Lipschitz but not differentiable $f(x,y) = x^2|y|$

Question

Prove that the function $f(x,y) = x^2|y|$ meets the Lipschitz condition with respect to $y$ in the rectangle $\{|x| \leq 1,|y| \leq1\} $ but f_y does not exist in many points of this rectangle.

Attempt :

Showing that the derivative with respect to $y$, $f_y(x,y)$ does not exist is not an issue, but I'd like to ask how I'd prove the Lipschitz condition part.

$|f(x,y_2) - f(x,y_1)| = |x^2|y_2| - x^2|y_1|| = |x^2(|y_2|-|y_1|)| = |x^2|||y_2| - |y_1||\leq |x^2||y_2 + y_1|$

But after that, how will I get the $|y_2 - y_1|$ needed for the Lipschitz condition ? It will be obvious once I can derive it, since $|x| \leq1 $ and is independent of $y_2,y_1$.

Answer 1

Hint: $$||y_2|-|y_1||\leq |y_2-y_1|$$ and $$|x|\leq 1 \Leftrightarrow x^2\leq1.$$

Answer 2

$$|f(x,y_1)-f(x,y_2)|=|x^2(|y_1|-|y_2|)|\leq ||y_1|-|y_2||\leq |y_1-y_2| $$ First inequality comes from the fact that $|x|\leq 1$

For partial derivative w.r.t $y$ consider the region $S=\{(x,y)\in R^2: y=0, |x|\leq 1\}$, $$(f(x,k)-f(x,0))/k= x^2 (|k|/k)$$ The limit $k$ tends to $0$ does not exists for the above expression, hence partial derivative of $f$ w.r.t $y$ does not exists in $S$.