# How can the Lipschitz continuity be shown with power iteration method?

I know that the definition of the Lipschitz continuity is defines as $$||f(y) - f(x)|| \leq L ||y-x||$$

My professor told me that by knowing $$f$$ we can find constant $$L$$ using the power iteration method, which essentially computes the maximum eigenvalues. Yet I don't see how definition above is related to searching for max eigenvalues. Can anyone please help me understand?

• Is $f$ a linear function(al)? Otherwise, eigenvalue analogy does not make sense. Assuming that this is related to the fixed point problem "find $x$ s.t. $x=f(x)$", you can approximate the constant as $L \approx \max_i \|f(x_{i+1})-f(x_i)\|/\|x_{i+1} - x_i\|$ which is equal to $\max_i \|x_{i+2}-x_{i+1}\|/\|x_{i+1} - x_i\|$ and $\max_i \|f(x_{i+1})-f(x_i)\|/\|f(x_{i}) - f(x_{i-1})\|$. This fairly looks like a power iteration and maybe that is what your professor meant. May 17, 2021 at 8:48
• I greatly appreciate your reply. This result is necessary as an assumption for the local convergence of primal-dual problem. Can you elaborate more why $\max_{i}$ looks like power iteration? From what I know, the power iteration algorithm is needed to find eigenvalues and I don't see any eigenvalue problem here. May 17, 2021 at 9:02
• There may not be an eigenvalue problem at all. That analogy may not be valid. Regarding why the maximum ratio over the iterations may approximate $L$, see cfdlab's answer. May 17, 2021 at 18:50

Define the iteration $$x_{n+1} = f(x_n)$$ Then $$e_n := \frac{\|x_{n+1} - x_n\|}{\| x_n - x_{n-1}\|} \le L$$ Compute $$e_n$$ and that should tell you something about L. If the iterations converge ($$L < 1$$), then $$e_n$$ should converge to $$L$$.