Abel Inequality (Sequence form) The sequence $a_n$ is monotonic then$$\left|\sum_{k=1}^n a_k b_k\right| \leq \max _{k=1, \ldots, n}\left|\sum_{i=1}^k b_i\right|\left(\left|a_n\right|+|a_n-a_1|\right)$$
Abel Inequality (Integral form) The Riemann integrable function $f(x)$ is monotonic then$$\left|\int_a^b f(x)g(x)dx\right| \leq \sup _{\xi\in[a,b]}\left|\int_a^\xi g(x)dx\right|\left(\left|f(b)\right|+|f(b)-f(a)|\right)$$
These two inequality should be equivalent. We can take $f$ and $g$ as some special step functions then derive the sequence form from the integral form.
But how can we derive the integral form from the sequence form? A possible way is proving the second mean value theorem for definite integrals firstly. But I think they're equivalent. So how can I prove the integral form by using sequence form with out second mean value theorem for definite integrals?