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Semi-Norm Proof

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [tex]C^{1}(\Re)[/tex] be the space of real-valued functions which have continuous first derivatives. Let

    [tex]||f||=sup_{-5 \leq x \leq 5} |f(x)|[/tex]

    Prove that ||.|| is a semi-norm, but not a norm on the space.


    2. Relevant equations
    As far as my understanding goes a semi-norm is a norm that allows some non-zero vectors to give a zero norm.

    Criteria for a norm:
    1. ||v|| >= 0
    2. ||v|| = 0 iff v = 0 (will have to prove this isnt true for our case)
    3. ||cv|| = |c| ||v||
    4. ||u + v|| <= ||u|| + ||v||



    3. The attempt at a solution

    The way I consider the norm is as follows:
    We find the supremum of the set {|f(x)|: -5 <= x <= 5}

    1. From the definition we know the norm will always be greater than zero (absolute value)

    2. The only way for this norm to be zero is if the function is zero on the interval (maybe this isnt the case but as far as my limited knowledge can tell it is...). This seems wrong though since if thats the case then v = 0 which would lead me to believe that it is indeed a norm and not a semi norm.

    3. Easy to prove, just algebra.

    4. Easy analysis proof.

    Where am I wrong? What haven't I considered?
     
  2. jcsd
  3. Aug 31, 2010 #2

    Dick

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    You haven't considered that the function space is defined on R. Not [-5,5].
     
  4. Sep 2, 2010 #3
    Well, why do I care about the whole space R if I am just looking at the supremum on [-5,5] ?
     
  5. Sep 2, 2010 #4

    Dick

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    Because if v is nonzero off of [-5,5] it's not the zero vector, but it's norm is still zero.
     
  6. Sep 2, 2010 #5
    Okay, I see your point... well does the supremum have to be in the interval ?
     
  7. Sep 2, 2010 #6
    Or do I just need to reword my statement for part 2? ie include the fact that the function isn't necessarily 0 but it is so on the interval
     
  8. Sep 2, 2010 #7

    Dick

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    You just need to find an example of a nonzero function with continuous derivatives which is zero on [-5,5].
     
  9. Sep 2, 2010 #8
    Maybe I'm 'thinking' too hard but I can't think of anything like that (that has continuous derivatives)

    I originally thought:
    f(x) = x - 5 when x>5 and 0 otherwise but it doesnt have a continuous derivative
     
  10. Sep 2, 2010 #9

    Dick

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    Think just a little harder. Yes, define the function piecewise. Just make sure the derivatives match where the pieces join. Can't you think of a nonzero function whose derivative is zero at x=5?
     
  11. Sep 2, 2010 #10
    f(x) = x^2 (x > 5), 0 otherwise

    For some reason I was thinking functions in C1 must have _constant_ derivatives >.<
     
  12. Sep 2, 2010 #11

    Dick

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    That example isn't differentiable or even continuous at x=5. Keep thinking.
     
  13. Sep 2, 2010 #12
    okay, to be continously differentiable the derivative has to be equal to zero at x = 5

    so how about x^2 - 5x

    this function is 0 at x = 5 so its continuous and the derivative is continuous at this point as well so this should be good
     
  14. Sep 2, 2010 #13

    Dick

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    The derivative of 0 at x=5 is 0. The derivative of x^2-5x at x=5 is -15. The derivative isn't continuous. The function you use to define the values for x>5 has to be zero at x=5 AND it's derivative has to zero at z=5. That's all you need. There's lots of them.
     
  15. Sep 2, 2010 #14
    Okay... I don't know why that was so difficult for me

    One such one could be:

    x^2 - 10x + 25


    Okay, so when writing my proving this I can just write this function as an example and not have to worry about that part anymore?
     
  16. Sep 3, 2010 #15

    Dick

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    Sure. Point out it's a nonzero function with zero norm.
     
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