Find Lipschitz constant for $F(x,y)=(x(1-ay),y(-1+bx))$

Question

Let $F(x,y)=(x(1-ay),y(-1+bx))$ on $0\leq x,y\leq 10$. Here $a$ and $b$ are assumed to be positive constants. Find the Lipschitz constant on this domain.

I have a couple of questions about the solution of this exercise, see below:

Q1. How do they arrive at $\sum_i(\sum_ja_j^iv^j)^2\leq\sum_{i,j}a_j^i$? The problem is, I don't even understand what they mean by the right hand side. I know that we arrive at the first equality by simply using $\Vert Av\Vert^2=(Av)^TAv$, and then using the definition of matrix-vector multiplication. But I don't see what happens in the inequality.

Q2. What norm could $\Vert A\Vert_2$ be? Is it $\max\{\sqrt \lambda:\lambda\text{ is an eigenvector of $A^TA$}\}$?

Q3. I also don't know how they get the last inequality. But I'm guessing once it's clear to me what $\Vert{.}\Vert_2$ means, I will understand what they're doing at the last step.

Answer 1

To avoid confusion with powers, I will write $a_{ij}$ instead of $a_i^j$.

Q1: Where does the first inequality come from?

In the first display Cauchy-Schwarz is used: $$ \|Av\|=\sum_i \left( \sum_j a_{ij} v_j \right)^2\ \overset{CS}{\leq}\ \sum_i \left( \sum_j a_{ij}^2 \cdot \sum_j v_j^2\right) = \sum_{i,j} a_{ij}^2 $$ because $\sum_j v_j^2=1$ by assumption. (There is a square missing in your screenshot; this is a mistake).

Q2: What is $\|A\|_2$?

If you would view a matrix as a vector of length $n\times n$ then $\sqrt{\sum_{i,j=1}^n a_{ij}^2}$ would be the 2-norm of this vector. Therefore we sometimes write $\|A\|_2 = \sqrt{\sum_{i,j} a_{ij}^2}$. In general this is not equal to the maximum you describe.

Q3: Where does the last inequality come from?

The total derivative of $F$ in $z = (x,y)$ equals $$ DF(z)=\begin{pmatrix} 1-ay & -ax\\ by & -1+bx \end{pmatrix} $$ so if we square all entries and sum them we obtain $$\begin{aligned} \|DF(z)\|^2 &= (1-ay)^2 + a^2x^2 + b^2y^2+(-1+bx)^2\\ &= 1 + a^2 y^2 - 2 a y + a^2 x^2 + b^2 y^2 + 1 + b^2 x^2 -2 bx\\ &= 2 + (x^2 + y^2)a^2 + (x^2 + y^2)b^2 - 2ay - 2bx\\ &\leq 2 + 200(a^2 + b^2) + 20(|a|+|b|) \end{aligned}$$ Here I ignored your assumption that $a$ and $b$ are positive. If we assume this the last term disappears entirely (and the absolute value in the answer wouldn't make sense).

Note that your approach gives a Lipschitz constant, but it might not be the best one.