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Problem 1.2.2 Suppose that $f$ maps compact interval $I$ into itself and that $$\mid f(x)-f(y) \mid < \mid x-y \mid$$ for all $x, y \in I, x\neq y$. Can one conclude that there is some constant $M<1$ such that, for all $x, y \in I$, $$\mid f(x)-f(y) \mid \leq M\mid x-y \mid$$?

If such function exists it is not continuously differentiable, otherwise $f'$ will attain supremum.

I tried to define $g: I \times I \rightarrow I$ s.t. $g(x,y)=\frac{\mid f(x)-f(y) \mid}{\mid x-y \mid}$ which is continuous and attains supremum. But $g$ is not defined for $x=y$

Please give hint. Please do not give solution. Thanks!

Vinay Deshpande
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  • Hint: the answer is negative. – uniquesolution Jul 24 '20 at 13:49
  • @uniquesolution Thanks. I found a counterexample – Vinay Deshpande Jul 24 '20 at 14:18
  • What is the problem of attaining supremum ? – EDX Jul 24 '20 at 14:19
  • You want a function which has derivative $1$ at one end of the interval, and less than $1$ on the rest of the interval. – tkf Jul 24 '20 at 14:19
  • @EDX case i: $sup { g(x, y) \mid x\neq y }<1$. Then $M=sup { g(x, y) \mid x\neq y }$. Case ii: $sup { g(x, y) \mid x\neq y }=1$. If $g$ were defined and continuous, its image is closed and $g$ attains its supremum which is a contradiction. So case ii is false. – Vinay Deshpande Jul 24 '20 at 14:58
  • Watch out because the condition doesn't tell you $M=\sup{g(x,y) | x \neq y}$ . Taking $g=0$, we can choose $M=\frac{1}{2}$ and we don't get $M=sup{g(x,y) | x \neq y}$ – EDX Jul 24 '20 at 15:01
  • @EDX your example falls in case i and $sup{ g(x,y) \mid x\neq y } (=0)$ also works as an $M$. So my argument holds. – Vinay Deshpande Jul 24 '20 at 15:09
  • The case is that $M$ is not unique in the definition given. So you've to precise that $M$ is unique or to precise it is defined as $M=sup(g)$ – EDX Jul 24 '20 at 15:16

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