Question about a continuous function on R2

In summary, the conversation discussed how to solve a problem involving a continuous function on a unit circle, where it was needed to show the existence of a point (x, y) such that f(x, y) = f(-x, -y). The conversation touched on using topology theorems and reducing the function to one variable, and ultimately ended with the problem being solved through the application of an intermediate value theorem.
  • #1
evalover1987
15
0

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?)
 
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  • #2
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.

Dick said:
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
evalover1987 said:
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
Dick said:
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
evalover1987 said:
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
Dick said:
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
evalover1987 said:
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.
 

Related to Question about a continuous function on R2

1. What is a continuous function on R2?

A continuous function on R2 is a mathematical function that has a domain of all ordered pairs (x, y) in the Cartesian plane and a range of real numbers. This means that for any point (x, y) in the domain, the function will have a corresponding output that is close to the function's value at that point.

2. How is a continuous function on R2 different from a continuous function on R?

A continuous function on R2 is different from a continuous function on R in that it has two independent variables (x and y) instead of just one (x). This means that the function's graph will be a surface instead of a curve in the case of a continuous function on R.

3. What are some examples of continuous functions on R2?

Some examples of continuous functions on R2 include the distance function, which gives the distance from a fixed point (such as the origin) to any point (x, y) in the plane, and the temperature function, which maps the temperature at any point (x, y) in a region. Other examples include polynomial functions, trigonometric functions, and exponential functions.

4. What is the importance of continuous functions on R2?

Continuous functions on R2 have many important applications in mathematics, science, and engineering. They are used to model real-world phenomena such as temperature, pressure, and velocity, and are essential for understanding concepts such as limits, derivatives, and integrals. They also play a crucial role in optimization and optimization problems.

5. How can we determine if a function is continuous on R2?

To determine if a function is continuous on R2, we can use the definition of continuity, which states that a function is continuous at a point if the limit of the function at that point is equal to the function's value at that point. This means that for a function to be continuous on R2, it must be continuous at every point in its domain, which can be tested using various continuity theorems and techniques.

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