Fix $s_{0}>0$ large enough so that $f(s_{0})^{\alpha}<1/2B$. We define a sequence $(s_{j})\in\mathbb{R}_{+}^{\mathbb{N}}$ by induction in the following way. If $f(s_{0})=0$ we stop here, otherwise we set

$$s_{1}:=\sup\left\{s>s_{0}\,/\,f(s)>\frac{1}{2}f(s_{0})\right\}.$$

Observe that $s_{1}\leq 1+s_{0}$ thanks to $H(\alpha,B)$ and by definition of $s_{0}$.

Since $f$ is right-continuous we get $f(s_{1})\leq f(s_{0})/2$. If $f(s_{1})=0$ we stop here, otherwise we go on by induction, setting

$$s_{j+1}:=\sup\left\{s>s_{j}\,/\,f(s)>\frac{1}{2}f(s_{j})\right\}.$$

At each step $f(s_{j+1})\leq f(s_{j})/2$ and $s_{j+1}\leq 1+s_{j}$. However the sequence $(s_{j})$ does not grow too fast. It follows indeed from $H(\alpha,B)$ that if $s\in]s_{j},s_{j+1}[$,

$$(s-s_{j})f(s)\leq Bf(s_{j})^{1+\alpha}\leq 2Bf(s)f(s_{j})^{\alpha},$$

since $f(s_{j})/2\leq f(s)\leq f(s_{j})$ on the interval $[s_{j},s_{j+1}]$. We infer

$$s_{j+1}-s_{j}\leq 2Bf(s_{j})^{\alpha}\leq 2B2^{-j\alpha}f(s_{0})^{\alpha}\leq 2^{-j\alpha}.$$

Thus the sequence $(s_{j})$ is bounded from above, with limit

$$S_{\infty}=s_{0}+\sum_{j\geq 0}(s_{j+1}-s_{j})\leq s_{0}+\frac{2Bf(s_{0})^{\alpha}}{1-2^{-\alpha}}\leq s_{0}+\frac{1}{1-2^{-\alpha}}.$$

∎