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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 locally Lipschitz, then $g$ is not necessarily of class $C^1$.

I've thought about functions $f:\mathbb{R}\rightarrow\mathbb{R}$. Functions like $f(x)=x^a$ for $a\geq 1$ are locally Lipschitz, but they're also continuously differentiable, so don't quite work.

Mika H.
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    Consider the integral of a bounded but not continuous function. A step function, maybe. – Daniel Fischer Oct 20 '13 at 22:06
  • @leo That's not locally Lipschitz, I believe – Mika H. Oct 20 '13 at 23:56
  • True. ${}{}{}{}$ – leo Oct 21 '13 at 02:38
  • @Mika H: Compute integral of a step-function and see if you can identify it with a familiar function. Then check that it is globally Lipschitz. – Moishe Kohan Oct 21 '13 at 04:02
  • @studiosus I agree with Daniel Fischer's function (and voted it up). I was responding to leo's function, which he already deleted upon realizing that it is not locally Lipschitz :) – Mika H. Oct 21 '13 at 04:16
  • The absolute value function obviously works when $m=n=1.$ Can you think of something analogous to use when $m$ and $n$ are arbitrarily specified positive integers? (The norm function in $\mathbb R$ is $\dots).$ – Dave L. Renfro Oct 22 '13 at 20:43
  • https://math.stackexchange.com/questions/2609765/is-a-lipschitz-function-differentiable/2609787#2609787 – Guy Fsone Jan 28 '18 at 13:42

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The Euclidean norm $f(x)=\|x\|$ is probably the simplest example: the Lipschitz condition comes straight from the triangle inequality, and the non-differentiability is seen from the fact that the restriction of $f$ to the first coordinate axis is $|x_1|$.

Post No Bulls
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By Rademacher's theorem any Lipschitz continuous function is differentiable a.e.

The derivative of a lipschitz function is a bounded measurable function, but this derivative need not be continuous.

Hence the locally lipschitz function need not be $C^1$.

For example $f(x) = \frac {x^i x^j}{ |x|} $ on the unit $n$-ball.

Rudolf L
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  • What does dividing by $|x|$ on the unit $n$-ball contribute to the example? Isn't $|x|=1$ in that setting? – hardmath Feb 18 '16 at 23:47
  • dividing by |x| give a singularity to show the derivative is not continuous. btw: |x|=1 is only on the boundary of the ball. – Rudolf L Feb 19 '16 at 19:17