Challenge Math Challenge - August 2020

[wrobel]

Ideas for problem #5:

Make the problem simpler: take B = identity map, so b =1, and take the constant a = 1, so we want to prove the map f taking x to x^2 + x, hits every point C with |C| ≤ 1/4, where |x^2| ≤ |x|^2
...

I can only say that my idea was completely different and I can not understand whether your argument eventually hits the aim.Perhaps it was not a good idea to post this problem for the Math Challenge. All my previous problems were easily solved and I decided to post something more interesting. And I posted an example from my article. I feel this was too much.

Spoiler: 5

https://arxiv.org/abs/1804.10501

 

[mathwonk]

Of course I am just pointing out that the inverse function theorem implies that the map f(x) = x^2 + x does map the ball of radius 1/2 around zero onto some open ball around 0, and the goal is to give a concrete radius for such an image ball, namely 1/4. My homotopy argument apparently works as given, in equal finite dimensions, (oh yes, and by restriction to a subspace also in the unequal finite dimensional case), and I did not pursue whether the existing proofs of the inverse function theorem, or surjective mapping theorem can be made explicit in the Banach space case. My intent was to leave it for someone else at least for some more weeks, since August is not over. In fact I think your problem is quite interesting, and I am very glad you posted it. Not all problems have to be, nor should be, easy. Thank you for the rather nice and instructive problem.

 

[Buzz Bloom]

Spoiler: Number 2

f(x) = g(x) = x
a = 1.5
b = .5
g(a) - g(b) = 1
f(a) - f(b) = 1
QED

 

[Office_Shredder]

Spoiler: Number 2

f(x) = g(x) = x
a = 1.5
b = .5
g(a) - g(b) = 1
f(a) - f(b) = 1
QED

Spoiler

You can actually do it when f(x)=x pretty easily. If g(1)=1, a=0 and b=1. Otherwise the function g(x+1)-g(x) is smaller than 1 for x=0 or 1, and larger than 1 for they other. So somewhere in between it just be equal to 1

 

[Buzz Bloom]

Hi @Office_Shredder:

I think I now understand that my answer was incomplete in that the problem wants a pair a and b for each of any pair of functions f and g where both are continuous and satisfy the end conditions at 0 and 2. I should have got that sooner, since using just a single pair f and g was very simple.

I guess I do not read the math language of symbols very well. The language has changed a lot since I was in college. If I had been writing the problem in the old days, I would have used:
"... for any pair f,g of continuous functions...".

I will now think a bit harder about the problem.

Regards,
Buzz