Working on lipschitz function and contraction map

[simo1]

if you given a function f from R^2 to R^2 f(x)=<f_1(x),f_2(x)>, x in R^2

with f_1 and f_2 from R^2 to R being differentiable on R. if there is contants K_1 and K_2 greater than or equal to 0 so the 2-norm of (gradient f_1(x)) is less than or equal to K_1 and 2-norm of (gradient f_2(x)) is less than or equal to K_2 for x in R^2.
show that the 2-norm of (f(x)-f(y)) is less than or equal to [square root of (k_1^2 -k_2^2)] multipy by 2-norm of (x-y) for all x and y in R^2

can i get hints on how to start

 

[Opalg]

if you given a function f from R^2 to R^2 f(x)=<f_1(x),f_2(x)>, x in R^2

with f_1 and f_2 from R^2 to R being differentiable on R. if there is contants K_1 and K_2 greater than or equal to 0 so the 2-norm of (gradient f_1(x)) is less than or equal to K_1 and 2-norm of (gradient f_2(x)) is less than or equal to K_2 for x in R^2.
show that the 2-norm of (f(x)-f(y)) is less than or equal to [square root of (k_1^2 – k_2^2)] multipy by 2-norm of (x-y) for all x and y in R^2 (That – should be a +.)

can i get hints on how to start

Hi simo, and welcome to MHB! By the definition of the 2-norm, $\|f(x) - f(y)\|_2^2 = |f_1(x) - f_1(y)|^2 + |f_2(x) - f_2(y)|^2.$ Now use the result from http://mathhelpboards.com/analysis-50/converging-maps-9673.html (applied to the functions $f_1$ and $f_2$) to deduce that $\|f(x) - f(y)\|_2^2 \leqslant (K_1^2 + K_2^2)\|x-y\|^2.$ Then take the square root of both sides.