# If $f$ is a contraction then it is continuous.

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

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

• 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. Jan 1, 2021 at 23:21
• 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... Jan 1, 2021 at 23:26
• 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) Jan 1, 2021 at 23:29
• I would prefer to use the definition using the epsilon-delta-criteria. However, I am fine with every (widely know) equivalent definition. Jan 1, 2021 at 23:33

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 ^_^

• 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... Jan 1, 2021 at 23:43
• 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. Jan 1, 2021 at 23:52
• I finally think I have come to a solution. Could you take the minute to check it? Feb 25, 2021 at 23:16

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.