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Infinity Norms and One Norms

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine constants c and C that do not depend on vector x but may depend on the dimension n of x, such that

    [itex]c ||x||_{∞} ≤ ||x||_{1} ≤ C||x||_{∞}[/itex]

    Use this result and the definition of matrix norms to find k and K that don't depend on the entries of matrix A (but depend on dimension n of A) such that

    [itex]k ||A||_{∞} ≤ ||A||_{1} ≤ K||A||_{∞}[/itex]

    3. The attempt at a solution

    Firstly i'm still trying to wrap my head around what an infinity norm actually is. According to my lecture notes,

    [itex] ||x||_{∞} \equiv 1 ≤ k ≤ n, max|x_{k}|[/itex]

    Does this mean the infinity norm is just the largest absolute value in x? There is no summation sign so my best guess is that "max" refers to the highest possible value contained in x.

    So in other words i need to figure out constants which satisfy the inequality. My first guess would be to take c=0 and C=1, assuming that infinity norms are always greater than one-norms.
     
  2. jcsd
  3. Sep 21, 2013 #2

    Mark44

    Staff: Mentor

    Sort of. It's the largest component xkof a vector x.
     
  4. Sep 22, 2013 #3

    pasmith

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    If [itex]x = (1,1)[/itex], what are [itex]\|x\|_1[/itex] and [itex]\|x\|_{\infty}[/itex]?

    Taking [itex]c = 0[/itex] doesn't tell you anything; norms are non-negative by definition.

    Hint for [itex]C[/itex]: given that [itex]\|x\|_1 = \sum_{i=1}^n {|x_i|}[/itex], what happens if you replace each summand with [itex]\|x\|_\infty = \max\{|x_j| : j = 1, \dots, n\}[/itex]?
     
  5. Sep 25, 2013 #4
    [itex]\|x\|_1[/itex] would be 2 since you're just summing the absolute values contained in x (right?) and [itex]\|x\|_{\infty}[/itex] would be 1, i think.

    I'm not sure i follow your second bit about C. If i were to replace every instance of |X|i with |X|∞ then, say for example, if i had x={1,2,3) I would be doing 3+3+3 instead of 1+2+3.
     
  6. Sep 25, 2013 #5

    pasmith

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    And is [itex]3 + 3 + 3 = 3\|(1,2,3)\|_{\infty}[/itex] greater than or less than [itex]1 + 2 + 3 = \|(1,2,3)\|_1[/itex]?
     
  7. Sep 25, 2013 #6
    The left hand side evaluates to 9 and the right hand side evaluates to 6. I find the question confusing because you could hypothetically find any 2 values of C and c which satisfy the question if you know what x contains. If x=(1,2,3) then you just need to pick C = 3, c = 1.

    I assumed picking c=0 would make sense because the left hand side would always be zero if you're multiplying the infinity norm by zero. That would satisfy the left hand side, and the right hand side would just require a sufficiently large value for C to be greater than the one norm.
     
  8. Sep 25, 2013 #7

    Office_Shredder

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    Suppose that my vector x = (M,x2,...,xn) where M is larger than all the other entries (so is the infinity norm of x). Then:
    [tex] ||x||_1 = |M| + |x_2| + ... + |x_n| \leq ???? [/tex]

    The ????? should be writeable in terms of only M and the dimension of the space. Think about doing the same kind of procedure to this sum as you did with that (1,2,3) vector.

    That is the harder direction. The easier one (I think) is to figure out
    [tex] ???? \leq |M| + |x_2| + ... + |x_n|[/tex]
    where again ??? is something that depends only on M and the dimension of the space
     
  9. Sep 25, 2013 #8
    In terms of n (where n is the size of x), would this be correct for the first part?

    [tex] ||x||_1 = |M| + |x_2| + ... + |x_n| \leq n * |M| [/tex]

    For the second part, would i simply use

    [tex] |M| * n-1 \leq |M| + |x_2| + ... + |x_n|[/tex]
     
  10. Sep 25, 2013 #9

    Office_Shredder

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    This looks good (and should tell you what C is in the first part)

    What if M = 1, x2 = x3 = 0 and n=3? You just told me that 3-1 = 2 < 1+0+0
     
  11. Sep 26, 2013 #10
    Hmmm, so if C=n then could c just be 1? There isn't much that the infinity norm could be multiplied by that would make it less than the one norm, right?
     
    Last edited: Sep 26, 2013
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