I have the following linear function $f(x,y,z) = ax + by + cz.$
I need to prove that f() is (p-q) Lipschitz continuous where $p=1$ and $q=\infty$. For a given two points $(x_1, y_1, z_1)$ and $(x_0, y_0, z_0)$, I should find an L that satisfies
$|| f(x_1, y_1, z_1) ||_{\infty} \leq L||(x_1, y_1, z_1)- (x_0, y_0, z_0)||_{1}$
This means
$|| a(x_1 - x_0) + b(y_1 - y_0) + c(z_1 - z_0)||_{\infty} \leq L ||(x_1, y_1, z_1) - (x_0, y_0, z_0)||_1 = ||x_1 -x_0 || + ||y_1- y_0|| + || z_1 - z_0||$.
As suggested, let $(x_0,y_0, z_0)=(0,0,0)$. We then have
$|| a(x_1 ) + b(y_1 ) + c(z_1)|| \leq L(||x_1 || + ||y_1|| + || z_1 ||)$.
If $a=b=c=1$, by using triangle inequality, can I conclude that L is at least 1? How about other values of $a, b$, and $c$?
But I can not move forward from this point. Would anyone here help me to find L? thanks.
First prove it for $P_0=(x_0,y_0,z_0)=(0,0,0)$. From the 2nd formula you missed the constant $L$, which can be chosen $L=3\cdot\max(|a|,|b|,|c|)$ in this case. You also exchanged the $\infty$ and $1$ norms in one of your lines. ($\|.\|_\infty=\|.\|_\max$.) $$|ax_1+by_1+cz_1| \le |a|\cdot|x_1|+|b|\cdot|y_1|+|c|\cdot|z_1|\le L/3\cdot |x_1|+L/3\cdot|y_1|+L/3\cdot|z_1|\le \\ \le L/3\cdot 3\cdot\|(x_1,y_1,z_1)\|_\max\,.$$ And, as $\|\_\|_\max \le \|\_\|_1$, the same $L$ works for $\|\_\|_1$ norm as well. (Note that on $\Bbb R$ all these norms are the same as $|\_|$.)
Finally, use linearity for the general case with $(x_0,y_0,z_0)$: we have $$f(x_1,y_1,z_1)-f(x_0,y_0,z_0)=f\left((x_1-x_0),(y_1-y_0),(z_1-z_0)\right)\,.$$