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Let $f$ be a function on $[0,1]$ into $\Bbb{R}$. Suppose that if $x\in[0,1],$ there exists $K_x$ such that \begin{align}|f(x)-f(y)|\leq K_x |x-y|,\;\;\forall\;\;y\in[0,1].\end{align} Prove or disprove that there exists $K$ such that \begin{align}|f(x)-f(y)|\leq K |x-y|,\;\forall\;\;x,y\in[0,1].\end{align}

DISPROOF

Consider the function \begin{align} f:[0&,1]\to \Bbb{R}, \\&x\mapsto \sqrt{x} \end{align}

Let $x=0$ and $y\in (0,1]$ be fixed. Then, \begin{align} \left| f(0)-f(y) \right|&=\left|0-\sqrt{y} \right| \end{align} Take $y=1/(4n^2)$ for all $n.$ Then, \begin{align} \left| f(0)-f\left(\dfrac{1}{4n^2}\right) \right|&=\left|0-\dfrac{1}{\sqrt{4n^2}} \right| \\&=2n^{3/2}\left|\dfrac{1}{4n^2} -0 \right| \end{align} By assumption, there exists $K_0$ such that \begin{align} \left| f(0)-f\left(\dfrac{1}{4n^2}\right) \right|&=2n^{3/2}\left|\dfrac{1}{4n^2} -0 \right|\leq K_0\left|\dfrac{1}{4n^2} -0 \right| \end{align} Sending $n\to\infty,$ we have \begin{align} \infty \leq K_0<\infty,\;\;\text{contradiction}. \end{align} Hence, the function $f$ is not Lipschitz in $[0,1]$.

QUESTION: Is my disproof correct?

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    $\begingroup$ I would suggest that before you start the computation you actually write explicitly the claim whose truth you need to decide. Is $f$ supposed to be continuous? Are there any other assumptions? Once that is explicit, say what you are going to do: "We will show that the statement fails in general by providing a counterexample. Note that it is enough to show that there is a differentiable $f$ with unbounded derivative, because ..." Only after you've done that, proceed with your counterexample. People don't want to put up with a wall of symbols if they don't know its purpose. $\endgroup$ – Andrés E. Caicedo Jan 5 '19 at 15:21
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    $\begingroup$ It would be much simpler to just define $f:[0,1]\to\Bbb R$ by $f(x)=\sqrt{x}.$ There is no need for a piecewise definition. $\endgroup$ – Cameron Buie Jan 5 '19 at 15:24
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    $\begingroup$ Nice proof, but you forgot to mention the question. I assume the question was: if $f$ is continuous, then prove that there exists $K$ such that $|f(x) - f(y)| \le K |x - y|$ for all $x, y$, or else find a counterexample $f$. $\endgroup$ – 6005 Jan 5 '19 at 15:26
  • $\begingroup$ Also, $f$ is not differentiable at $0$. So you should say that $f|_{(0,1)}$ is differentiable but the derivative is unbounded, thus a constant $K$ cannot exist for $f|_{(0,1)}$, thus a constant $K$ cannot exist for $f$ either. $\endgroup$ – 6005 Jan 5 '19 at 15:31
  • $\begingroup$ related $\endgroup$ – Omnomnomnom Jan 5 '19 at 16:42
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You are correct, the function $f$ is not Lipschitz in $[0,1]$, but your argument should be modified. You may simply say that $$\frac{f(1/n)-f(0)}{\frac{1}{n}-0}=\sqrt{n}\to +\infty$$ which contradicts the fact that $|f(x)-f(y)|/|x-y|$ is bounded by a constant $K$.

On the other hand this is not a counterexample for your statement. For the same reason as above (Just take $y=1/n$), for $x=0$ there is no constant $K_0$ such that $$|f(0)-f(y)|\leq K_0 |0-y|,\;\;\forall y\in[0,1].$$ Instead consider the function $$f(x)=\cases{x\sin(1/x)& if $x\not=0$\\0& if $x=0$,}$$ If $x_n=1/(2\pi n)$ and $y_n=1/(2\pi n+\pi/2)$. then $|f(x_n)-f(y_n)|/|x_n-y_n|$ is unbounded which implies that $f$ is not Lipschitz in $[0,1]$, but for any $x\in[0,1],$ there exists $K_x$ such that $$|f(x)-f(y)|\leq K_x |x-y|,\;\;\forall\;\;y\in[0,1].$$ In fact take $K_0=1$ and for $x\in(0,1]$ the existence of $K_x$ follows from $f'\in C^1((0,1])$.

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  • $\begingroup$ @Mike I added a few lines. Your disproof is incorrect. $\endgroup$ – Robert Z Jan 5 '19 at 16:48
  • $\begingroup$ You might mention why the correct counterexample does satisfy the hypothesis. (If $f$ is differentiable at $$ then $K_x$ exists...) $\endgroup$ – David C. Ullrich Jan 5 '19 at 23:06
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This isn't incorrect per se, but its unnecessarily convoluted. To show that $f'$ is unbounded, just observe that $f'(x)=\frac{1}{2\sqrt{x}}$ so that $\lim_{x\to 0^+}f'(x)=\infty$. Hence, $f$ is not Lipschitz on $[0,1]$, which is precisely what you want to prove. That's all you need to say.

The last part is a little hand-wavy though. Technically, you should say something like this: $f$ is Lipschitz on $[\epsilon,1]$ with constant $K_\epsilon:=\sup_{\epsilon\leq x\leq 1}|f'(x)|$. Since $K_\epsilon\to\infty$ as $\epsilon\to 0^+$, $f$ is not Lipschitz on $[0,1]$.

Better still just to do an explicit calculation, as the other fellow did.

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  • $\begingroup$ Thanks a lot, Ben W! I really learnt a lot from you! Kindly check my post, I edited it! (+1) $\endgroup$ – Omojola Micheal Jan 5 '19 at 16:36

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