# Distance between fixed points of contracting maps

#### joshmccraney

1. The problem statement, all variables and given/known data
Let $V$ be a Banach space. Let $f:V\to V$ and $g:V\to V$ be two $q$-contracting maps, $q\in(0,1)$. Assume they are uniformly close to each other. Show the distance between fixed points of $f,g$ is at most $\epsilon/(1-q)$.

2. Relevant equations
Definitions:

Uniformly close implies $\forall \, \epsilon>0, v\in V$ we have $\| f(v) - g(v) \| < \epsilon$.

A map $f$ is $q$-contracting if $\exists \, q \in (0,1):\forall x,y\in\mathbb R^n, \| f(x) - f(y) \| \leq q\| x-y\|$.

3. The attempt at a solution
My idea is to iterate one map starting at the fixed point of the other map and show after sufficient amount of iterations, you will be within $\epsilon$ of the other fixed point. Since we are proving the distance between fixed points, the Banach Fixed Point Theorem seems relevant: $$\| x-x_n \| \leq \frac{q^n}{1-q}\| x_1-x_0 \|$$ where $x$ is the fixed point and $x_n$ is the $n$th iteration.

So trying to put all this together, if $x_f$ is the fixed point of $f$, then consider
$$\| g(x_f) - f(x_f) \| = \| g(x_f) - x_f \| \implies\\ \| g(x_f) - x_f \| < \epsilon$$

But this can't be right. Can someone help me smooth it out?

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#### fresh_42

Mentor
2018 Award
I don't understand uniformly close. If $||f(v)-g(v)|| < \varepsilon$ for any $\varepsilon>0$, why isn't $f \equiv g?$

#### joshmccraney

I don't understand uniformly close. If $||f(v)-g(v)|| < \varepsilon$ for any $\varepsilon>0$, why isn't $f \equiv g?$
Ok, so my conclusion isn't too crazy after all? I'll check and ask the professor when I next see him and post what I find. Thanks for responding!

#### fresh_42

Mentor
2018 Award
Ok, so my conclusion isn't too crazy after all? I'll check and ask the professor when I next see him and post what I find. Thanks for responding!
It could be $\forall\,\varepsilon>0 \,\exists\,v_\varepsilon$ but you haven't quantified $v$, so it reads as if it is still under the for all quantifier.

In general, you should start with $||x_f-y_f||=||f(x_f)-g(y_f)||$, insert some terms and use the triangle inequality.

#### joshmccraney

It could be $\forall\,\varepsilon>0 \,\exists\,v_\varepsilon$ but you haven't quantified $v$, so it reads as if it is still under the for all quantifier.
Can you clarify what $v_\epsilon$ is?

In general, you should start with $||x_f-y_f||=||f(x_f)-g(y_f)||$, insert some terms and use the triangle inequality.
Isn't that how I started (begin with a fixed point from one function)?

#### WWGD

Gold Member
Can't you use the general form of the fixed point? It comes from repeated iterations of the map , i.e., $Lim _{ n \rightarrow \infty} f^{n}(x)=x$ for any x in the space. Then do something to show ff(x) is very close to g(g(x)), etc?

Last edited:

#### fresh_42

Mentor
2018 Award
Can you clarify what $v_\epsilon$ is?
In my notation, the choice of $v$ depends on $\varepsilon$, because I wrote $\forall\,\varepsilon \, \exists\,v = v(\varepsilon)$, it varies with $\varepsilon$. However, what we need is $||f(x_f)-g(x_f)||<\varepsilon$.
Isn't that how I started (begin with a fixed point from one function)?
You started with the $||f(x_f)-g(y_f)||$ whereas I started with $||x_f-y_f||$. They are equal, but at the end of the day, it makes the difference!

#### WWGD

Gold Member
Can you clarify what $v_\epsilon$ is?

Isn't that how I started (begin with a fixed point from one function)?
Isn't this assuming
Can you clarify what $v_\epsilon$ is?

Isn't that how I started (begin with a fixed point from one function)?
Where does this come from? We know each map has a fixed point, but where does this equality come from?
1. The problem statement, all variables and given/known data
Let $V$ be a Banach space. Let $f:V\to V$ and $g:V\to V$ be two $q$-contracting maps, $q\in(0,1)$. Assume they are uniformly close to each other. Show the distance between fixed points of $f,g$ is at most $\epsilon/(1-q)$.

2. Relevant equations
Definitions:

Uniformly close implies $\forall \, \epsilon>0, v\in V$ we have $\| f(v) - g(v) \| < \epsilon$.

A map $f$ is $q$-contracting if $\exists \, q \in (0,1):\forall x,y\in\mathbb R^n, \| f(x) - f(y) \| \leq q\| x-y\|$.

I think the key issue here is that we have the same q for both maps, and using the relation:

$\exists \, q \in (0,1):\forall x,y\in\mathbb R^n, \| f(x) - f(y) \| \leq q\| x-y\|$.maybe with #x_1,x_2 $and$ y_1, y_2 $and the triangle inequality. This would not be true if we had , say a q- and p- contraction s with$ p \neq q $. 3. The attempt at a solution My idea is to iterate one map starting at the fixed point of the other map and show after sufficient amount of iterations, you will be within$\epsilon$of the other fixed point. Since we are proving the distance between fixed points, the Banach Fixed Point Theorem seems relevant: \| x-x_n \| \leq \frac{q^n}{1-q}\| x_1-x_0 \| where$x$is the fixed point and$x_n$is the$n$th iteration. So trying to put all this together, if$x_f$is the fixed point of$f$, then consider \| g(x_f) - f(x_f) \| = \| g(x_f) - x_f \| \implies\\ \| g(x_f) - x_f \| < \epsilon  But this can't be right. Can someone help me smooth it out? I think the key issue is that these are both q-contractions and not p-, q- contractions, in which case the result would not hold up. #### joshmccraney Sorry, I'm very confused by post 7; perhaps WWGD was viewing from phone app? Regarding post 6, what do you add? I thought about \| f(x_f) - g(x_f) + x_g - x_g \| \leq \| f(x_f) -x_g\| + \|g(x_f) - x_g \| =\\ \| f(x_f) -g(x_g)\| + \|g(x_f) - x_g \| \leq \epsilon + \|g(x_f) - x_g \| \leq \epsilon + \frac{q}{1-q}\| g(g(x_f)) - g(x_f) \| but am now stuck. #### fresh_42 Mentor 2018 Award Sorry, I'm very confused by post 7; perhaps WWGD was viewing from phone app? Regarding post 6, what do you add? I thought about \| f(x_f) - g(x_f) + x_g - x_g \| \leq \| f(x_f) -x_g\| + \|g(x_f) - x_g \| =\\ \| f(x_f) -g(x_g)\| + \|g(x_f) - x_g \| \leq \epsilon + \|g(x_f) - x_g \| \leq \epsilon + \frac{q}{1-q}\| g(g(x_f)) - g(x_f) \| but am now stuck. Why is$\| f(x_f) -g(x_g)\|<\varepsilon\,?$And I do not understand the rest either. Try my hint:$||x_f - x_g|| = ||f(x_f) -g(x_g)|| = ||f(x_f) - g(x_f) + g(x_f) - g(x_g)|| \leq \ldots$#### joshmccraney Why is$\| f(x_f) -g(x_g)\|<\varepsilon\,?$And I do not understand the rest either. Try my hint:$||x_f - x_g|| = ||f(x_f) -g(x_g)|| = ||f(x_f) - g(x_f) + g(x_f) - g(x_g)|| \leq \ldots$\leq \|f(x_f) - g(x_f)\| + \|g(x_f) - g(x_g)\| \leq \epsilon + \frac{q}{1-q}\| g(g(x_f)) - g(x_f) \|  where the final term comes from Remark 1, from the following wiki page: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem But now I'm unsure what to do. Is this not what you had in mind? #### fresh_42 Mentor 2018 Award \leq \|f(x_f) - g(x_f)\| + \|g(x_f) - g(x_g)\| \leq \epsilon + \frac{q}{1-q}\| g(g(x_f)) - g(x_f) \|  where the final term comes from Remark 1, from the following wiki page: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem But now I'm unsure what to do. Is this not what you had in mind? This is way too complicated. We have$\|f(x_f) - g(x_f)\|<\varepsilon$by assumption of uniformly close functions, and$\|g(x_f) - g(x_g)\| < q ||x_f-x_g||$since the functions are contractions. All together there is a term$||x_f-x_g||$at the beginning of the LHS, which is why I wanted to start with it, and a term$||x_f-x_g||$at the end on the RHS. But this is exactly the distance we want to estimate. The rest is a little algebra and attention on possibly negative factors in our inequality. Last edited: #### joshmccraney This is way too complicated. We have$\|f(x_f) - g(x_f)\|<\varepsilon$by assumption of uniformly close functions, and$\|g(x_f) - g(x_g)\| < q ||x_f-x_g||$since the functions are contractions. All together there is a term$||x_f-x_g||$at the beginning of the LHS, which is why I wanted to start with it, and a term$||x_f-x_g||$at the end on the RHS. But this is exactly the distance we want to estimate. The rest is a little algebra and attention on possibly negative factors in our inequality. Oooooohhhhh! Sorry, I was thinking the only way to introduce the$q$component was through iterative estimation. You invoke the definition of a$q##-contacting map, which sucks for me because I even wrote the definition in the first post but spaced using it!

Thanks so much! Marking this as solved!

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