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Let $A$ be open in $\mathbb{R}^m$; let $g:A\rightarrow\mathbb{R}^n$. If $S\subseteq A$, we say that $S$ satisfies the Lipschitz condition on $S$ if the function $\lambda(x,y)=|g(x)-g(y)|/|x-y|$ is bounded for $x\neq y\in S$. We say that $g$ is locally Lipschitz if each point of $A$ has a neighborhood on which $g$ satisfies the Lipschitz condition.

Show that if $g$ is of class $C^1$, then $g$ is locally Lipschitz.

If $g$ is of class $C^1$, that means the component derivatives of $g$ are continuous. The $i$th-component derivative ($i=1,2,\ldots,m$) at a point $r\in A$ is $D_ig(r)=\lim_{t\rightarrow 0}\dfrac{g(r+te_i)-g(r)}{t}$. So this function is continuous over all points $r\in A$. How can I deduce local Lipschitz from here?

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Mika H.
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    Hint: $C^1$ means the derivatives are locally bounded. Apply the mean value theorem. – Anthony Carapetis Oct 18 '13 at 05:58
  • @AnthonyCarapetis How can you conclude that the derivatives are locally bounded from the definition as I mentioned in the post? – Mika H. Oct 19 '13 at 18:19
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    The derivatives are continuous; so if you restrict to a closed ball around any point they are bounded. Thus the interior of this ball is a neighbourhood of the point on which the derivatives are bounded. – Anthony Carapetis Oct 20 '13 at 00:56
  • @AnthonyCarapetis I could use that $g(a+h)-g(a)=\sum_{j=1}^{m}D_jg(q_j)h_j$ where $a$ is a point of $A$, $h$ is a point of $\mathbb{R}^m$ with $0 < |h| < \delta$, each point $q_i$ lies in the cube $C$ of radius $|h|$ centered at $a$, $\delta$ arises from the continuity of $Dg$ and so $|g(a+h)-g(a)|=|\sum_{j=1}^{m}D_jg(q_j)h_j|\leq \sum_{j=1}^{m}|D_jg(q_j)||h_j|\leq \sum_{j=1}^{m}|Dg(q)||h_j|\leq n|Dg(x_0)+1||h_j|$ For the answer in https://math.stackexchange.com/questions/200637/locally-continuously-differentiable-implies-locally-lipschitz? – user402543 Apr 24 '18 at 04:20

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