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I am reading Griffiths-Harris book on algebraic geometry and in "Theorem B", where he proves (the analogue of Serre's vanishing theorem) that

Let $M$ be a compact, complex manifold, $L\rightarrow M$ be a positive line bundle. Then for any holomorphic vector bundle $E$, there exists $\mu_{0}$ such that for all $q>0$ and $\mu\geq\mu_{0}$, $H^{q}(M,\mathcal{O}(L^{\mu}\otimes E))=0$.

Here, if we let $\Lambda$ be the dual of the Lefschetz operator, $\Theta_{E}$ be the curvature of $E$, he asserts that $[\Lambda,\Theta_{E}]$ is bounded on $A^{0,*}(L^{-\mu}\otimes E)$. Why is this so?

Thanks!

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All they are saying is that if $V\to M$ is a metric vector bundle on some compact manifold and $T$ is an endomorphism of $V$ (a section of $\text{Hom}(V,V)$) then $T$ is uniformly bounded; i.e. there exists a $C>0$ such that $\|T(x)v\|\leq C\|v\|$ for all $x\in M$ and $v\in V_x$. Proof: Let $\|T(x)\|^2= \text{tr}[ T(x)T^*(x)]$ (sum of squares of entries of $T(x)$ wrt some orthonormal frame of $V_x$). Then by Cauchy-Schwarz $\|T(x)v\|\leq \|T(x)\|\|v\|$. Now $\|T(x)\|$ is a continuous function on a compact space hence bounded.

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