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

1. Oct 1, 2013

Felafel

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. Oct 1, 2013

jbunniii

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. Oct 1, 2013

Felafel

I'm using the 1-norm
any suggestions on how to proceed then?

4. Oct 2, 2013

brmath

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. Oct 6, 2013

Felafel

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. Oct 6, 2013

brmath

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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Continuity mathbb_{R}^3 mathbb_{R}## Date
How can I find 544 mod 3 using congruence Feb 19, 2018
Continuity and discrete space Nov 17, 2017
Local Formulation of Continuity Oct 15, 2017