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Determine if the following function satisfies a local or a a uniform Lipschitz condition. The definition of locally Lipschitz and globally lipschitz are as follows:

(i) We say that f is (uniformly) Lipschitz with respect to y, or simply in y, on $A\subset U$ if there is a constant L such that $||f(t,x)-f(t,y)||\leq L ||x-y||$, whenever $(t,x)$ and $(t,y)$ are in A.

(ii) f is locally Lipschitz in y if for every $(t_0, y_0) \in (c,d) \times U$, there exists a neighborhood V of $(t_0, y_0)$, (i.e. $V = \{f(t,y) \in(c,d) \times U : ||t-t_0|<a \ \text{and} \ |y-y_0|\leq b\}$) and a constant K = K(V) such that $||f(t,x)-f(t,y)||\leq K||x-y||$ for any $(t,x),(t,y) \in V$

1) $t^2|y|$.

So for this problem, I did the following: $|f(t,x)-f(t,y)|=|(t^2|x|-t^2|y|)|=t^2||x|-|y||\leq t^2|x-y|$. This shows from the second definition that this function is locally lipschitz because as we vary t, our bound changes (the $t^2$ is the K from the definition). Is this correct? How would I go about showing that this is\is not uniformly lipschitz?

2) $\frac{y}{1+y^2+t^2}$

For this one, I tried to follow a similar method.

$|\frac{x}{1+x^2+t^2}-\frac{y}{1+y^2+t^2}|=|\frac{t^2(x-y)-x^2y+x(y^2+1)-y}{(1+x^2+t^2)(1+y^2+t^2)}| \leq |t^2(x-y)-x^2y+x(y^2+1)-y|$.

Then I got stuck.

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In the first case, you need to specify the set $A$. In general 1) is not uniformly Lipschitz.

Suppose $A$ is such that there exists an unbounded sequence $t_n$, and open sets $U_n$ such that $\{t_n\} \times U_n \subset A$. Then the function $f(t,x) = t^2|x|$ is not uniformly Lipschitz. To see this, choose $L>0$ and choose $t_n$ such that $t_n^2>L$. Choose $x,y \in U_n$ such that $x,y$ are different and have the same sign ($U_n$ is open, so this can be done). Then $|f(t_n,x)-f(t_n,y)| = t_n^2||x|-|y|| = t_n^2 |x-y|> L |x-y|$.

Case 2) is more straightforward. If $f(t,x) = \frac{x}{1+x^2+t^2}$, then $\frac{\partial f(t,x)}{\partial x} = \frac{1-t^2-x^2}{(1+t^2+x^2)^2}$, from which we easily see that $|\frac{\partial f(t,x)}{\partial x}| \leq 1$ for any $x,t$. Then a quick application of the mean value theorem gives $|f(t,x)-f(t,y)| \leq |x-y|$ for any $t,x,y$. Hence this is uniformly Lipschitz.

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  • $\begingroup$ Thanks a lot. Your approach for case 2 really helped a lot. $\endgroup$ – user52932 Feb 24 '13 at 23:26
  • $\begingroup$ You are welcome. $\endgroup$ – copper.hat Feb 24 '13 at 23:27

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