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Let $H$ be a very ample effective Cartier divisor on a projective scheme $X$, and suppose $L{|}_H$ is ample. Then $H^i(X,L^m)=0$ for $i\ge 2$ and sufficiently large $m$. By uniform Serre vanishing for two ample twists, cohomology of $L^m(tH){|}_H$ vanishes uniformly for $t\ge 0$. The restriction sequence identifies the higher groups of $L^m(tH)$ and $L^m((t+1)H)$ for $i\ge 2$. For fixed $m$, large $t$ kills them by Serre vanishing for $H$; descend to $t=0$. No $H^1$ vanishing is claimed.

We prove (b)$\Rightarrow$(a) by induction on dimension, using the independently proved (c)$\Rightarrow$(a) argument in the next section. Work on an integral projective variety $V$ of dimension $d>0$. The hypothesis is inherited by all its integral closed subvarieties. Induction therefore makes $D$ ample on every strictly lower-dimensional closed reduced subvariety, and hence on every proper closed subscheme of dimension less than $d$, by ampleness on reduced components.

Choose an effective Cartier divisor $H$ which is a very ample hyperplane section of $V$. Then ${\mathcal{O}}_H(D)$ is ample. We need a vanishing statement uniform in extra positive $H$-twists:Here is a justification using Castelnuovo–Mumford regularity. Embed $H$ by ${\mathcal{O}}_H(H)$. For each of the finitely many positive cohomology degrees $j$, Serre vanishing for the ample bundle ${\mathcal{O}}_H(D)$ makes $H^j(H,{\mathcal{O}}_H(mD-jH))$ vanish for $m\ge M$. Thus the pushed-forward sheaf ${\mathcal{O}}_H(mD)$ is zero-regular. Persistence of regularity makes it $t+j$-regular for every $t\ge 0$, giving exactly the displayed vanishing. This is the uniform Serre vanishing for two ample twists lemma. It does not assume that $D$ is ample on $V$.

Apply the divisor restriction exact sequence to $H$ with twists $mD+tH$. For $i\ge 2$, both neighbouring cohomology groups on $H$ vanish, soFor a fixed $m\ge M$, sufficiently large $t$ kills the right-hand cohomology by Serre vanishing for $H$ on $V$. The higher cohomology vanishing from an ample hyperplane restriction argument gives $H^i(V,{\mathcal{O}}_V(mD))=0$ for all $i\ge 2$. ConsequentlyThis step is essential: divergence of a polynomial does not by itself prove that its top-degree coefficient is positive.

For $m$ large enough, $h^0(V,mD)>1$. Evaluation at a closed point has a one-dimensional target, so its kernel contains a nonzero section vanishing there. We have obtained condition (c) on $V$. Lower-dimensional subvarieties already have the same property by induction, so the next section's implication (c)$\Rightarrow$(a) applies. ThusFor $d=1$, higher groups with $i\ge 2$ vanish automatically and the same evaluation argument starts the induction. The regularity facts used above are stated in the Stacks Project, regularity lemmas.