I am having trouble understanding/proving the jump from $$\int_0^{T_i}g(u) du=T_i g(0)+\int_0^{T_*}(T_i-u)^+g'(u)du$$ to $$\int_0^{T_i}g(u) du=<g,h>_H$$

In the course presentation deck I see the following statement that I have been unable to prove.

One of my course tutors responded to my question with this hint/ answer to how I should interpret $$<g,h>_H=g(0)h(0)+\int_0^{T_*}g'(u)h'(u)du$$...

I posted a question to the

**Calculus**forum (http://mathhelpforum.com/calculus/280897-inner-product-integration-parts.html) but the more I dig into this, the more advanced it seems to become.