Consider $g$ continuous on $[a,b]\times \mathbb{R}^n$, $[a,b]\subset \mathbb{R}$, and $x\mapsto g(t,x)$ continuously differentiable for all $t\in [a,b]$.
I wonder if $g$ is Lipschtz wrt to $x$ that is for every compact $K\subset [a,b]\times \mathbb{R}^n$, there exists a constant $L_K\ge 0$ such that for all $(t,x)\in K$, $(t,y)\in K$, we have $$ |g(t,x) - g(t,y)|\le L_K|x-y| $$
I wrote $$ g(t,x) - g(t,y) = (x-y)\cdot \int_0^1 \frac{\partial g}{\partial x}(t,y+s(x-y))d s $$ and then using Cauchy-Schwarz, $$ |g(t,x) - g(t,y)|\le |x-y| |\int_0^1 \frac{\partial g}{\partial x}(t,y+s(x-y))d s| $$ and from the continuity of $x\to \partial_x g$, there exists $C(t)$ such that : $$ |g(t,x) - g(t,y)|\le C(t)|x-y| $$
However it seems that nothing ensures that $C$ is bounded for $t$ such that $(t,x),(t,y)\in K$ or am I missing something ?
EDIT : I am looking for an counter-example.
How about $g(t,x) = tx^3\sin(1/(tx))$ ?
we have for $0<y<x$: $$ C(t)\ge\frac{tx^3\sin(1/(tx)) - ty^3\sin(1/(ty))}{x-y} \ge \frac{x^2-y^2-1/(6t^2)}{x-y} = x+y-\frac{1}{6t^2(x-y)} $$ Clearly $C(t)$ is not bounded.
However this doesn't show that my question is wrong.
Take $g:[-1,1]\times \mathbb{R}\to \mathbb{R}$ defined by $$ g(t,x)=\sqrt{t} x^2\sin(\frac{1}{tx}) $$
Clearly $g$ is continuous on $[-1,1]\times \mathbb{R}$ and $x\mapsto g(t,x)$ is continously differentiable ($\partial_x g(t,x) = 3 \sqrt{t} x^2 \sin \left(\frac{1}{t x}\right)-\frac{x \cos \left(\frac{1}{t x}\right)}{\sqrt{t}}$).
Take $K=[-1,1]\times [1,2]$ and $(t,x), (t,y)\in K$. By the mean value theorem that there exists $c\in [1,2]$ such that $g(t,x)−g(t,y)=\partial_x g(t,c)=3 \sqrt{t} c^2 \sin \left(\frac{1}{t c}\right)-\frac{c \cos \left(\frac{1}{t c}\right)}{\sqrt{t}}$. Clearly this diverges for $t>0$ small enough and any value of $c$ in $[1,2]$. Hence the statement cannot be true.