Let $f,g: V \to \mathbb{R}^{n}$ differentiable in open $V \subset \mathbb{R}^{n}$, $\delta$ a real positive number and $X \subset V$. Write $``|f-g|_{1} \leq \delta\; \text{in} \;X"$ to mean that $|f(x) - g(x)| \leq \delta$ and $|f'(x) - g'(x)| \leq \delta$ for all $x \in X$. Prove that if $\varphi: U \to V \in C^{1}$ in the open $U \subset \mathbb{R}^{k}$, then, given $K \subset U$ compact and $\eta >0$, there is $\delta >0$ such that $|f-g|_{1} < \delta$ in $\varphi(K)$ implies $|f\circ \varphi - g\circ \varphi|_{1}<\eta$ in $K$.
$``|\cdot|"$ can be any norm in $\mathbb{R}^{n}$.
I don't know how to relate $\varphi \in C^{1}$ to $f,g$ and the others hypothesis. I appreciate any hint!
Given $K\subset U$ and $\eta>0$, define $M:= \max\{1,\max\{ \|\varphi'(x)\|,x \in K \} \}$ and $\delta := \frac{\eta}{M}$.
Then, if $|f - g|_1 < \delta$ in $\varphi(K)$, we conclude
$$|f\circ \varphi(x) - g\circ\varphi(x)| < \delta = \frac{\eta}{M}\leq \eta,$$
and
\begin{align*} |(f\circ\varphi)'(x) -(g\circ\varphi)'(x) |&=|f'(\varphi(x)) \cdot \varphi'(x) - g'(\varphi(x)) \cdot \varphi'(x) | \\ &\leq |f'(\varphi(x)) - g'(\varphi(x)) | \cdot |\varphi'(x)|\\ &\leq \delta\cdot M \\ &\leq \frac{\eta}{M}\cdot M\\ &= \eta \end{align*}
Then $|f\circ \varphi - g\circ\varphi|_1 < \eta$ in $K$.