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In the book "Markov Chains and Stochastic Stability" (page 171, http://probability.ca/MT/) of Meyn and Tweedie there is used the following condition of equicontinuity:

Assume that functions $f_n:\mathbb{R}^d\rightarrow\mathbb {R}$, $n\in\mathbb{N}$ are continuous and they have partial derivative due to each variable. If there exists constant M such that \left\| f_n'(x)\right\|\leq M, for $n\in\mathbb{N}$ and $x\in\mathbb{R}^d$, then the family $\{f_n: n\in\mathbb{N}\}$ is equicontinuous.

Do anybody know how to proof that or where can I find the proof?

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    I guess that the functions from that family should occur to be lipschitz continuous with the same constant? Am I right?2011-08-31

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This is really just the mean value theorem.

One way to see it is this. Suppose $f : \mathbb{R}^d \to \mathbb{R}$ has $||\nabla f|| \le M$ everywhere (I'm writing $\nabla f$ instead of your f' to remind us that it is a vector). Pick any $x,y \in \mathbb{R}^d$ and set $\gamma : [0,1] \to \mathbb{R}^d$ to be a straight-line path between them: $\gamma(t) = (1-t)x + ty$. Note \gamma'(t) = y-x.

Now write $g(t) = f(\gamma(t))$, so that $f(y) - f(x) = g(1) - g(0)$. By the chain rule and the Cauchy-Schwarz inequality, |g'(t)| = |\nabla f(\gamma(t)) \cdot \gamma'(t)| \le ||\nabla f(\gamma(t))||\cdot ||\gamma'(t)|| \le M ||y-x||. So by the (one-variable) mean value theorem, $|g(1) - g(0)| \le |1-0|M ||y-x||$, and we have that $f$ is Lipschitz with constant at most $M$.

Of course, it should be easy to see that any collection of functions which is Lipschitz with the same constant must be equicontinuous.