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

  • Thread starter Felafel
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  • #1
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Homework Statement



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

Homework 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##

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 :)
 

Answers and Replies

  • #2
jbunniii
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and by choosing ##C=max\{|yz|,|x_0y|,|x_0z_0|\}## I have my inequality.
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}##.
 
  • #3
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I'm using the 1-norm
any suggestions on how to proceed then?
 
  • #4
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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?
 
  • #5
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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.
 
  • #6
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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.
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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