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Thread: Inner product from integration by parts

  1. #1
    Junior Member mkmath's Avatar
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    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
    $\displaystyle g$?
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  2. #2
    Junior Member mkmath's Avatar
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    Re: Inner product from integration by parts

    Step 3 should read
    $\displaystyle T_ig(0)+\int_{0}^{T_*}(T_i-u)^+g'(u)du=<h_i,g>_H$
    for all $\displaystyle g\:\epsilon\:H$.

    Further, the $\displaystyle +$ exponent in the second term means "positive values only": $\displaystyle (T_i-u)=max((T_i-u),0)$.
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  3. #3
    Junior Member Walagaster's Avatar
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    Re: Inner product from integration by parts

    Quote Originally Posted by mkmath View Post
    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.
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