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Homework Help: ? p-norm <= (p-1)norm <= <= 2-norm <= 1-norm?

  1. Jun 19, 2010 #1
    the p-norm for a vector x in R^n is defined usually:

    || x ||_p = (x_1^p + x_2^p + ... + x_n^p)^{1/p}

    the question is to verify:

    || x ||_p <= || x ||_{p-1}

    2. Relevant equations

    I guess even more generally p-norm is a decreasing function in p for "any" x?

    3. The attempt at a solution

    Neither Cauchy, nor the more general Holder doesn't seem to apply.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 19, 2010 #2


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    Try using Jensen's inequality:

    If [itex]\phi[/itex] is a convex function, and [itex]y_1, \ldots, y_N[/itex] lie in its domain, and [itex]\lambda_1, \ldots \lambda_N[/itex] are positive real numbers such that

    [tex]\sum_{n=1}^{N} \lambda_n = 1[/tex]


    [tex]\phi\left(\sum_{n=1}^{N} a_n y_n\right) \leq \sum_{n=1}^{N} a_n \phi(y_n)[/tex]

    For [itex]p = 2[/itex], try using [itex]y_n = x_n^2[/itex], [itex]\phi(x) = x^{1/2}[/itex], and [itex]a_n = 1/N[/itex]. You will have to modify these choices appropriately for general [itex]p[/itex].

    Also, your proposed generalization is correct: if [itex]1 \leq q \leq p \leq \infty[/itex] then [itex]||x||_p \leq ||x||_q[/itex]. The [itex]p = \infty[/itex] case is handled separately (Jensen's inequality doesn't apply) but it's easy.
  4. Jun 20, 2010 #3
    I think it's a proof for norm 2 and 3. For any n, n-1 it's the same essentially.

    First consider R2.
    Note (x^2+y^2)^1/2 > (x^3+y^3)^1/3 is equivalent to
    [1+(y/x)^2]^1/2 > [1+(y/x)^3]^1/3
    Assume y/x < 1 without losing generality, then from [1 + ...] > 1 one sees
    [1+(y/x)^2]^1/2 > [1+(y/x)^3]^1/2 > [1+(y/x)^3]^1/3.

    Next do an induction toward Rn.
    (x1^2+x2^2+...+xn^2)^1/2 > (x1^3+x2^3+...+xn^3)^1/3 is equivalent to
    (y^2+xn^2)^1/2 > (z^3+xn^3)^1/3, where y and z stands for (x1^2+...+xn-1^2)^1/2 and (x1^3+...+xn-1^3)^1/3 respectively.
    By induction y > z. Hence (y^2+xn^2)^1/2 > (z^2+xn^2)^1/2 > (z^3+xn^3)^1/3. QED.
  5. Jun 21, 2010 #4
    thanks, guys, it helps!
  6. Sep 5, 2010 #5
    ||x||_p isn't a norm for p<1 ?

    Hello people!! I have a question about the p-norm.
    We know that the p-norm is a true norm for p >= 1.
    But I dont'n understand why it is not a true norm for 0 < p < 1.
    I think it hold the three conditions for a norm include when 0 < p < 1.

    Somebody can helpme??
    Thanks in advance!!!
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