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Let $D_1$ and $D_2$ be Chern connections on the same holomorphic vector bundle, possibly for different Hermitian metrics, and put $a=D_1-D_2$. Then $a$ has type $(1,0)$ andIndeed, the curvature difference formula has only types $(2,0)$ and $(1,1)$; both Chern curvatures have type $(1,1)$, so the $(2,0)$ part vanishes and the remaining part is $D_2^{0,1}a={\bar{\partial }}_{\mathrm{End}E}a$.

Let $D_1$ and $D_2$ each be a connection on a vector bundle. Their difference is tensorial, so $D_1=D_2+a$ with $a\in {\Omega }^1(\mathrm{End}E)$. Extend $D_2$ to the endomorphism bundle connection. Expanding $(D_2+a{)}^2$ with the supplied graded Leibniz rule gives the curvature difference formula

Because the two Chern connections have the same $(0,1)$ part, $a=D_1-D_2$ has type $(1,0)$. The curvature difference formula has only $(2,0)$ and $(1,1)$ parts. Both Chern curvatures have type $(1,1)$, so the total $(2,0)$ part vanishes. The remaining part is obtained from $D_2^{0,1}={\bar{\partial }}_{\mathrm{End}E}$, proving the curvature difference of two Chern connections