Curvature estimate in Hamilton's Ricci flow paper for traceless $\operatorname{Rm}$ on $4$-dimensional manifold 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!
 A: After a lot more time thinking about it, I think I've figured it out. Let $\Gamma$ be a constant such that $ \Gamma \|g \odot g \| = 1$ (where $\odot$ denotes the Kulkarni Nomizu product, and the only reason I don't make $\Gamma$ explicit here is because it depends on the conventions for the definitions of $g \odot g$ and the inner product, but they all agree up to a factor). Clearly, we can control $\| \mathrm{Rm} \|^2$ once we control $A, B, C$, i.e
$$\| \mathrm{Rm} \|^2 \leq\beta( \|A \|^2 + 2\|B \|^2 + \|C \|^2)$$
where $\beta$ is some constant. For convenience, I'll commit the (harmless) abuse of denoting both the operators on $\Lambda^{2}_{\pm}(M)$ and its associated $(0, 4)$ tensors by the same letters. Recalling that $R = \mathrm{tr}(A) = \mathrm{tr}(C)$, by a very similar argument to the one seen in the end of Chapter 9 of Peter Topping's lectures on Ricci flow , we see that the estimates provided by Hamilton imply that for all $\varepsilon > 0$ (however small) there exists a constant $C_{\varepsilon}$ such that $$2 \max\left\{\left\|A -  \frac{2}{3}\Gamma R (g \odot g)\right\| , \left\|C -  \frac{2}{3} \Gamma R (g \odot g)\right\| \right\} \leq \varepsilon R+C_{\varepsilon}$$
The norm of $B$ is controlled by $b_3^2$ which is controlled by $(b_3 + b_2)^{2 + \delta}$ which in turn is controlled by $K a_1 c_1$. The latter is controlled by the estimate above. Putting all this together we get almost exactly the desired following inequality (it's worth noting this may not exactly be $\mathring{\mathrm{Rm}}$ and some undesirable $\varepsilon^2 R^2$ terms may appear, but since we'll just use the fact that $\varepsilon$ can be made arbitrarily small in the end, this will not matter).
$$\left\|\mathrm{Rm} -  \frac{2}{3}\Gamma R (g \odot g)\right\| \leq \varepsilon R+C_{\varepsilon}$$
