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# Contractions watch

1. If I have a function , does this mean it's a contraction because it maps the interval back to itself? If so how would I show that this function is a contraction? Would it be sufficient to show that the absolute value of the gradient on the function is strictly less than 1? Thanks.
2. Now here was me thinking this was all about the later stages of pregnancy when I read the title...
3. (Original post by JBKProductions)
If I have a function , does this mean it's a contraction because it maps the interval back to itself? If so how would I show that this function is a contraction? Would it be sufficient to show that the absolute value of the gradient on the function is strictly less than 1? Thanks.
It's not necessarily true: take f on rational to be 1 and f on irrational numbers to be 0.
Also with just the information you've provided it doesn't follow that f has a gradient.
4. (Original post by IrrationalNumber)
It's not necessarily true: take f on rational to be 42 and f on irrational numbers to be 0.
Also with just the information you've provided it doesn't follow that f has a gradient.
Sorry I forgot to add that the function is
5. my mind went to pregnancy when I saw this title..
6. (Original post by JBKProductions)
Sorry I forgot to add that the function is
A gradient less than 1 will suffice in this case. If f is C^1 and |f'(x)|<1 then |f'(x)|<d for some d<1 (why?) and then you can use the mean value theorem.
7. (Original post by IrrationalNumber)
A gradient less than 1 will suffice in this case. If f is C^1 and |f'(x)|<1 then |f'(x)|<d for some d<1 (why?) and then you can use the mean value theorem.
I'm not sure why, can you explain it a bit further? Also I'm not sure why I'm supposed to bring the mean value theorem into this. Thanks.
8. (Original post by JBKProductions)
I'm not sure why, can you explain it a bit further? Also I'm not sure why I'm supposed to bring the mean value theorem into this. Thanks.
The MVT tells you that if f is differentiable on (0,2) and continuous on [0,2] then for all x,y in [0,2] we can find a c in (x,y) such that

In particular then
Can you see why that might be useful now?

Now if f belongs to C^1, the derivative is continuous. The modulus of any continuous function is still continuous. What can we say about the maximum value of a continuous function on a compact set(you may know this result as the maximum value of a continuous function on a closed interval)?
9. (Original post by IrrationalNumber)
The MVT tells you that if f is differentiable on (0,2) and continuous on [0,2] then for all x,y in [0,2] we can find a c in (x,y) such that

In particular then
Can you see why that might be useful now?

Now if f belongs to C^1, the derivative is continuous. The modulus of any continuous function is still continuous. What can we say about the maximum value of a continuous function on a compact set(you may know this result as the maximum value of a continuous function on a closed interval)?
So the maximum value of the continuous function on the closed interval is when is at the maximum?
10. (Original post by JBKProductions)
So the maximum value of the continuous function on the closed interval is when is at the maximum?
Right, in particular there exists a d such that
So if you can show that |f'|<1 then you get that |f'|<C<1 for some C. Now suppose that there are x and y in [0,2] such that and argue by contradiction.
11. I'm not sure why we're worrying about compactness or where f'(x) attains a maximum here. f'(x) = x/16 - (pi/4) sin( x pi/4). |sin x| <=1 and |x/16| <=2/16. Since pi/4+2/16 < 1, you're done. (Still need to use MVT afterwards, of course).
12. (Original post by DFranklin)
I'm not sure why we're worrying about compactness or where f'(x) attains a maximum here. f'(x) = x/16 - (pi/4) sin( x pi/4). |sin x| <=1 and |x/16| <=2/16. Since pi/4+2/16 < 1, you're done. (Still need to use MVT afterwards, of course).
Is it possible to construct a function f for which |f'|<1 everywhere but f is not a contraction? I haven't thought about it too hard but I think it should be possible by getting the derivative to approach 1. It certainly won't be possible to do that on a compact space.
13. (Original post by IrrationalNumber)
Right, in particular there exists a d such that
So if you can show that |f'|<1 then you get that |f'|<C<1 for some C. Now suppose that there are x and y in [0,2] such that and argue by contradiction.
Thanks!
14. (Original post by DFranklin)
I'm not sure why we're worrying about compactness or where f'(x) attains a maximum here. f'(x) = x/16 - (pi/4) sin( x pi/4). |sin x| &lt;=1 and |x/16| &lt;=2/16. Since pi/4+2/16 &lt; 1, you're done. (Still need to use MVT afterwards, of course).

(Original post by IrrationalNumber)
Is it possible to construct a function f for which |f'|&lt;1 everywhere but f is not a contraction? I haven't thought about it too hard but I think it should be possible by getting the derivative to approach 1. It certainly won't be possible to do that on a compact space.
Yes, if we consider f(x)=x^2 on (0,1/2) and consider and then which cannot be always less than or equal to for some fixed K<1.
15. (Original post by IrrationalNumber)
Is it possible to construct a function f for which |f'|<1 everywhere but f is not a contraction?
Yes.

E.g. g(x) = x - exp(-x) on [0,\infty).

But in the original case it's easy to show |f'| < k < 1 directly, so no need to do anything more.
16. (Original post by DFranklin)
Yes.

E.g. g(x) = x - exp(-x) on [0,\infty).

But in the original case it's easy to show |f'| < k < 1 directly, so no need to do anything more.
Ah, you're quite right.

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