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Show that this inner product is positive definite

  1. Nov 5, 2015 #1
    1. The problem statement, all variables and given/known data

    MathProblemSheet4#1.PNG

    2. Relevant equations


    3. The attempt at a solution

    I was able to do the second part of part a using integration by parts. But I am having no luck for the first part, proving that the inner product is positive definite. Pointers are appreciated!
     
  2. jcsd
  3. Nov 5, 2015 #2

    Orodruin

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    Hint: what does it mean for an inner product to be positive definite?
     
  4. Nov 5, 2015 #3
    Is this acceptable ?

    ##g(f,f)=\int_0^1 f(-p'f'-pf''+qf)dx\\~~~~~~~~~~~=\int_0^1-fp'f'-fpf''+qf^2dx\\~~~~~~~~~~~=\int_0^1-ff'p'dx-\int_0^1pff''dx+\int_0^1qf^2dx\\~~~~~~~~~~~=-[fpf']_0^1+\int_0^1p(f')^2dx+\int_0^1pff''dx-\int_0^1pff''dx+\int_0^1qf^2dx\\~~~~~~~~~~~=\int_0^1p(f')^2dx+\int_0^1pf^2dx##
    Which is positive definite due to the square and p,q being positive definite and positive or zero.
     
  5. Nov 5, 2015 #4

    Orodruin

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    Well, you do not have to do all of the partial integrations as this was the form the inner product was given on from the beginning, and the last p should be a q, but otherwise fine.
     
  6. Nov 5, 2015 #5
    Oh of course, my mistake!

    Do you think you would be able to give me some advice on part b ?

    MathProblemSheet4#2.PNG
     
  7. Nov 6, 2015 #6
    No, in the first part of the question, you are being asked to show that for all ##f\in V_0##

    (a) ## g(f,f) \ge 0 ##
    (b) ## g(f,f) = 0 \Rightarrow f = 0 ##

    Point (a) is easy, point (b) is more subtle and continuity plays a central role.

    EDIT:
    I realized I didn't help much in my post: there is a theorem that says that if ##f: [a,b] \rightarrow \mathbb{R}## is continuous, non-negative, and such that ##\int_a^b f(x) \ dx = 0 ##, then ## f = 0 ##
     
    Last edited: Nov 6, 2015
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