# Confused about Lipschitz functions

Given a function $f(x,y)$, I want to show whether the function satisfies a uniform Lipschitz condition with respect to $y$, and determine the Lipschitz constant $L$.

Questions:

1.) I know that $f(x,y)$ is Lipschitz with respect to $y$ if $$|f(x,y_1) - f(x,y_2)|\le L|(x,y_1)-(x,y_2)|$$ But what does uniform Lipschitz condition mean?

2.) What is a good strategy to determine whether a function $f(x,y)$ is uniformly Lipschitz with respect to $y$? (Maybe somebody can explain it by using a few examples)....

Thank you!

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"Uniform with respect to y" means only that it is Lipschitz with the same constant L for any x –  bonext Dec 7 '12 at 14:13
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## 1 Answer

If $f$ is continuous and $\frac{\partial f}{\partial y}$ exists and is uniformly bounded on a convex region $\Omega$, then $f$ is uniformly Lipschitz on $\Omega$. To see this, apply the mean value theorem at points $(x,y_1), (x,y_2)\in\Omega$, with $y_1<y_2$. The mvt tells us that there exists $y_3\in(y_1,y_2)$ such that $$\frac{\partial f}{\partial y}(x,y_3)=\frac{f(x,y_2)-f(x,y_1)}{y_2-y_1}.$$ If the partial derivative is uniformly bounded on $\Omega$ by $M$, we can then say that $$\frac{|f(x,y_2)-f(x,y_1)|}{|y_2-y_1|}\leq M\quad\hbox{for all}\quad (x,y_1), (x,y_2)\in\Omega.$$

Consider your example $$f(x,y)=x\sin y,$$ and lets look at this on the rectangle $$\Omega=\{(x,y): -a\leq x\leq a, -b\leq y\leq b\}$$ where $a$ and $b$ are positive. We have $$\frac{\partial f}{\partial y} = x\cos y$$ and so $$\left|\frac{\partial f}{\partial y}\right| = |x\cos y|\leq |x| \leq a \quad \hbox{for all}\quad (x,y)\in\Omega.$$ Thus $\frac{\partial f}{\partial y}$ is uniformly bounded by $M=a$ on $\Omega$.

(Definition: The function $g$ is uniformly bounded on the set $A$ if there is a number $K$ such that $|g(p)|\leq K$ for all points $p\in A$.)

A set $\Omega$ is convex if for every pair of points $p,q$ in the set, the line joining $p$ and $q$ is also contained in the set. In $\mathbb{R}^2$, examples include rectangles, parallelograms, discs, ellipses and $\mathbb{R}^2$ itself - but not stars or a disc with a bite taken out of it.

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Tnx. So for example $f(x, y) = x sin(y)$, which is continuous, df/dy exists. But how can I say if it's uniformly bounded on a convex region? What does that mean exactly? –  MSKfdaswplwq Dec 7 '12 at 15:05
Good questions - I've edited my answer to try to address them. –  user12477 Dec 7 '12 at 15:23
Is the following set convex? $$\text{The rectangle of }[x_0,x_0+a], (-\infty, +\infty)$$ –  MSKfdaswplwq Dec 7 '12 at 21:05
@Hempo I wouldn't trust that book! $\frac{y}{(1+x)^2}$ is undefined on the line $x=-1$, and so its domain cannot be $\mathbb{R}^2$. The rectangle you mention is indeed convex: if you think about what it looks like - it's an infinite vertical strip between the vertical lines $x=x_0$ and $x=x_0+a$ - it should be clear that any pair of points in the set are connected by a segment that stays inside the set. It is also fairly straightforward to prove this formally. –  user12477 Dec 8 '12 at 12:22
To find \underline{a} domain on which a function $f(x,y)$ satisfies the Lipschitz condition is to choose a number $M>0$ and then find the largest convex domain on which the inequality $|\frac{\partial f}{\partial y}|<M$ is satisfied. If you are interested in the largest domain on which the function is Lipschitz, then ask: what is the largest value that $M$ may have? This just describes one possible and non-exhaustive approach: there isn't a general prescription for doing what you ask. –  user12477 Dec 8 '12 at 12:27
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