Is $f(x)=(x_1 x_2, x_1^2+x_2^2) $ well-conditioned?

Question

Is $f(x)=(x_1 x_2, x_1^2+x_2^2) $ well-conditioned $\forall x$ (use $|| \ ||_\infty$ to calculate Condition number) ? $$f'(x)=\begin{pmatrix} x_2 \ \ x_1 \\ 2x_1 \ \ 2x_2 \end{pmatrix}$$

$$||f'(x) ||_\infty=\max(\lvert x_2 \lvert+ \lvert x_1 \lvert, \lvert 2x_1 \lvert+\lvert 2 x_2 \lvert)= \lvert 2x_1 \lvert+\lvert 2 x_2 \lvert$$

$$|| x ||_\infty=\max(\lvert x_1 \lvert , \lvert x_2 \lvert)$$

$$|| f ||_\infty=\max(\lvert x_1 x_2 \lvert , x_1^2+x_2^2 )$$

$$K(f,x)=\frac{2(\lvert x_1 \lvert+\lvert x_2 \lvert) \max(\lvert x_1 \lvert , \lvert x_2 \lvert)}{\max(\lvert x_1 x_2 \lvert , x_1^2+x_2^2 )}$$

I consider: $x_{max}=\max(\lvert x_1 \lvert , \lvert x_2 \lvert)$ and $x_{min}=\min(\lvert x_1 \lvert , \lvert x_2 \lvert)$

$$K(f,x) \le \frac{2(\lvert x_{min} \lvert +\lvert x_{max} \lvert) \lvert x_{max}\lvert}{\lvert x_{max} \lvert^2} = \frac{2(\lvert x_{min} \lvert+\lvert x_{max} \lvert)}{\lvert x_{max} \lvert} $$

$$\lim_{\lvert x_{max} \lvert \rightarrow +\infty} \frac{2(\lvert x_{min} \lvert+\lvert x_{max} \lvert)}{\lvert x_{max} \lvert}=2 $$

$f$ is well-conditioned for $(x_1,x_2) \ne (0,0)$

Is it correct?

Thanks!

Answer 1

The function is symmetric with respect to $x_1$ and $x_2$, so wlog assume $|x_1| \geq |x_2| \neq 0$. Then

\begin{align} K(f,x) &= \frac{2(|x_1|+|x_2|)\max(|x_1|,|x_2|)}{\max(|x_1||x_2|,x_1^2+x_2^2)}\\ &= \frac{2(|x_1|+|x_2|)|x_1|}{x_1^2+x_2^2}\\ &\leq \frac{4|x_1|^2}{x_1^2}\\ &= 4 \end{align}