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Continuity ##f:\mathbb_{R}^3 \to \mathbb_{R}## with Lipschitz

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

    Prove
    ## f(x,y,z)=xyw## is continuos using the Lipschitz condition

    2. Relevant equations

    the Lipschitz condition states:
    ##|f(x,y,z)-f(x_0,y_0,z_0)| \leq C ||(x,y,z)-(x_0,y_0,z_0)||##
    with ##0 \leq C##

    3. The attempt at a solution

    ##|xyz-x_0y_0z_0|=|xyz-x_0y_0z_0+x_0yz-x_0yz|\leq|(x-x_0)(yz)|+|x_0(yz-y_0z_0)|##
    ##\leq|(x-x_0)(yz)|+|x_0(yz-y_0z_0+yz_0-yz_0)| \leq |yz(x-x_0)|+|x_0[y(z-z_0)+z_0(y-y_0)]|##
    ## \leq |yz(x-x_0)|+|x_0y(z-z_0)|+|x_0z_0(y-y_0)|##
    and by choosing ##C=max\{|yz|,|x_0y|,|x_0z_0|\}## I have my inequality.
    I'm not sure I can do this, though. Are all the passages logic?
    Thank you in advance :)
     
  2. jcsd
  3. Oct 1, 2013 #2

    jbunniii

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    I didn't check the rest, but this part won't work. The constant needs to be independent of ##x##,##y##, and ##z##. But yours depends on ##y## and ##z##.

    Also, which norm are you using for ##\|(x,y,z) - (x_0,y_0,z_0)\|##? From your work, it would appear that you are using ##\| (a,b,c)\| = |a| + |b| + |c|## (the 1-norm), but unless otherwise specified, in ##R^{n}## it's usual to assume that the 2-norm is intended: ##\|(a,b,c)\| = \sqrt{a^2 + b^2 + c^2}##.
     
  4. Oct 1, 2013 #3
    I'm using the 1-norm
    any suggestions on how to proceed then?
     
  5. Oct 2, 2013 #4
    Here is a different approach. Start in one dimension with h(x) = x. This is clearly Lipschitz continuous with C = 1.

    Now do an induction step. Suppose we assume that g(x,y) = xy is Lip continuous. Does it follow that f(x,y,z) = xyz is Lip continuous?

    You can certainly write f(x,y,z) = g(x,y)h(z).

    Can you finish it from here?
     
  6. Oct 6, 2013 #5
    Here's my attempt:
    assuming g(x,y)=xy Lipschitz
    then
    g(x,y)*h(z) is the product of two Lipschitz functions, which is Lipschitz itself. In particular,
    f(x,y,z)=g(x,y)*h(z)=##C \cdot |xy - x_0y_0|\cdot1\cdot|z-z_0|## but ##|xy-x_0y_0|## is equal to ##(x-x_0)(y-y_0)## with scalar product (i'm not sure this passage actually works, but i don't know how to rewrite ##|xy-x_0y_0|## in an "useful" manner, such as ##|x-x_0| |y-y_0|##, so:
    ##f(x,y,z)=|z\cdot xy|\leq 1\cdot |z-z_0| \cdot C (|x-x_0|\cdot|y-y_0|)##
    anf therefore f is lipschitz with constant C.
    then by induction g is lipschitz too, as assumed.
     
  7. Oct 6, 2013 #6
    Stop right there. You've just said g(x,y)h(z) is Lipschitz continuous, so f(x,y,z) = xyz = g(x,y)h(z) is Lipschitz continuous. You don't have to prove another thing ; if it is Lip continuous, it is continuous.
     
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