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?

Addenda 1: "Lipschitz-Quadratic" functions are 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.


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

(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:

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

  • 1
    $\begingroup$ 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. $\endgroup$ Apr 23 '21 at 11:54
  • $\begingroup$ @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. $\endgroup$
    – cfp
    Apr 23 '21 at 12:15
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 23 '21 at 12:34
  • $\begingroup$ @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"? $\endgroup$
    – cfp
    Apr 27 '21 at 16:47
  • $\begingroup$ @MartinVäth It would be great if you could also take a look at this related (but harder) question: math.stackexchange.com/questions/4118772/… $\endgroup$
    – 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 $X\times Y$.


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