Question

Show that f(x y)=xy^(2) satisfies a Lipschite condition on any rectangle a<=x<b and c<=y<=d.

Answer 1

A function f(x, y) is said to satisfy Lipschitz condition on a domain $D\sube\R^2$ , if there exists $L>0$ such that $|f(x, y_1)-f(x, y_2)|\leq L|y_1-y_2|$ for all $(x, y_1), (x, y_2)\in D.$

On closed rectangle $[a, b]\times[c, d],$ the partial derivative $f_y$ is continuous and hence bounded and hence $f$ satisfies Lipschitz condition. 

Note that $|x|\leq max(|a|, |b|)=A$ and $|y|\leq max(|c|, |d|)=C.$

$$|\dfrac{f(x, y_1)-f(x, y_2)}{y_1-y_2}|=|\dfrac{xy_1^2-xy_2^2}{y_1-y_2}|$$

$$=|x(y_1+y_2)|\leq|x||y_1+y_2|\leq2A\cdot C$$

is bound.

The function $f(x, y)=xy^2$ satisfies Lipschitz condition on any rectangle $[a, b]\times[c, d].$