On Sat, Oct 3, 2020 at 5:57 PM Dan Asimov <dasimov@earthlink.net> wrote:
Allan and Andy have pointed out the ridiculousness of my original suggestion. But maybe a slight change could make it possible:
Could there be a continuous function on the disk
F : D^2 —> R
such that for every continuous function
g : [a, b] —> [0, 1]
on an interval, there is a curve C = C_g
C : [a, b] —> D^2
such that
F(C(t)) = g(t)
for all a ≤ t ≤ b
This is trivial. Let F be the identity function on [0,1] We don't care what values F takes on the rest of the disk, since the image of C will always ie in the interval [0,1] Just let C(t) = g(t) Andy
???
—Dan
----- Or even simpler than that, only 1/L constant functions, out of uncountably many, are possible.
Andy
On Sat, Oct 3, 2020 at 1:29 PM Allan Wechsler <acwacw@gmail.com> wrote:
I am feeling strongly that Dan's universal circular function is impossible. For any L, let N be ceil(1/L). Consider just the family of N+1 functions f[i](x) = ix, i ranging from 0 to N. It seems obvious to me that R/Z isn't big enough to fit them all.
On Sat, Oct 3, 2020 at 12:24 PM Dan Asimov <dasimov@earthlink.net> wrote:
I wonder if analogous continuous objects can exist.
For instance, can there be a continuous function on the circle R/Z
f : R/Z —> R
and some fixed L > 0, such that f contains all continuous functions
g : [0, L] —> R
as represented on an arc [c, c+L] ⊂ R/Z of the circle, via
g(x) = f(x+c)
(addition modulo 1) ???
Or something along these lines, maybe for a restricted class of functions f and g ?
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