"Doubly Lipschitz" functions

Let $$X,Y,Z$$ be normed spaces. Let $$P\subseteq X$$ and $$Q\subseteq Y$$ be bounded. Endow the product space $$X\times Y$$ with the $$\max$$ norm.

Suppose $$f:P\times Q\rightarrow Z$$ is Lipschitz continuous with Lipschitz modulus $$L$$.

By way of background, note that by the triangle inequality and $$f$$'s Lipschitz continuity, for $$x_1,x_2\in P,y_1,y_2\in Q$$:

\begin{align}& \|f(x_1,y_1)+f(x_2,y_2)-f(x_1,y_2)-f(x_2,y_1)\| \\ &\le \min{\{\|f(x_1,y_1)-f(x_1,y_2)\|+\|f(x_2,y_2)-f(x_2,y_1)\|,\\\|f(x_1,y_1)-f(x_2,y_1)\|+\|f(x_2,y_2)-f(x_1,y_2)\|\}} \\ &\le 2L \min{\{\|x_1-x_2\|,\|y_1-y_2\|\}}.\end{align}

This bound seems surprisingly weak. For example, if $$f$$ is actually linear, then the left hand side is always equal to $$0$$.

Suppose we call $$f$$ "Doubly Lipschitz" with modulus $$K$$ if it is Lipschitz and for all $$x_1,x_2\in P,y_1,y_2\in Q$$:

$$\|f(x_1,y_1)+f(x_2,y_2)-f(x_1,y_2)-f(x_2,y_1)\| \le K \|x_1-x_2\|\|y_1-y_2\|.$$

This seems to be a stronger requirement than just Lipschitz continuity, as for $$\|x_1-x_2\|\le 1,\|y_1-y_2\|\le 1$$, $$\|x_1-x_2\|\|y_1-y_2\|\le\min{\{\|x_1-x_2\|,\|y_1-y_2\|\}}$$, but the converse does not hold in general.

What else is known about "Doubly Lipschitz" functions? What are they usually called? Do they have a simple characterisation? Does being Doubly Lipschitz relate to membership of a certain Sobolev space?

Answered: Is there a simple example of a function that is Lipschitz but not Doubly Lipschitz?

For $$n\in\mathbb{N}$$, let $$W_n$$ be an inner product space, all defined over the same field ($$\mathbb{R}$$ or $$\mathbb{C}$$), and suppose $$Z$$ is this field.

Let $$a:X\rightarrow Z$$, $$b:Y\rightarrow Z$$, $$C_n:X\rightarrow W_n$$ and $$D_n:Y\rightarrow W_n$$ for $$n\in\mathbb{N}$$, all be Lipschitz continuous, with $$\sum_{n=0}^\infty{\langle C_n(x_0),D_n(y_0)\rangle}<\infty$$ for some $$x_0\in X$$ and $$y_0\in Y$$. Suppose the Lipschitz moduli of $$C_n$$ and $$D_n$$ are $$L_{C_n}$$ and $$L_{D_n}$$ respectively, with $$\sum_{n=0}^\infty{L_{C_n} L_{D_n}}\le L$$ for some $$L<\infty$$.

Suppose further that:

$$f(x,y)=a(x)+b(y)+\sum_{n=0}^\infty{\langle C_n(x),D_n(y)\rangle}.$$

We call this a "Lipschitz-Quadratic" function.

Then:

\begin{align}& \|f(x_1,y_1)+f(x_2,y_2)-f(x_1,y_2)-f(x_2,y_1)\| \\ & \le \sum_{n=0}^\infty{ |\langle C_n(x_1)-C_n(x_2),D_n(y_1)-D_n(y_2)\rangle| } \\ & \le \sum_{n=0}^\infty{ \|C_n(x_1)-C_n(x_2)\|\|D_n(y_1)-D_n(y_2)\| } \\ & \le L \|x_1-x_2\|\|y_1-y_2\|,\end{align}

(by the Cauchy Schwarz inequality), as required.

This class encompasses many discontinuous examples, e.g. $$f(x,y)=\|x\| \|y\|$$.

It also includes smooth examples such as $$f(x,y)=\exp{(xy)}$$.

Addenda 2: Functions that are uniformly Frechet differentiable in one argument, with the derivative being uniformly Lipschitz continuous in the other are Doubly Lipschitz.

Suppose $$f(x,y)$$ is Frechet differentiable in $$x$$ everywhere, uniformly over $$y$$. I.e. there exists a function $$G:P\times Q \rightarrow B(X,Z)$$ such that for all $$x\in X$$:

$$\lim_{x_1\rightarrow x, x_2\rightarrow x}\sup_{y\in Y}{\frac{\| f(x_1,y)-f(x_2,y)-G(x,y)(x_1-x_2)\|}{\|x_1-x_2\|}}=0.$$

Suppose further that $$G$$ is Lipschitz in $$y$$, uniformly over $$x$$. I.e., there exists a constant $$K$$ such that for all $$x\in X$$ and $$y_1,y_2\in Y$$:

$$\|G(x,y_1)-G(x,y_2)\|\le K\|y_1-y_2\|.$$

Given that $$\|f(x_1,y_1)+f(x_2,y_2)-f(x_1,y_2)-f(x_2,y_1)\|\le 2L \min{\{\|x_1-x_2\|,\|y_1-y_2\|\}}$$, it is sufficient to prove that for all $$x\in X$$ and $$y\in Y$$:

$$\lim_{x_1\rightarrow x, x_2\rightarrow x,\\y_1\rightarrow y, y_2\rightarrow y}{\frac{\|f(x_1,y_1)+f(x_2,y_2)-f(x_1,y_2)-f(x_2,y_1)\|}{\|x_1-x_2\|\|y_1-y_2\|}}<\infty.$$

Now, for $$x\in X$$ and $$y_1,y_2\in Y$$:

\begin{align}& \lim_{x_1\rightarrow x, x_2\rightarrow x}{\frac{\|f(x_1,y_1)+f(x_2,y_2)-f(x_1,y_2)-f(x_2,y_1)\|}{\|x_1-x_2\|}} \\ & = \lim_{x_1\rightarrow x, x_2\rightarrow x}{\left\|\frac{f(x_1,y_1)-f(x_2,y_1)-G(x,y_1)(x_1-x_2)}{\|x_1-x_2\|}\\-\frac{f(x_1,y_2)-f(x_2,y_2)-G(x,y_2)(x_1-x_2)}{\|x_1-x_2\|}\\+\frac{(G(x,y_1)-G(x,y_2))(x_1-x_2)}{\|x_1-x_2\|}\right\|} \\ & \le {\lim_{x_1\rightarrow x, x_2\rightarrow x}{\frac{\|f(x_1,y_1)-f(x_2,y_1)-G(x,y_1)(x_1-x_2)\|}{\|x_1-x_2\|}}\\+\lim_{x_1\rightarrow x, x_2\rightarrow x}{\frac{\|f(x_1,y_2)-f(x_2,y_2)-G(x,y_2)(x_1-x_2)\|}{\|x_1-x_2\|}}\\+\lim_{x_1\rightarrow x, x_2\rightarrow x}{\frac{\|(G(x,y_1)-G(x,y_2))\|\|x_1-x_2\|}{\|x_1-x_2\|}}} \\ & = \|G(x,y_1)-G(x,y_2)\| \end{align}

and the convergence here is uniform over $$y_1, y_2$$, so by the Moore-Osgood theorem:

\begin{align}& \lim_{x_1\rightarrow x, x_2\rightarrow x,\\y_1\rightarrow y, y_2\rightarrow y}{\frac{\|f(x_1,y_1)+f(x_2,y_2)-f(x_1,y_2)-f(x_2,y_1)\|}{\|x_1-x_2\|\|y_1-y_2\|}} \\ & = \lim_{y_1\rightarrow y, y_2\rightarrow y}\lim_{x_1\rightarrow x, x_2\rightarrow x}{\frac{\|f(x_1,y_1)+f(x_2,y_2)-f(x_1,y_2)-f(x_2,y_1)\|}{\|x_1-x_2\|\|y_1-y_2\|}} \\ & \le \lim_{y_1\rightarrow y, y_2\rightarrow y}{\frac{\|G(x,y_1)-G(x,y_2)\|}{\|y_1-y_2\|}} \\ & \le K.\end{align}

• I did not do the calculation, but I would expect that every $C^1$-function with a bounded derivative is “doubly Lipschitz”. Probably even already Gateaux differentiability with a bounded derivative is sufficient. Apr 23 '21 at 11:54
• @MartinVäth Yes, this seems plausible. I am more interested in these non-differentiable cases though. The question now contains a second addenda with a function that is non-differentiable but Doubly Lipschitz. Your answer is non-differentiable but not Doubly Lipschitz. Your answer suggests a criteria for dividing between the two cases, it would be great if this was the complete answer. I wonder if being Doubly Lipschitz relates to membership of a Sobolev space.
– cfp
Apr 23 '21 at 12:15
• The smoothness is certainly only sufficient: As I have written in the answer, my intuition is that for a counterexample the “nondifferentiable peak” must not be parallel along one of the axes. For instance $f(x,y)=g(x)+h(y)$ is “doubly Lipschitz” independent of $g$ and $h$ ($f,g,h$ need not even be Lipschitz!), because the non-smoothness is parallel along the axes, only. Apr 23 '21 at 12:34
• @MartinVäth The question now has a broader class of examples. It seems to be as broad as is possible. Can you think of a doubly lipschitz function with scalar Z that is not "Lipschitz-Quadratic"?
– cfp
Apr 27 '21 at 16:47
• @MartinVäth It would be great if you could also take a look at this related (but harder) question: math.stackexchange.com/questions/4118772/…
– cfp
Apr 27 '21 at 18:01

An example of a Lipschitz function for which the first bound is best possible is in case $$X=Y$$ the function $$f(x,y)=\lVert x-y\rVert$$.
Indeed, in case $$x_1=y_1=a$$, $$x_2=y_2=b$$, there holds $$\lVert f(x_1,y_1)+f(x_2,y_2)-f(x_1,y_2)-f(x_2,y_1)\rVert=2\lVert a-b\rVert=2\min\{\lVert x_1-x_2\rVert,\lVert y_1-y_2\rVert\}.$$
In particular, this rather non-exotic function already fails to be “doubly Lipschitz”. My intuition is that this is always the case whenever the function has a non-differentiable “peak” along a direction which is not “parallel” to one of the 2 coordinate axes of $$X\times Y$$.