# Find the Lipschitz constant for the following function

$f(t, \begin{bmatrix}x_1\\x_2\end{bmatrix}, u)=e^{-|u|^2}\begin{bmatrix}x_1\\x_2\end{bmatrix},$ 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.

• What theory can you use?, for instance, do you know the mean value theorem? Feb 16, 2017 at 11:35
• with mean value theorem how to find? Feb 16, 2017 at 11:46

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.

• I'm not sure what you mean when you say that this $f$ is not jointly Lipschitz...in the third argument it's $C^1$ on a compact and in the second argument it's linear with a bounded coefficient matrix.
– Ian
Feb 16, 2017 at 12:23
• The problem is that the Lipschitz constant w.r.t. the third argument is lineary dependent on the second. There is no one single constant that then will work for all values of the second parameter. Considered as a function from the metric space $[0,3]\times\mathbb R^2\times\mathbb R$ to the metric space $\mathbb R^2$ it is simply not Lipschitz continuous. Feb 16, 2017 at 14:19
• Now I see your point, it is more delicate: the point is that $\frac{\partial f_1}{\partial u}$ is large if $x_1$ is large and $\frac{\partial f_2}{\partial u}$ is large if $x_2$ is large.
– Ian
Feb 16, 2017 at 16:19

The dependence on $t$ is irrelevant, so we can suppress that. Then the Jacobian is

$$J(u,x)=\begin{bmatrix} -2ue^{-u^2} x_1 & e^{-u^2} & 0 \\ -2ue^{-u^2} x_2 & 0 & e^{-u^2} \end{bmatrix}.$$

Your problem is to extremize the operator norm of this matrix as $u,x_1,x_2$ each range over $\mathbb{R}$. If I take the norm to be the Euclidean norm (which is not exactly what you did in the question, but is more conventional), then this is the square root of the maximum eigenvalue of $J J^T=\begin{bmatrix} -2ue^{-u^2} x_1 & e^{-u^2} & 0 \\ -2ue^{-u^2} x_2 & 0 & e^{-u^2} \end{bmatrix} \begin{bmatrix} -2ue^{-u^2} x_1 & -2ue^{-u^2} x_2 \\ e^{-u^2} & 0 \\ 0 & e^{-u^2} \end{bmatrix}=\begin{bmatrix} 4 u^2 e^{-2u^2} x_1^2 + e^{-2u^2} & 4u^2 e^{-2u^2} x_1 x_2 \\ 4u^2 e^{-u^2} x_1 x_2 & 4u^2 e^{-2u^2} x_2^2 + e^{-2u^2} \end{bmatrix}.$ So if $|x_i|$ is the bigger of $|x_1|$ and $|x_2|$ then this can be bounded by $4u^2 e^{-2u^2} (x_i^2+|x_1 x_2|) + e^{-2u^2}$ (this is Gerschgorin's theorem). This is not uniformly bounded over $\mathbb{R}^3$, which is because the first column of $J$ itself is not uniformly bounded over $\mathbb{R}^3$. It is bounded if the vector $x$ is confined to any bounded set in $\mathbb{R}^2$, in which case we would recover a Lipschitz constant of $\sqrt{4 M_1 M_2 + 1}$, where:

• $M_1$ is the maximum of $u^2 e^{-2u^2}$ over $\mathbb{R}$
• $M_2$ is the maximum of $\max \{ |x_1|,|x_2| \}^2+|x_1 x_2|$ over the bounded set in question.

Exactly computing the eigenvalues would give you the optimal Lipschitz constant, but this would be annoying/tedious (and people almost never do this).

You can imitate this procedure to rephrase this in terms of separated Lipschitz constants, but the point is that your "$b$" cannot be universal if $x$ is allowed to vary over all of $\mathbb{R}^2$, because then the $u$ partial derivatives of $f$ are not bounded.

• so you mean we cannot find the smallest possible a and b values? Feb 16, 2017 at 16:35
• @Vijay What I mean is that there is no admissible $b$ value at all, much less a smallest one, unless you restrict $x$ to a bounded set.
– Ian
Feb 16, 2017 at 16:36
• then is there any example of such type of other nonlinear function for which we can find a and b? Feb 16, 2017 at 16:46
• @Vijay Sure, you can make up all sorts of weird functions with bounded derivatives. The problem with your example was that you had that linear dependence on $x_1,x_2$ which amplified changes in the $u$ variable. But if you make the whole thing nonlinear then you can make it work.
– Ian
Feb 16, 2017 at 16:48
• if i take $x_1^2$ and $x_2^2$ instead of $x_1$ and $x_2$ then? Feb 16, 2017 at 16:54