{ij}$, $C_{ij} = c_i \delta_{ij}$).

In his "Four-manifolds with positive curvature operator" paper, Hamilton proved the following estimates:

If we choose successively positive constants $G$ large enough, $H$ large enough, $\delta$ small enough, $J$ large enough, $\varepsilon$ small enough, $K$ large enough, $\theta$ small enough, and $L$ large enough, with each depending on those chosen before, then the closed convex subset $X$ of $\{M_{\alpha \beta} \geq 0 \}$ defined by the inequalities

- $(b_2 + b_3)^2 \leq G a_1 c_1$
- $a_3 \leq H a_1$ and $c_3 \leq H c_1$
- $(b_2 + b_3)^{2 + \delta} \leq J a_1 c_1 (a - 2b + c)^{\delta}$
- $(b_2 + b_3)^{2 + \varepsilon} \leq K a_1 c_1$
- $a_3 \leq a_1 + L a_1^{1 - \theta}$ and $c_3 \leq c_1 + L c_1^{1 - \theta}$
is preserved under the Ricci flow for a suitable ODE.

For further context, one can see page $165$ of this book, where it is then claimed without any proof that an immediate consequence of the above estimates is the following estimate on the traceless Riemannian curvature operator:

$$\| \mathring{\mathrm{Rm}} \| \leq \varepsilon R + C_{\varepsilon}$$

where $\varepsilon$ can be arbitrarily small and $C_{\varepsilon} < \infty$ is a constant. How does this estimate follow from the previous ones? I've tried to express $\| \mathring{\mathrm{Rm}} \|$ in terms of the $a_i$'s, $b_i$'s and $c_i$'s in order to use the estimates proved for them, but that led me nowhere. How can one conclude this estimate from the previous ones? I can't fill this hole in the paper. I'd appreciate any help!