Finding a common Lipschitz constant for a family of contractions Let $(X,d)$ be a complete metric space and $f:[0,1] \times X \to X$ a family of functions such that $f(t,\cdot)$ is a contraction for every $t \in [0,1]$. Further assume that $f(\cdot,x)$ is continuous for every fixed $x \in X$.
Can we conclude that there exists a common Lipschitz constant $C$ which is less than $1$ in the sense that

$$
d(f(t,x),f(t,y)) \leq Cd(x,y) \ \ \forall t \in [0,1] \ \ \forall x,y \in X \ \ ? 
$$

I think an equivalent statement would be to ask if the Lipschitz constant of the above family of functions depends continuously on $t$ because then we could map the compact interval $[0,1]$ to another compact interval under this continuous map which would yield the result.
 A: Take $X=\mathbb{R}$ with the usual metric. Let $h:\mathbb{R} \rightarrow \mathbb{R}$ be a compactly supported Lipschitz function with Lipschitz constant 1. Then
$$ f(t, x) = \begin{cases} t \cdot h\big( x - \frac{1}{1-t} \big), &t\neq 1 \\
0, &t=1
\end{cases}
$$
is a counterexample. Indeed, we have (as the Lipschitz constant does not care about translating the argument)
$$ Lip(f(t, \cdot)) = \begin{cases}
Lip(t\cdot h(\cdot - \frac{1}{1-t})), &t\neq 1,\\
0, &t=1
\end{cases} 
= \begin{cases}
t \cdot Lip(h), &t\neq 1, \\
0, & t=1
\end{cases}
= \begin{cases}
t, & t\neq 1, \\
0, & t=1.
\end{cases}$$
Hence, $Lip(f(t, \cdot))<1$, but $\sup_{t\in [0,1]} Lip(f(t,\cdot))=1$.
If you want a counterexample on a compact metric space, you might take $X=[0,1]$ and
$$ g: [0,1]\times [0,1] \rightarrow \mathbb{R}, \ g(t,x) = \begin{cases} t \cdot \sin\big( \frac{1}{t+t^2} x\big), 
&t\neq 0, \\ 0, &t=0.
\end{cases}$$
Then we get the Lipschitz constant by taking the derivative and see that this yields a counterexample. Namely, we have
$$\partial_x g(t,x)=
\begin{cases}
\frac{t}{t+t^2} \cos \big( \frac{1}{t+t^2} x\big),&t\neq 0, \\
0,& t=0
\end{cases}$$
Thus,
$$ Lip(g(t, \cdot)= \sup_{x\in [0,1]} \partial_x g(t, \cdot)
= \begin{cases} \frac{t}{t+t^2}, & t\neq 0,\\
0, & t=0.
\end{cases} $$
Hence, $Lip(g(t,\cdot))<1$ for all $t\in [0,1]$, but
$$ sup_{t\in [0,1]} Lip(g(t,\cdot)) = \sup_{t\in (0,1]} \frac{t}{t+t^2} =1. $$
