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Let $f: U \rightarrow \mathbb{R}^{m}$ differentiable in the open $U \subset \mathbb{R}^{m}$. If $f': U \rightarrow \mathcal{L}(\mathbb{R}^{m}, \mathbb{R}^{n})$ is continuous and $K \subset U$ a compact show that there is $a>0$ such that if $x,y \in K$ then $\|f(x)-f(y)\| \leq a\|x-y\|$.

We have to:

Let $X$ and $Y$ be metric spaces. If $f: X \rightarrow Y$ is continuous and $K \subset X$ is compact then $f (K) \subset Y$ is compact.

Generalized Weierstrass theorem. Let $V$ be a normed vector space, $X$ a topological space, $ K \subset X$ a compact set. If $f: X \rightarrow V$ is continuous, then there exist $x_1$, $x_2$ such that $||f(x_1)|| \leq ||f(x)|| \leq ||f(x_2)||$ for any $x \in K$.

How to use the hypothesis about $f '$ and get to what you want to get? Any help please

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Recall that for an open $W \subset \mathbb{R}^n$ and $h: W \to \mathbb{R}^m$ differentiable with continuous derivable, then for all $x$, $y \in W$ such that $\{ty+(1-t)x \, | \, t \in [0,1] \} \subset W$ one has $$ h(y)-h(x)=\int_0^1 h^\prime_{ty+(1-t)x}(y-x) dt .$$ Indeed, define $g(t):=h(ty+(1-t)x)$ for $t \in [0,1]$. Then $g$ is $C^1$ and $g(1)-g(0)=\int_0^1 g^\prime(t)dt$ but $g^\prime(t)=h^\prime_{ty+(1-t)x}(y-x)$.

Returning to your question, take $K \subset V$ such that $V$ is open in $U$ and $\bar{V} \subset U$. There exists a smooth function $g:\mathbb{R}^n \to \mathbb{R}^m$ such that $g=1$ on $K$ and $g=0$ outside $V$. Define $h :\mathbb{R}^n \to \mathbb{R}^m$ to be $gf$ on $U$ and $0$ outside $U$. $h$ is differentiable and has a continuous derivative. Note that $h=f$ on $K$. Take $B(0,r)$ such that $K \subset B(0,r)$. For $x$, $y \in B(0,r)$, $h(y)-h(x)=\int_0^1 h^\prime_{ty+(1-t)x}(y-x) dt$, so $$ |h(y)-h(x)| \leqslant \int_0^1 |h^\prime_{ty+(1-t)x}(y-x)| dt \leqslant |y-x| \cdot \int_0^1 ||| h^\prime_{ty+(1-t)x} ||| dt $$ but $||| h^\prime_{z} |||$ is bounded by a real $a$ for all $z \in B(0,r)$ because $h^\prime$ is continuous on the compact set $\bar{B}(0,r)$, whence $$ |h(y)-h(x)| \leqslant a|y-x| . $$ Since $f=h$ on $K$ and $K \subset B(0,r)$ we get $$ |f(y)-f(x)| \leqslant a|y-x| $$ for all $x$, $y \in K$.

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  • $\begingroup$ I hope it is not too complicated. But actually the integral and the function $g$ (bump function) are common things and very useful, so you don't waste time learning about them $\endgroup$
    – user598294
    Commented Sep 16, 2020 at 0:59

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