# Does locally Lipschitz in one variable imply uniformly Lipschitz in another?

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 $t \in D=[t_0,t_1]\subset \mathbb{R}$. Here $D$ is a closed finite interval. Does this imply that $f$ is locally Lipschitz in $x$ uniformly in $t \in D$?

I understand uniformly here means that for all $t \in D$ the Lipschitz constant is independent of $t$. Intuitively I don't think the implication holds. However, I have not been able to come up with a counter example.

• I am either looking for proof that the implication holds, or for a valid counter example. – ITA Feb 11 '16 at 5:44
• So what you are asking is: In considering $\frac{|f(t,x_1)-f(t,x_2)|}{|x_1-x_2|}$, for each pair $(x_1,x_2)$ this is a continuous function. This family is point-wise bounded for every fixed $t$. Is then the supremum over this family continuous? Which would imply boundedness. – LutzL Feb 11 '16 at 8:04
• @LutzL Yes in a manner of speaking I suppose. For every fixed $t$ there exists some local Lipschitz constant $L$ (local to $x_0$, i.e $x_1, x_2$ are in some neighborhood of $x_0$). Does mere continuity in $t$, imply there is some maximal Lipschitz constant that works for all $t \in [t_0, t_1]$? – ITA Feb 11 '16 at 13:17

$$f:\left[ {0,1} \right] \times \mathbb{R} \to \mathbb{R}{\text{ such that }}f\left( {t,x} \right) = \left\{ {\begin{array}{*{20}{c}} {t\sin \left( {x/{t^2}} \right)}&{t \ne 0} \\ 0&{t = 0} \end{array}} \right.$$
This function is continuous in $t$ and even at zero since the limit as $t \to 0$ is indeed zero. This function is also continuously differentiable as the partial derivative with respect to $x$ given by $\cos \left(x/t\right) /t$ is also continuous except for exhibiting a pathology at $t=0$ (though it remains continuous as it grows unbounded).
Then by the Lipschitz condition, $$\left| {f\left( {t,x} \right) - f\left( {t,y} \right)} \right| = \left| {t\sin \left( {x/{t^2}} \right) - t\sin \left( {y/{t^2}} \right)} \right| \leqslant \left| t \right|\left| {\sin \left( {x/{t^2}} \right) - \sin \left( {y/{t^2}} \right)} \right|$$ The last term is bounded above by 2. Hence, $$\left| {f\left( {t,x} \right) - f\left( {t,y} \right)} \right| \leq 2\left| t \right| \Rightarrow \left| {x - y} \right|\frac{{\left| {f\left( {t,x} \right) - f\left( {t,y} \right)} \right|}}{{\left| {x - y} \right|}} \leqslant \frac{{2\left| t \right|}}{{\left| {x - y} \right|}}\left| {x - y} \right|$$ Then, $$\left| {f\left( {t,x} \right) - f\left( {t,y} \right)} \right| \leqslant L\left| {x - y} \right|{\text{ where }}L = \frac{{2\left| t \right|}}{{\left| {x - y} \right|}}$$ Clearly the value of $L$ depends on $t$ and also will grow arbitrarily large to satisfy the local Lipschitz continuity as $x \to y$. Thus there is no single value of $L$ that will serve as the Lipschitz constant for the entirety of the chosen domain.