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Math Help - Inner Product Involving Integrals

  1. #1
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    Inner Product Involving Integrals

    Let V denote the vector space of all continuous functions :

    f:[0, 1] -> R

    Where the 1st and 2nd derivatives exist and are continuous. Suppose further that:

    f(0)=f(1)=0

    Now let L: V -> V be defined by

    L(f) = f''

    Define the inner product <f, g> as follows:

    <f, g> = \int^1_0 f(t)g(t)dt

    Prove that <L(f), g> = <f, L(g)>

    (I've already proved that L is a linear map. The hint on this part says to use integration by parts twice, but I'm not seeing it. Help? If I can get some guidance on the left-hand side, I can do the right hand side as well since it will be similar. Thanks.)
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  2. #2
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    Quote Originally Posted by mathematicalbagpiper View Post
    Let V denote the vector space of all continuous functions :

    f:[0, 1] -> R

    Where the 1st and 2nd derivatives exist and are continuous. Suppose further that:

    f(0)=f(1)=0

    Now let L: V -> V be defined by

    L(f) = f''

    Define the inner product <f, g> as follows:

    <f, g> = \int^1_0 f(t)g(t)dt

    Prove that <L(f), g> = <f, L(g)>

    (I've already proved that L is a linear map. The hint on this part says to use integration by parts twice, but I'm not seeing it. Help? If I can get some guidance on the left-hand side, I can do the right hand side as well since it will be similar. Thanks.)
    Starting point: \langle L(f),g\rangle = \int^1_0 f''(t)g(t)\,dt = \Bigl[f'(t)g(t)\Bigr]_0^1 - \int_0^1f'(t)g'(t)\,dt (integration by parts). Now use the fact that g(0) = g(1) = 0.
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  3. #3
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    Quote Originally Posted by mathematicalbagpiper View Post
    Let V denote the vector space of all continuous functions :

    f:[0, 1] -> R

    Where the 1st and 2nd derivatives exist and are continuous. Suppose further that:

    f(0)=f(1)=0

    Now let L: V -> V be defined by

    L(f) = f''

    Define the inner product <f, g> as follows:

    <f, g> = \int^1_0 f(t)g(t)dt

    Prove that <L(f), g> = <f, L(g)>

    (I've already proved that L is a linear map. The hint on this part says to use integration by parts twice, but I'm not seeing it. Help? If I can get some guidance on the left-hand side, I can do the right hand side as well since it will be similar. Thanks.)
    <L(f),g>=\int_{0}^{1}f''gdx

    Now as the hint says use integration by parts with
    u=g \implies du=g'dx \text{ and } dv=f''dx \implies v=f'

    Then you get

    <L(f),g>=\int_{0}^{1}f''gdx=f'g\bigg|_{0}^{1}-\int_{0}^{1}f'gdx

    So the middle term is zero becuase g(1)=g(0)=0

    So you end up with

    <L(f),g>=\int_{0}^{1}f''gdx=-\int_{0}^{1}f'gdx=<f',g>

    Now just integrate by parts one more time to finish the job.
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