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The proof might seem intuitive if just has one or more jump points which have a distance $d$ from each other.

But I am struggling, with the following problem:

If f is not continuous, such that for all intervals: $$[a,b], \text{with } a < b < c < d: f \text{ is not continuous in } [b,c]$$ where $a,d$ are constant and $b,c$ freely selectable.

Can it then be a contraction?

Edit:

$$\exists a,d: \forall b,c: (a < b < c <d \implies f \text{ is not continuous in } [b,c])$$

I would prefer to use the definition of continuity using the epsilon-delta-criteria. However, I am fine with every (widely know) equivalent definition.

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    $\begingroup$ Try to write down the definition of a contraction and the definition of a continuous function and try to see how one implies the other. $\endgroup$ Jan 1, 2021 at 23:21
  • $\begingroup$ Tried it for about 2 hours. I always end up having the same problem... If I assume that f is continuous at the interval on the right of the jump point; easy. But if not, no Idea... $\endgroup$
    – Niclas
    Jan 1, 2021 at 23:26
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    $\begingroup$ What is your definition of continuity? There's several (equivalent) ones that exist - and it matters a lot to how your question would be answered. (And if your definition of continuity involves a limit, might as well give us the definition of limits too) $\endgroup$ Jan 1, 2021 at 23:29
  • $\begingroup$ I would prefer to use the definition using the epsilon-delta-criteria. However, I am fine with every (widely know) equivalent definition. $\endgroup$
    – Niclas
    Jan 1, 2021 at 23:33

2 Answers 2

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Welcome to MSE!

Hint:

If $f$ is a contraction, then $d(fx,fy) < d(x,y)$ for every $x,y$.

To show $f$ is continuous, we want to show we can make $d(fx,fy) < \epsilon$ small by controlling $d(x,y)$... Of course, this is exactly the flavor of control that contractibility buys us.

Formally, say $\epsilon > 0$. We want to find a $\delta$ so that whenever $d(x,y) < \delta$, we're guaranteed $d(fx,fy) < \epsilon$... By contractibility, we're guaranteed $d(fx,fy) < d(x,y) < \delta$.

Do you see where to go from here? What's a good choice of $\delta$?


I hope this helps ^_^

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  • $\begingroup$ This is the exact point I stopped. What is the good choice of $\delta$? I do not anything about $f$... Certainly, $\delta$ and $epsilon$ must have a relationship? There are some $x,y$ for which the contraction holds true... Why not all... $\endgroup$
    – Niclas
    Jan 1, 2021 at 23:43
  • $\begingroup$ You want $d(fx,fy) < \epsilon$. You're guaranteed $d(fx,fy) < \delta$. You can choose $\delta$ to be any function of $\epsilon$ you want. $\endgroup$ Jan 1, 2021 at 23:52
  • $\begingroup$ I finally think I have come to a solution. Could you take the minute to check it? $\endgroup$
    – Niclas
    Feb 25, 2021 at 23:16
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Let $c: (M,d) \to (M,d)$ with $d(c(x_1),c(x_2)) \leq \gamma d(x_1,x_2)$ with $\gamma \in [0,1)$ be a contraction.

If $c$ is continuous, then

$$\forall x_0 \in M: \forall \epsilon > 0: \exists \delta > 0: c(B(x_0, \delta)) \subset B(c(x_0),\epsilon)$$

Choose $\delta = \epsilon$

Then: $c(B(x_0, \delta)) = c(B(x_0, \epsilon)) \subset B(c(x_0), \gamma\epsilon) \subset B(c(x_0),\epsilon)$

Therefore all contractions are continous.

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