# $f: U \rightarrow \mathbb{R}^{m}$ differentiable. if $f’$ is continuous, show that $f$ is locally Lipschitz

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

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 $$x$$, $$y \in K$$.
• 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