$\mathcal{C}^1$ implies locally Lipschitz in $\mathbb{R}^n$ [duplicate]

This question already has an answer here:

According to wikipedia a function $f\colon \mathbb{R}^n\to\mathbb{R}^n$ that is continuously, is also locally Lipschitz.

I there someone who knows a good reference which contains a proof of this statement?

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marked as duplicate by Najib Idrissi, Harish Chandra Rajpoot, USER91500, Jendrik Stelzner, Claude LeiboviciJan 12 at 11:36

Beware: Wikipedia reads "continuously differentiable", not "continuous differentiable in all directions" (which could be interpreted so as to make the statement false). – D. Thomine May 10 '12 at 20:18
Now it just says "continuously" which makes even less sense. Please clarify. – Nate Eldredge May 10 '12 at 21:22

The proof on $\mathbb{R}^n$ is fairly straightforward.

Choose some ball $B(\hat{x},\epsilon)$. The closure is compact, so the derivative $\frac{\partial f}{\partial x}$ is bounded by some $L$ on the ball. Now suppose $x,y \in B(\hat{x},\epsilon)$, then using Taylor's formula, we have: $$f(x)-f(y) = \int_o^1 \frac{\partial f (y+t(x-y))}{\partial x}(x-y)\;dt.$$ Hence we can get the bound: $$\|f(x)-f(y) \| \leq \int_o^1 \|\frac{\partial f (y+t(x-y))}{\partial x}\|\; \|(x-y)\| \;dt \leq L \|x-y\|.$$

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The exact quote on wiki was:

In particular, any $C^1$ function is locally Lipschitz, as continuous functions on a locally compact space are locally bounded so its gradient is.

The logic here is we would like to show the gradient of a $C^1$-function is locally bounded on a locally compact space, thus to obtain the Lipschitz continuity. Hence your question boils down to how to prove:

A continuous function on a locally compact space is locally bounded.

While I am not one hundred percent sure which real analysis textbook has the exact result, you may refer to this Wikipedia entry, we could use Stone-Weierstrass theorem to prove: roughly speaking, since $\mathbb{R}^n$ is locally compact, for any point $x\in \mathbb{R}^n$, choose a compact neighborhood $S$ of $x$, we could use S-W theorem to prove that, for a continuous function on $S$, there exists a sequence of vector-valued polynomials $\{p_n\}$ that converge uniformly to the function of interest $f$ on $S$, and use the triangle inequality in the supreme norm, we could see $f$ is bounded on $S$. Thus the local boundedness of $f$ is proved.

To sum up: Gradient of $C^1$-function is continuous $\to$ A continuous function on a locally compact space(notice the gradient of $f$ you gave is the Jacobian matrix but the argument still applies) is locally bounded $\to$ Gradient of $C^1$-function is locally bounded $\to$ By mean value theorem $C^1$-function is locally Lipschitz.

EDIT: As Neal suggested, it suffices to argue as follows: For any $x_0\in \mathbb{R}^n$, on a compact set within some neighborhood $S \subset B(x_0,\epsilon)$, for $x,y\in S$: $$\frac{\|f(x)-f(y)\|}{\|x-y\|} \leq \sup_{\xi\in S} \| Df\|<L(x_0)$$ again we rely on continuous functions are bounded on compact set, and the Lipschitz constant depends on $x_0$, this is the local Lipschitz property you need.

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You might add that being locally Lipschitz is just a slight weakening of differentiability: all you need is $|\frac{f(x)-f(y)}{x-y}|<C$ for all $x,y$ in small neighborhoods of each other. – Neal May 10 '12 at 23:43
@Neal Thanks for the heads up, edited. – Shuhao Cao May 11 '12 at 0:29