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Math Help - [SOLVED] bilinear mapping

  1. #1
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    [SOLVED] bilinear mapping

    Assume a<b. C[a,b] denotes the set of all continuous functions [a,b] \rightarrow \mathbb{R} and that all continuous functions [a,b] \rightarrow \mathbb{R} are integrable. So we have a function I: C[a,b] \times C[a,b] \rightarrow \mathbb{R} given by I(f,g)= \int^b_a f(t)g(t)dt.

    (I have already proved and concluded that I is a bilinear form).

    Here's the question:

    Is this bilinear form non-degenerate? Justify your answer by stating explicitly any results you use from calculus.

    My Attempt

    The bilinear form I is non-degenerate

    1. If I(f,g)=0 \forall f \in C[a,b] then g=0
    2. If I(f,g)=0 \forall g \in C[a,b] then f=0

    I personally think this bilinear form is non-degenerate. I will try to give a proof by contradiction. For a contradiction suppose I(f,g) = \int_a^b f(t)g(t)\, dt = 0 but g \neq 0.

    I'm not sure how to carry on and finish this proof from this point. I don't know how to use calculus here... I know that integrals represent areas under the graphs. And in this case we know know g(t) \neq 0.
    And f is not defined, if we define f = sin(t) and a=-1, b=1 we get

     <br />
\int^1_{-1} sin(t) =0<br />

    But this is a counter example to what we are trying to prove!

    I have to submit this tomorrow, so I appreciate it if anyone could please help me complete this proof...urgently
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  2. #2
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    Quote Originally Posted by Roam View Post
    Assume a<b. C[a,b] denotes the set of all continuous functions [a,b] \rightarrow \mathbb{R} and that all continuous functions [a,b] \rightarrow \mathbb{R} are integrable. So we have a function I: C[a,b] \times C[a,b] \rightarrow \mathbb{R} given by I(f,g)= \int^b_a f(t)g(t)dt.

    (I have already proved and concluded that I is a bilinear form).

    Here's the question:

    Is this bilinear form non-degenerate? Justify your answer by stating explicitly any results you use from calculus.

    My Attempt

    The bilinear form I is non-degenerate

    1. If I(f,g)=0 \forall f \in C[a,b] then g=0
    2. If I(f,g)=0 \forall g \in C[a,b] then f=0

    I personally think this bilinear form is non-degenerate. I will try to give a proof by contradiction. For a contradiction suppose I(f,g) = \int_a^b f(t)g(t)\, dt = 0 but g \neq 0.

    I'm not sure how to carry on and finish this proof from this point. I don't know how to use calculus here... I know that integrals represent areas under the graphs. And in this case we know know g(t) \neq 0.
    And f is not defined, if we define f = sin(t) and a=-1, b=1 we get

     <br />
\int^1_{-1} sin(t) =0<br />

    But this is a counter example to what we are trying to prove!

    I have to submit this tomorrow, so I appreciate it if anyone could please help me complete this proof...urgently
    You may use the following easy

    Lemma: if f is a continuous function in (a,b) and f(w) > 0 for some w there then there exists a neighborhood J= (x-e, x+e) in (a,b), with e > 0, s.t. f(x) > 0 fo all x in J

    Now, supose I(f,g) = 0 for all f in C[0,1] ==> in particular, I(g,g) = 0 ==>
    INT{0,1}[g^(x) dx] = 0 ; if there's some point w in (0,1) s.t. g(w) is not zero then g^2(w) > 0 in some neighborhood J = (c,d) of w ==> 0 < INT{c,d}[g^2(x) dx] <= INT{0,1}[g^2(x)dx] since we know that the Riemann integral of a positive function in an interval of positive length is positive.

    Tonio
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  3. #3
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    Is this bilinear form even non-degenerate?
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  4. #4
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    Quote Originally Posted by Roam View Post
    Is this bilinear form even non-degenerate?


    Read again carefully my answer: I think the solution is clear there.

    Tonio
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  5. #5
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    Quote Originally Posted by Roam View Post

    1. If I(f,g)=0 \forall f \in C[a,b] then g=0
    This is known as the fundamental lemma of calculus of variations.
    Fundamental lemma of calculus of variations - Wikipedia, the free encyclopedia
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  6. #6
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    Quote Originally Posted by tonio View Post
    Read again carefully my answer: I think the solution is clear there.

    Tonio
    So it IS non-degenerate since the integral is equal to zero?
    Last edited by Roam; October 15th 2009 at 10:59 AM.
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