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Are the p-norm and and the p'-norm equivalent?

  1. Jun 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Are the p-norm and and the p'-norm equivalent?

    With
    [tex]1 \leq p < \infty[/tex] and [tex]1 \leq p' < \infty[/tex]


    2. Relevant equations

    Two norms p and q are equivalent if there is a M>0 and m>0 such that [tex] ||x||_p \leq M ||x||_q [/tex] and

    [tex] ||x||_q \leq m ||x||_p [/tex]

    3. The attempt at a solution
    I do not seem to get started. How can I find the constants?
     
  2. jcsd
  3. Jun 2, 2014 #2

    LCKurtz

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    Before expecting help you need to
    1. Tell us what vector space ##x## is in.
    2. Show us what you have tried.
     
  4. Jun 2, 2014 #3
    [tex] x \in \mathbb{R} ^k [/tex]

    Suppose I take k = 2 and I presume p>q (I do not like the accent so I choose to use q) then I need to prove that:


    [tex] (|x_1|^p + |x_2|^p)^{1/p} \leq M (|x_1|^q + |x_2|^q)^{1/q} [/tex]


    This is not correct but if I assume the vector elements to be larger than one then:

    I know that [itex] |x_i|^p \geq |x_i|^q [/itex] with i=1,2.

    I also know that [itex] |x_i|^{1/p} \leq |x_i|^{1/q} [/itex] with i=1,2.

    Then I'm stuck.
     
    Last edited: Jun 2, 2014
  5. Jun 2, 2014 #4

    jbunniii

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    Suppose ##1 \leq p < q < \infty##.

    If ##\|x\|_p = 1##, can you find an upper bound for ##|x_i|^p##? What does this imply about ##\|x\|_q## in this case?
     
  6. Jun 3, 2014 #5
    if [itex] ||x||_p =1 [/itex] then [itex] |x_i|^p \leq 1 [/itex] which implies then that [itex] |x_i|^q \leq |x_i|^p [/itex] for i = 1,...,k but then:

    [tex] |x_1|^q +... + |x_k|^q \leq |x_1|^p +... + |x_k|^p [/tex]

    taking root at both sides

    [tex] (|x_1|^q +... + |x_k|^q)^{1/p} \leq (|x_1|^p +... + |x_k|^p)^{1/p} [/tex]

    but the RHS is 1

    [tex] (|x_1|^q +... + |x_k|^q)^{1/p} \leq 1 [/tex]

    taking root at both sides

    [tex] ||x||_q^{1/p} \leq 1 [/tex]

    This has an lower bound

    [tex] ||x||_q \leq ||x||_q^{1/p} [/tex]


    So the conclusion is that:

    [itex] ||x||_q \leq ||x||_p [/itex] with [itex] ||x||_p=1[/itex]


    Suppose the vector x has length A (in the p-norm) then we can adapt this to:

    [itex] ||x||_q \leq ||x||_p \cdot A^{1/q} [/itex]


    Is this correct?
     
    Last edited: Jun 3, 2014
  7. Jun 3, 2014 #6

    jbunniii

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    OK so far.

    OK, here you're using the fact that if you take the right hand side to the ##p##'th power, then the result is smaller since ##\|x\|_q^{1/p} \leq 1##.

    You can save a few steps if you avoid taking ##p##'th roots. If ##\|x\|_p = 1##, then ##|x_1|^p +... + |x_k|^p = 1##, so
    $$|x_1|^q +... + |x_k|^q \leq |x_1|^p +... + |x_k|^p = 1$$
    and therefore
    $$\|x\|_q = (|x_1|^q +... + |x_k|^q)^{1/q} \leq 1 = \|x\|_p$$

    No. If ##\|x\|_p = A##, then we can write ##x = Au## where ##\|u\|_p = 1##. Then:
    $$\|x\|_q = \|Au\|_q = A\|u\|_q \leq ???$$
    Now apply the result obtained above.



    For the other inequality ##||x||_p \leq M ||x||_q##, I recommend that you once again start with the case ##\|x\|_p = 1##. In that case, you need to prove that there is some ##M## for which
    $$\|x\|_q \geq \frac{1}{M}$$
    The key idea here is that if ##\|x\|_p = 1##, it's not possible for ALL of the elements of ##x## to be small. Can you find a lower bound for ##\max |x_i|##?
     
  8. Jun 5, 2014 #7
    [tex]\|x\|_q = \|Au\|_q = A\|u\|_q \leq A\|u\|_p = \|Au\|_p= ||x||_p [/tex]

    So m=1 in this case?

    I think that the lower bound for [itex] \mbox{max}|x_i| = (\frac{1}{k})^{1/p}[/itex]

    But I don't understand how to proceed.
     
    Last edited: Jun 5, 2014
  9. Jun 5, 2014 #8

    jbunniii

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    Yes.

    Yes, this is correct. Sketch of proof: if ##|x_i| < (1/k)^{1/p}## for all ##i##, then ##|x_i|^p < 1/k## for all ##i##, so ##|x_1|^p + \cdots + |x_k|^p < 1##, contradicting the fact that ##\|x\|_p = 1##. So for at least one ##i## we must have ##|x_i| > (1/k)^{1/p}##.

    For the next step, note that the above implies (for the same ##i##) that ##|x_i|^q > (1/k)^{q/p}##. What does this imply about ##\|x\|_q##?
     
  10. Jun 10, 2014 #9
    I think it implies that [itex] ||x||_q \geq \left(\frac{1}{k} \right)^{q/p} [/itex]


    Now suppose [itex]||x||_p=A [/itex]


    We can write [itex]x=Au [/itex] with [itex]||u||_p= 1[/itex] then we get


    [itex]||x||_q =||Au||_q =A ||u||_q \geq A \left(\frac{1}{k} \right)^{q/p} = M\cdot A||u||_p = M\cdot ||Au||_p = M ||x||_p[/itex] with [itex]M= \left(\frac{1}{k} \right)^{q/p} [/itex]

    Is this correct?
     
    Last edited: Jun 10, 2014
  11. Jun 10, 2014 #10

    jbunniii

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    Close but not quite right:
    $$\|x\|_q^q = |x_1|^q + \cdots + |x_k|^q \geq |x_i|^q > (1/k)^{q/p}$$
    where ##i## is as in my previous post. Taking ##q##'th roots of both sides, we get
    $$\|x\|_q > (1/k)^{1/p}$$
    The rest of your post is fine if you make the corresponding adjustment.
     
  12. Jun 10, 2014 #11
    Ok got it thanks.
     
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