$f(t, \begin{bmatrix}x_1\\x_2\end{bmatrix}, u)=e^{-|u|^2}\begin{bmatrix}x_1\\x_2\end{bmatrix},$ where $f:[0, 3]\times \mathbb{R}^2\times \mathbb{R}\rightarrow \mathbb{R}^2$ is a given nonlinear function. I want to find the Lipschitz constants for this function w.r.t second and third arguments. Lipschitz continuity definition w.r.t second and third arguments is $\|f(t, \begin{bmatrix}x_1\\x_2\end{bmatrix}, u)-f(t, \begin{bmatrix}y_1\\y_2\end{bmatrix}, v)\|\leq a\|\begin{bmatrix}x_1\\x_2\end{bmatrix}-\begin{bmatrix}y_1\\y_2\end{bmatrix}\|+b|u-v|, $ where a and b are the Lipschitz constants. I want to find the smallest of these.
[Math] Find the Lipschitz constant for the following function
lipschitz-functions
Best Answer
First of all it's a Lipschitz constant since the constant in Lipschitz continuity is not uniquely determined. In fact if $K$ is a Lipschitz constant then all values larger that $K$ are also Lipschitz constants.
Second $f$ is not Lipschitz continuous if you allow variation in all variables.
Now we can use the mean value theorem since $g(a)-g(b) = g'(\xi)(a-b)$ for some $\xi\in]a,b[$. This means that $|g(a)-g(b)| = |g'(\xi)||a-b|$. And if $g'(\xi)$ is bounded an upper bound of $g'(\xi)$ would be a Lipschitz constant.
For given values of two of the arguments the partial derivate of $f$ is bounded which is easy to see since one can easily find an upper bound of the derivate.
It's only in that sense $f$ is Lipschitz continuous. We can see that if you consider variation in all variables the derivate is not bounded and with unbounded derivates we can find secants with arbitrarily high slope which makes it non-Lipschitz continuous.