$$ f(x,y)=\frac{2x^2y+y^3}{\sqrt{x^2+y^2}} $$ $$ I \ know \ that \ it \ is \ continuous \ on\ \overline{B_1(0)} \ . \\ I\ think \ I \ have \ to \ show \ ||f(x_1,x_2)-f(y_1,y_2)||\leq L||(x_1,x_2)-(y_1,y_2)|| \ , \ for \ x_i,y_i\in \overline{B_1(0)} \ ,\\i=1,2\ and \ L\in \mathbb{R}_+ $$
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What have you tried so far? Note that $f(x,y) = y\sqrt{x^2+y^2}$. – amsmath Sep 09 '18 at 13:14
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is it correct what I have to show ? $$ y\sqrt{x^2+y^2}* \sqrt{x^2+y^2} =yx^2+y^3 . $$ – Matillo Sep 09 '18 at 13:37
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Yes, that's correct by definition of Lipschitz continuity. Note that it's enough to show that $f$ is differentiable and that its derivative is bounded in the disk. – amsmath Sep 09 '18 at 19:37