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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 x2012-12-07

1 Answers 1

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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. 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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    uh $y/(1+x^2)$2012-12-09