# Thread: Inner product from integration by parts

1. ## Inner product from integration by parts

I am trying to work through a proof of a smoothing spline approach in Hilbert Space. The approach is named Lorimier's Theorem.

Under the section titled proof I see the following steps:

$\displaystyle \int_{0}^{T_i}g(u)du = T_ig(T_i)-\int_{0}^{T_i}ug'(u)du$
$\displaystyle = T_ig(0)+T_i\int_{0}^{T_i}g'(u)du-\int_{0}^{T_i}ug'(u)du$
$\displaystyle = T_ig(0)+\int_{0}^{T_*}(T_i-u)^+g'(u)du=<h_i,g>_H$

I do not understand the mathematics between step 2 and 3.

Here is my derivation so far:

Begin with $\displaystyle \int_{0}^{T_i}g(u)du$

Integrate by parts $\displaystyle (\int_{}^{}x dy = xy-\int y dx)$ by setting $\displaystyle x=g(u)$, $\displaystyle dx=g'(u)du$, $\displaystyle y=u$ and $\displaystyle dy=du$:
$\displaystyle \int_{0}^{T_i}g(u)du = [ug(w)]_0^{T_i}-\int_{0}^{T_i}ug'(u)du = T_ig(T_i)-\int_{0}^{T_i}ug'(u)du$
Recognize that $\displaystyle T_ig(T_i)=T_ig(0)+T_ig(T_i)-T_ig(0)=T_ig(0)+T_i\int_{0}^{T_i}g'(u)du$ and substitute into the previous equation:
$\displaystyle \int_{0}^{T_i}g(u)du = T_ig(0)+T_i\int_{0}^{T_i}g'(u)du-\int_{0}^{T_i}ug'(u)du$

What do I do next to arrive at step 3 and how is it that the author automatically recognizes that the result is equal to the inner product of $\displaystyle h_i$ and
$\displaystyle g$?

2. ## Re: Inner product from integration by parts

$\displaystyle T_ig(0)+\int_{0}^{T_*}(T_i-u)^+g'(u)du=<h_i,g>_H$
for all $\displaystyle g\:\epsilon\:H$.

Further, the $\displaystyle +$ exponent in the second term means "positive values only": $\displaystyle (T_i-u)=max((T_i-u),0)$.

3. ## Re: Inner product from integration by parts

Originally Posted by mkmath
I am trying to work through a proof of a smoothing spline approach in Hilbert Space. The approach is named Lorimier's Theorem.

Under the section titled proof I see the following steps:

$\displaystyle \int_{0}^{T_i}g(u)du = T_ig(T_i)-\int_{0}^{T_i}ug'(u)du$
$\displaystyle = T_ig(0)+T_i\int_{0}^{T_i}g'(u)du-\int_{0}^{T_i}ug'(u)du$
$\displaystyle = T_ig(0)+\int_{0}^{T_*}(T_i-u)^+g'(u)du=<h_i,g>_H$

I do not understand the mathematics between step 2 and 3.
It isn't stated, but I assume that $T_i > 0$. Since in the integral $0 \le u \le T_i$ then the expression $T_i - u \ge 0$ on $[0,T_i]$ so $T_i - u = (T_i - u)^+$ on that interval.