Question about a continuous function on R2

  • #1

Homework Statement



Let S denote a unit circle centered at origin in xy plane, and
f is a continuous function that sends S to R (no need to be 1 to 1 or onto).
show that there's (x, y) such that f(x, y) = f(-x, -y)

Homework Equations



have a feeling it has something to do with theorems related to topology.


The Attempt at a Solution



by re-writing S as (cos z, sin z) for 0 <= z < 2 pi
i can reduce f to a function of one variable, f(z), such that
f(z) = - f(z + pi), but then I'm stuck what to do after that.

(or maybe I shouldn't have reduced f to a function of one variable in the first place?)
 

Answers and Replies

  • #2
Dick
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Reducing it to one variable is fine. And, yes, (x(z+pi),y(z+pi)) is then (-x(z),-y(z)) but how does that tell you anything about the value of f? Try considering g(z)=f(z)-f(z+pi). Can g(z) be nonzero for all values of z? This is more about continuous functions than it is about topology.
 
  • #3
yeah, i did that, but i failed to show that g(z) must be zero for some value of z, and
that was exactly the problem I had.
If I can show that if g(z) > 0 for some value of z, then g(x) < 0, for some value of x that's
not equal to z, then I can apply an intermediate value theorem to solve it, but
I also forgot how to apply an intermediate value theorem in 2 dimensional case.

Reducing it to one variable is fine. And, yes, (x(z+pi),y(z+pi)) is then (-x(z),-y(z)) but how does that tell you anything about the value of f? Try considering g(z)=f(z)-f(z+pi). Can g(z) be nonzero for all values of z? This is more about continuous functions than it is about topology.
 
  • #4
Dick
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yeah, i did that, but i failed to show that g(z) must be zero for some value of z, and
that was exactly the problem I had.
If I can show that if g(z) > 0 for some value of z, then g(x) < 0, for some value of x that's
not equal to z, then I can apply an intermediate value theorem to solve it, but
I also forgot how to apply an intermediate value theorem in 2 dimensional case.

You aren't in the two dimensional case anymore, you reduced it to a function of one variable, z, didn't you? Start thinking in one dimension.
 
  • #5
You aren't in the two dimensional case anymore, you reduced it to a function of one variable, z, didn't you? Start thinking in one dimension.

true. but with all due respect, you're not adding anything to what i've done so far.
my problem was that I am not quite sure how to prove the existence of z, x such that
they are not equal to each other, and
g(z) > 0 and g(x) < 0 (or g(z) <= 0 and g(x) > 0)
so that I can apply intermediate value theorem.
 
  • #6
Dick
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true. but with all due respect, you're not adding anything to what i've done so far.
my problem was that I am not quite sure how to prove the existence of z, x such that
they are not equal to each other, and
g(z) > 0 and g(x) < 0 (or g(z) <= 0 and g(x) > 0)
so that I can apply intermediate value theorem.

Alright, so you need a stronger hint. How is g(z) related to g(z+pi)?
 
  • #7
Alright, so you need a stronger hint. How is g(z) related to g(z+pi)?

you know what? i solved it. you don't have to continue to reply to this.
plus, it's not a "hint" if it's something that I already proved on my own.
 
  • #8
Dick
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you know what? i solved it. you don't have to continue to reply to this.
plus, it's not a "hint" if it's something that I already proved on my own.

Very welcome.
 

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