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Consider a transformation $T(x,y)=(u,v)$ defined on (the unit square $E\ $): $0\leq x\leq 1$, $0\leq y\leq 1$. Where $$u=2x^2+6xy-4x^3/3-3xy^2, \ \ \ v=x^3-y^2 $$ Show that an estimate for the Lipschitz constant $M$ on $E$ is $M=\sqrt{65}$

By Lipschitz constant $M$ we mean $$ |T(p)-T(q)|\leq M|p-q|,\ \ p,q\in E$$ Since $T$ is of class $C'$ we know that $T$ is Lipschitz, I tried doing $|(u,v)-(u',v')|$ but it doesn't seems to be a good way. The hint says maximize the absolute value of derivatives arriving at the matrix $$\left(\begin{array}{rc} 4 & 6\\ 3 & 2 \end{array}\right) $$ But I don't see how to arrive to this matrix nor how this can help. Please any guide.

Luis Valerin
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1 Answers1

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The Jacobian of $T$ at $(x,y)$ is $$J(T)(x,y) = \begin{bmatrix}4x+6y-4x^2-3y^2 & 6x - 6xy \\ 3x^2 & -2y\end{bmatrix}.$$

Maximizing the absolute value of each entry (over $x,y \in [0,1]$) individually leads to the matrix in the hint.

In general, we have the linear approximation $T(q) \approx T(p) + [J(T)(p)](q-p)$ when $q$ is close to $p$, so the idea is that $$|T(q)-T(p)| \approx |[J(T)(p)](q-p)|.$$

The right-hand side is bounded by $M |q-p|$ where $M$ is the operator norm (largest singular value) of the Jacobian. The operator norm is bounded by the Frobenius norm (square root of sum of squares of entries, or equivalently, square root of sum of squares of singular values). Using the matrix in the hint, the Frobenius norm is $\sqrt{4^2+6^2+3^2+2^2}=\sqrt{65}$.

angryavian
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