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Inner Product Spaces

  1. Mar 6, 2005 #1
    Question)
    Define [tex] (f|g) = \int_0^2 (1+x^2)f(x)g(x)dx[/tex] on [tex] V = C([0,2],\mathbb{R})[/tex].

    Then this is an inner product space because it obeys the four axioms

    1. [tex](f|f) = \int_0^2 (1+x^2)f^2(x)dx \geq 0[/tex] since [tex] f^2(x) \geq 0 [/tex] for all [tex] x \in [0,2][/tex]

    2. [tex](f|g) = \int_0^2 (1+x^2)f(x)g(x)dx = \int_0^2 (1+x^2)g(x)f(x)dx = (g|f)[/tex]

    3. [tex](f+g|s) = \int_0^2 (1+x^2)((f(x)+g(x))s(x)dx = \int_0^2(1+x^2)f(x)s(x)dx + \int_0^2(1+x^2)g(x)s(x)dx = (f|s) + (g|s)[/tex]

    4. [tex](\lambda f|g) = \int_0^2 \lambda(1+x^2)f(x)g(x)dx = \lambda\int_0^2(1+x^2)f(x)g(x)dx = \lambda(f|g)[/tex]

    Further, because [tex] f^2(x) \geq 0[/tex] and [tex]f(x)[/tex] is continuous on [tex][0,2][/tex], it follows that
    [tex]\int_0^2f^2(x)dx = 0 \Leftrightarrow f(x) = 0 \forall x \in [0,2][/tex]
    This further proves that axiom 1 holds.


    My query is, I have only ever had to prove things like
    [tex] (f|g) = \int_a^b f(x)g(x)dx [/tex]
    I don't exactly know what do do with the [tex](1+x^2)[/tex] bit in there.
    Can anyone tell me where this comes into the calculations?
     
    Last edited: Mar 6, 2005
  2. jcsd
  3. Mar 6, 2005 #2
    Question 2.

    I was thinking that
    [tex] (f|g) = \int_0^2(1-x^2)f(x)g(x)dx [/tex]
    is NOT an inner product space because axiom 1 fails.

    1. [tex] (f|f) = \int_0^2(1-x^2)f^2(x)dx[/tex] is not guaranteed to be greater than or equal to zero on the interval [tex][0,2][/tex] because at 2 the integral is negative due to the [tex](1-x^2)[/tex] bit.


    Question 3.

    [tex] (f|g) = \int_1^2(1+x^2)f(x)g(x)dx [/tex]

    should also be an inner product space. My reasoning is because this one is identical to the one in Question 1. with only the interval of integration changing. The only axiom which deals with the range of integration is axiom 1 which states that
    [tex](f|f) \geq 0 \,\forall \, x\in[a,b][/tex]
    by inspection, so long as a or b remain positive, then [tex](f|g) \geq 0[/tex] and thus remain an inner product space.
     
  4. Mar 6, 2005 #3

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    Maybe the experts here will say I am not justified in using this terminology, but I would consider the [tex](1+x^2)[/tex] factor to be playing the role of a metric in the space. The more common metric would simply be 1.

    I agree with the points you make in your second post.
     
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