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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.

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

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

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

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  • $\begingroup$ 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

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