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Consider this question. I have given a proof below which only uses the openness of the domain and the continuity of the function $f$; however, this result should not be true, but still I cannot find where is the mistake in my proof.

Proof:

Let $f: A \to \mathbb{R}^n$, where $A \subseteq \mathbb{R}^m$ and $A$ is open, and $f \in C^1$. To show that $f$ is locally lipschitz, let $x\in A$ be arbitrary and $U_x$ be a open neighbourhood of $x$ in $A$. Then take a cube around $x$ that is contained in $U_x$. Since $f$ is cont. $$m_x \leq f(y) \leq M_x \quad \exists m_x , M_x \in \mathbb{R} \quad \forall y \in C_x.$$

Moreover, the function $h:C_x \to \mathbb{R}^n$ defined by $h(y) = y-x$ is also bounded, say $a_x \leq y-x \leq b_x$, hence we have either

$$1/|y-x| \leq max \{1/|a_x|, 1/|b_x|\}.$$

Also we have either $$|f(y) - f(x)/ |(y-x)| \leq max \{|M_x - f(x)|/|(y-x)|, |m_x - f(x)|/|(y-x)|\}$$ $ \quad \forall y (\not = x) \in C_x.$

Hence $$|f(y) - f(x)/ |(y-x)| \leq max \{|M_x - f(x)|/|(y-x)|, |m_x - f(x)|/|(y-x)|\} * max \{1/|a_x|, 1/|b_x|\}$$ But this implies

Hence $f$ is locally lipschitz. QED

Question:

What is wrong with the above proof ?

Our
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    In your inequality, you are seemingly stating that $|f(y)-f(x)|\leq |M_x-f(x)|$, but this kind of manipulation with absolute values is untrue. To give an example, imagine $f(y)=2, f(x)=3, M_x = 3$. Then you are stating $1\leq 0$. Moreover, note that you would like to show that the quotient is less or equal some constant to prove that the function is Lipschitz. But there still is some $y$ on the right hand side. – Suzet Jul 16 '18 at 12:27
  • @Suzet Good point; however, there since $f(y)$ is also bounded from below, we can get around that problem, see my edit in a second. – Our Jul 16 '18 at 12:30
  • @Suzet Yes, there is a $y$ but, since we are taking those $y$s from $C_x$, which is bounded, aren't they also bounded ? I mean the function $h(y) = y-x$ on $C_x$ has also max and min value, so I can find a min & max value for that part also. – Our Jul 16 '18 at 12:34
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    $|y|$ may be a bounded function, but surely not $\frac{1}{|y-x|}$. As $y$ gets closer to $x$, this quotient just goes straight to infinity. – Suzet Jul 16 '18 at 12:40
  • @Suzet Yeah, I have noticed that while editing :) – Our Jul 16 '18 at 12:41
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    @Suzet Thanks a lot for the comments. – Our Jul 16 '18 at 12:41

2 Answers2

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Even if the estimates by $\frac{\lvert M_x-f(x)\rvert}{\lvert y-x\rvert}$ and $\frac{\lvert m_x-f(x)\rvert}{\lvert y-x\rvert}$ were justified:

Hence $f$ is locally Lipschitz.

Is it? You did not prove, for instance, that $\sup_{y\in C_x\setminus\{x\}}\frac{\lvert M_x-f(x)\rvert}{\lvert y-x\rvert}<\infty$ (and for good reason, since it's false unless $f(x)=M_x$)

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Locally Lipschitz means that each $x \in A$ has a neighborhood $N(x) \subset A$ such that $f \mid_{N(x)}$ is Lipschitz, i.e. such that there exists $L \ge 0$ with the property $\lVert f(y) - f(z) \rVert \le L \lVert y - z \rVert$ for all $y, z \in N(x)$. I cannot see how your computations are related to that.

Your argument is this: $A$ is locally compact, therefore it suffices to consider a compact neighborhood $N(x)$. $f$ is bounded on $N(x)$ and you try to use this single fact to show that $f$ is Lipschitz on $N(x)$. But if your argumentation were sound, then each bounded function would be Lipschitz, in particular continuous, which is obviously not true.

In fact it is not true that all $f \in C^0(A)$ are locally Lipschitz. An example is $f : \mathbb{R} \to \mathbb{R}, f(x) = \sqrt{\lvert x \rvert}$: There is no neighborhood of $0$ on which it is Lipschitz. See http://www.mathcounterexamples.net/function-that-is-uniformly-continuous-but-not-lipschitz-continuous/.

Paul Frost
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  • Yes, I do know that the result is false, as I have stated in the question, so that was the reason why I was asking why the proof is wrong. – Our Jul 16 '18 at 13:02
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    The essential mistake is that $1/ \lVert y - x \rVert$ is not bounded on $C_x$. In fact $min { \lVert h(y) \rVert \mid y \in C_x } = 0$ so that $a_x = 0$. – Paul Frost Jul 16 '18 at 13:14
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    You can more generally take $f : \mathbb{R}^n \to \mathbb{R}, f(x) = \sqrt{\lVert x \rVert}$. Note that $0$ is the only point where it is not differentiable. – Paul Frost Jul 16 '18 at 13:26