1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inner Product spaces proof

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data

    We consider P2 the vector space of all real polynomials of degree at most 2.
    Show that

    [itex]
    <f,g> = \int_{-1}^{1}f(x)g(x)dx
    [/itex]

    defines an inner product space
    2. Relevant equations

    I'm Using one of the Axioms of Inner product spaces IP1. which states that.

    [itex]
    <u,u>[/itex] [itex] \geq 0 [/itex] with equality when and only when [itex]u=0_{v}[/itex]

    3. The attempt at a solution

    For IP1

    let [itex]a,b,c,d,e,f,x \in R[/itex]

    then

    [itex]
    <f,g> = \int_{-1}^{1}f(x)g(x)dx
    [/itex]

    [itex]
    <f,g> = \int_{-1}^{1}(a+bx+cx^2)(d+ex+fdx^2)dx
    [/itex]

    [itex]
    <f,g> = \left[ \frac{cd+fx^3}{3}+\frac{bd+ex^2}{2}+adx\right]_{-1}^{1}
    [/itex]

    [itex]
    <f,g> = 2ad+\frac{2cd+f}{3}\
    [/itex]

    [itex]
    <f,g> \geq 0
    [/itex]




    To show that equality to 0 implies the zero polynomial can I use?


    let [itex]a,b,c,d,e,f \in R \neq 0[/itex]
    and
    [itex]x\in R [/itex]

    Consider

    [itex]
    <f,f> = \int_{-1}^{1}f(x)f(x)dx = \left[ \frac{cd+fx^3}{3}+\frac{bd+ex^2}{2}+adx\right]_{-1}^{1}

    [/itex]

    if we assume
    [itex]
    <f,f> = 0
    [/itex]

    then this implies

    [itex]
    \left[ \frac{cd+fx^3}{3}+\frac{bd+ex^2}{2}+adx\right]_{-1}^{1}
    = 0
    [/itex]

    but a,b,c,d,e,f [itex]\neq 0[/itex]

    therefore

    [itex]
    \left[ \frac{cd+fx^3}{3}+\frac{bd+ex^2}{2}+adx\right]_{-1}^{1}
    \neq 0
    [/itex]

    contradicting our assumption

    hence a,b,c,d,e,f [itex]= 0[/itex] the zero polynomial


    So we have

    [itex]
    <u,u>[/itex] [itex] \geq 0 [/itex]

    and

    [itex]
    <u,u>[/itex] [itex] = 0 \rightarrow f = 0_{v}[/itex]

    So IP1 holds


    Is that enough to prove IP1 ?
     
  2. jcsd
  3. Oct 9, 2009 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Could you explain the steps in your proof that
    [tex]\langle f, g \rangle \geq 0[/tex]?​

    P.S. don't you find that identity rather suspicious?
     
  4. Oct 9, 2009 #3

    Mark44

    Staff: Mentor

    You have a lot of extra work that you don't need or want. For the nonnegativity thing you want to show that <f, f> >= 0, with equality iff f = 0. You have a lot of work showing <f, g>, which isn't applicable for this axiom. You want the inner product of a function f with itself.
     
    Last edited: Oct 10, 2009
  5. Oct 9, 2009 #4
    I'll try again

    For IP1

    let
    [itex]a,b,c \in R[/itex]

    then


    [itex]<f,f>[/itex][itex] = \int_{-1}^{1}f(x)f(x)dx[/itex]


    [itex]<f,f>[/itex][itex] = \int_{-1}^{1}(a+bx+cx^2)(a+bx+cx^2)dx[/itex]


    [itex]<f,f>[/itex][itex]=\left[\frac{c^2x^5}{5}+\frac{bcx^4}{2}+\frac{(2ac+b^2)x^3}{3}abx^2+a^2x\right]_{-1}^{1}[/itex]



    [itex]<f,f>[/itex][itex] = 2a^2+\frac{4ac}{3}+\frac{2b^2}{3}+\frac{2c^2}{5}[/itex]


    [itex] \geq 0[/itex]

    Is that better?
     
  6. Oct 9, 2009 #5
    Still too complicated...by a mile. You're gonna feel like an idiot when you see it :P Happens to the best of us ;)
     
  7. Oct 9, 2009 #6

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Why should this be true? :confused:

    While this is certainly not the easy way to prove it, once you figure out the problem you should still try and make sure you understand how to do things this way.
     
  8. Oct 10, 2009 #7

    Mark44

    Staff: Mentor

    Is there another way to write this than the way you chose?
    [itex] <f, f> = \int_{-1}^{1}(a+bx+cx^2)(a+bx+cx^2)dx[/itex]
     
  9. Oct 10, 2009 #8
    Another way is

    [itex] = \int_{-1}^{1}(a+bx+cx^2)^2dx[/itex]
     
  10. Oct 10, 2009 #9

    Mark44

    Staff: Mentor

    It sure is. Now is there anything you can definitely say about the integrand, and therefore say about the integral?
     
  11. Oct 11, 2009 #10
    Could I say that the integrand is a non-negative functon therefore the integral will be a positive real number greater or equal to zero ?
     
  12. Oct 11, 2009 #11
    Yes but that doesn't make it definite. What can you say about the integral of a non-negative continuous function that's non-zero at at least one point?
     
  13. Oct 12, 2009 #12
    It is measurable?
     
  14. Oct 12, 2009 #13
    No, if you remember the definition of the integral as a limit of a riemannian sum, you will understand what aPhilosopher is trying to say.
     
  15. Oct 12, 2009 #14
    Are you talking about the fact that it is a definite integral because the riemann sum exists as the limit of sums as n approaches infinity and it's a bounded function?
     
  16. Oct 12, 2009 #15
    You have an integral whose integrand is [tex]|f(x)|^{2}[/tex]. If [tex]f(x) = 0[/tex] everywhere, the integral is obviously zero.

    If [tex]f(x) \neq 0[/tex] in at least one point, the integrand is [tex]|f(x)|^{2}>0[/tex] in at least that point, being zero in the others. Your interval of integration is positive, [tex]dx > 0[/tex] everywhere. So, your integral can be interpreted as an summation of products of positive values [tex]|f(x)|^{2} dx[/tex] and of zeros.

    This argument proves that [tex]\left< f , f \right> > 0[/tex] if [tex]f(x)[/tex] is not identically zero, and proves that [tex]\left< f , f \right> = 0[/tex] if [tex]f(x) = 0[/tex] everywhere in the interval.
     
  17. Oct 13, 2009 #16

    Landau

    User Avatar
    Science Advisor

    The formal argument is that if a continuous function equals some number strictly greater than zero at some point, then there is whole neighborhood of that point on which the function is strictly greater than zero. This is an essential property of continuous functions (and follows almost immediately from the definition). Since our integrand is nonnegative, a whole neighborhood (rather than just a single point) does contribute to the integral being greater than zero.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook