Does continuity in one variable and locally Lipschitz in another imply uniformity in the first? 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. 

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