I understand the definition of Lipschitz functions when talking of functions of single variables. However, I have trouble understanding it when it is a multivariable function.

Suppose $ f(t,x):D \times \mathbb{R} \to \mathbb{R}$ is continous in $t$ and locally Lipschitz in $x$ for each fixed $ t \in D $ where $D= [t_0,t_1] \subset \mathbb{R}$. Here $D$ is a closed finite interval. In other words, for every fixed $t$ there exists some local Lipschitz constant $L$ such that $|f(t,x_1) -f(t,x_2)| \leq L|x_1 - x_2|$. Local is understood to mean that for some $x_0 \in \mathbb{R}$ we have $ x_1,x_2 $ belong to a neighborhood of $x_0$.

Does this imply that $f$ is locally Lipschitz in $x$ uniformly in $t \in D$?

Here uniformly means that for all $t \in D$ the Lipschitz constant is independent of $t$ i.e given $x_0$ and some neighborhood of $x_0$ there exists some maximal Lipschitz constant $L_*$ that works for all $t \in D$.

Intuitively I don't think the implication holds. However, I have not been able to come up with a counter example with D defined to be a closed finite interval. With D defined as an open set, various counter examples exist.

Also, if you impose conditions that $D$ is compact and that $f(t,x)$ can be written as $f(t,x)=g(t)h(x)$ or $f(t,x)=g(t)+h(x)$ it can be shown than $g(t)$ achieves its maximum in $D$ and that $L$ defined as a function of this maximum works for all $t$.

However I am looking for proof that the implication holds in the general case (D is closed and finite, f(t,x) has no special form) OR for a valid general counter example.


migrated from mathoverflow.net Feb 12 '16 at 5:54

This question came from our site for professional mathematicians.


Taking $D = [0,1]$, I think $f(t,x) = t \sin(x/t^2)$ is a counterexample.

(Here of course $f(0,x)=0$.)

  • $\begingroup$ I am not so sure. The function is bounded above by 2 in $D$. For example, $$\left| {f\left( {t,x} \right) - f\left( {t,y} \right)} \right| = \left| {t\sin \left( {x/{t^2}} \right) - t\sin \left( {y/{t^2}} \right)} \right|$$ $$\left| {f\left( {t,x} \right) - f\left( {t,y} \right)} \right| \leq \left| t \right|\left| {\sin \left( {x/{t^2}} \right) - \sin \left( {y/{t^2}} \right)} \right|$$ $$\left| {f\left( {t,x} \right) - f\left( {t,y} \right)} \right| \leq 2\left| t \right| \leq 2\left| t \right|\left| {x - y} \right|$$ Then we can chose $L=2|t|$ which in $D$ is 2. $\endgroup$ – ITA Feb 11 '16 at 22:24
  • $\begingroup$ @IvanAbraham: Your last inequality $2|t| \le 2 |t| |x-y|$ isn't right - this inequality is supposed to hold for any $x$ and $y$, even when $|x-y|$ is very small. In fact, since $f(t, \cdot)$ is $C^1$ its best Lipschitz constant on a neighborhood is the supremum of its derivative (with respect to $x$) on that neighborhood - and you can directly compute that this goes to infinity as $t \to 0$. $\endgroup$ – Nate Eldredge Feb 11 '16 at 23:01
  • $\begingroup$ @IvanAbraham: Explicitly, take $t = 0.01$, $x=0$ and $y = 0.0001 \pi/2 $. Then $|f(t,x)-f(t,y)| = 0.01$ while $2 |t| |x-y| \approx 0.0000031415$. $\endgroup$ – Nate Eldredge Feb 11 '16 at 23:16
  • $\begingroup$ Yes you are right, the inequality would have needed to be multiplied by $|x-y|$ on both sides and not just on the right to hold. My bad. I am accepting your answer, because after looking at some plots I get the intuition. As $x,y$ get arbitrarily close to each other $f(.,.)$ doesn't. Is there any way to formally show that $L=L(t)$ along the lines I was proceeding? Or does one have to do an $\epsilon - \delta$ type of construction? $\endgroup$ – ITA Feb 12 '16 at 0:09
  • $\begingroup$ I think this works: $$\left| {f\left( {t,x} \right) - f\left( {t,y} \right)} \right| = \left| {t\sin \left( {x/{t^2}} \right) - t\sin \left( {y/{t^2}} \right)} \right| $$ $$ \leq \left| t \right|\left| {\sin \left( {x/{t^2}} \right) - \sin \left( {y/{t^2}} \right)} \right| $$ $$\left| {f\left( {t,x} \right) - f\left( {t,y} \right)} \right| \leq 2\left| t \right| \Rightarrow \left| {x - y} \right|\frac{{\left| {f\left( {t,x} \right) - f\left( {t,y} \right)} \right|}}{{\left| {x - y} \right|} } $$ $$ \leq \frac{{2\left| t \right|}}{{\left| {x - y} \right|}}\left| {x - y} \right| = L|x-y|$$ $\endgroup$ – ITA Feb 12 '16 at 0:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.