The Lipschitz continuous is by $||Ax-Ay|| \leq L||x-y||$, where A is a matrix. Here by cauchy-schwarz inequality, it is easy to have $||Ax-Ay||^2 \leq ||A|| \times ||x-y||$. But the problem is, what the exact norm is for $||A||$?
Normally we use Euclidean norm, thus both vector and matrix is under Euclidean norm, thus $||A||$ is under Frobenius norm by $\left\| A \right\|_{F}$ = $\sqrt{\sum_{i=1}^{m}\sum_{j=1}^{n}{|a_{ij}|^{2}}}$ = $\sqrt{\sum_{i=1}^{\min{(m,n)}}{\sigma_{i}^2(A)}}$, where $\sigma_i$ is the singular value of A.
However, some people say that the Lipschitz constant L for the matrix A is given by its spectral norm $\sigma(A)$ = largest $\sigma_i$, like the question in Lipschitz constant of matrix. I understand that $||A||_2$ = spectral norm $\sigma(A)$, under operator norm definition: $\left\| A \right\|_{p}$ = $\sup_{x \ne 0} \frac{\left\| Ax \right\|_{p}}{\left\| x \right\|_{p}}$.
However, could we directly use spectral norm in the equation $||Ax-Ay||^2 \leq ||A|| \times ||x-y||$? Because under Euclidean norm, ||A|| = $\sqrt{\sum_{i=1}^{\min{(m,n)}}{\sigma_{i}^2(A)}}$, is the sum of square of all singular values. But with operator norm, ||A|| = largest $\sigma_i$, simply one largest singular value.
So exactly is ||A||? Thanks a lot.