Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / F / F2 / b

Put $Q=1+x^2+y^2>0$. Differentiation of $Q^{-1}$ givesThe positive denominator also shows that the function and these derivatives are smooth throughout ${\mathbb{R}}^2$.

At $(-1,0)$ the gradient is $(1/2,0{)}^T$. The total derivative is therefore the linear functional

The derivative of the curve isIts image lies on the unit circle, so $f(g(t))=1/2$ and $\boxed{D(f\circ g)(t)=0}$. Meanwhileand multiplication by $J_g(t)$ gives $-\sin t\cos t/2+\cos t\sin t/2=0$. This verifies the chain rule explicitly.

Because $2|x|\le 1+x^2\le Q$, we have $|f_x(x,y)|=2|x|/Q^2\le 1/Q\le 1$; similarly $|f_y|\le 1$. Apply the one-variable mean value theorem on the horizontal and vertical segments joining $(x,y)$ to $(u,y)$ to $(u,v)$. The triangle inequality givesThus the function is globally one-Lipschitz for the ${\ell }^1$ metric on the plane.