Problem with Lipschitz function Given $a,b$ $\in$ $\mathbb{R}$ ($a \lt b$) and functions $y,z \in C\big([a,b],\mathbb{R}\big)$. Prove if $f \in C^{1}\big(\mathbb{R}^{2},\mathbb{R}\big)$, then $\exists K \gt 0$ as:
$$
|f(s,y(s)) - f(s,z(s))| \le K|y(s) - z(s)| \quad \forall s \in [a,b]
$$
 A: Note that $c(s)=\min\{y(s), z(s)\}$ and $d(s)=\max\{ y(s), z(s)\}$ are continuous  functions of $s$. Fix $I_s=[c(s),d(s)]$ and $I=\bigcup_{s\in [a,b]}I_s$. If $c=\inf_{s\in [a,b]}c(s)$ and $d=\sup_{s\in [a,b]}d(s)$ then $I\subset [c,d]$.
It suffices that you prove that the function $\mathbb{R}^2\ni (s,y)\mapsto f(s,x)\in \mathbb{R}$ is uniformly lipschitz in the second variable $x$ in the set
$
[a,b]\times I
$.
Here the constant $K$ will depend only on the functions $y(s)$ and $z(s)$ but not $s\in[a,b]$.
That is, there is a constant $K>0$ that does not depend on $s\in[a,b]$ such that
$$
|f(s,u) - f(s,v)| \le K|u - v|.
$$
Recall that a function $(s,x)\mapsto f(s,x) $ is a uniformly lipschitz in the second variable if, only if, 
$$
K=\sup_{s}\sup_{u,v} \frac{|f(s,u)-f(s,v)|}{|u-v|}<\infty.
$$
By mean value theorem there is $\tau_s\in [0,1]$, whit  $w_s= \tau_sc_s+(1-\tau_s)d_s\in [c_s,d_s]$  and $s=\tau_ss+(1-\tau_s)s$, such that 
\begin{align}
\sup_{u,v\in [c_s,b_s]} \frac{|f(s,u)-f(s,v)|}{|y(s)-z(s)|}
=&
\sup_{u,v\in [c_s,b_s]} \frac{\big| D_1 f(s,w_s)\cdot [s-s)]+ D_2 f(s,w_s)\cdot [d_s-c_s]\,\big|}{|y(s)-z(s)|}
\\
=&
\sup_{u,v\in [c_s,b_s]} \frac{\big|D_2 f(s,w_s)\cdot [d_s-c_s]\,\big|}{|y(s)-z(s)|}
\\
=&
\sup_{u,v\in [c_s,b_s]} \big|D_2 f(s,w_s)\big|
\\
\leq &
\sup_{u,v\in [c_s,b_s]} \sup_{w\in [c_s,d_s]}\big|D_2 f(s,w)\big|
\\
= &
\sup_{w\in [c_s,d_s]}\big|D_2 f(s,w)\big|
\\
\leq &
\sup_{w\in [c,d]}\big|D_2 f(s,w)\big| 
\end{align}
for $[c,d]=\bigcup_{s\in [a,b]}[a_s,b_s]$.
Since $[a,b]\ni s \mapsto \sup_{w\in [c,d]}\big|D_2 f(s,w)\big| $ is a continuous function then
$$
\sup_{s\in [a,b]}\sup_{w\in [c,d]}\big|D_2 f(s,w)\big|<\infty.
$$
