Continuous Functions: Uniform Continuity

[kingstrick]

Homework Statement

Let f be continuous on the interval [0,1] to ℝ and such that f(0) = f(1). Prove that there exists a point c in [0,1/2] such that f(c) = f(c+1/2). Conclude there are, at any time, antipodal points on the Earth's equator that have the same temperature.

Homework Equations

The Attempt at a Solution

I need help working through this problem. I am confused on how to start. First off the problem feels like it completely changes tempo from discussing real analysis to jumping to Earth's axis. What exactly are antipodal points? Are those the same as mid-points of subpoints in an interval? I assume so. Can someone tell me the first line in this proof after the assumptions so that i can get started?

 

[jgens]

Consider the function $g:[0,\frac{1}{2}] \rightarrow \mathbb{R}$ defined by $g(x)=f(x+\frac{1}{2})-f(x)$ and apply the Intermediate Value Theorem to $g$.

You usually talk about antipodal points on an $n$-sphere. In this case, you are talking about antipodal points on the circle $S^1$. If $x \in S^1$, then the antipodal point corresponding to $x$ is the point $-x$. So $x,-x$ form a pair of antipodal points. So for the second part, you are going to need to write your mapping as a function $S^1 \rightarrow \mathbb{R}$ and then apply the result above.

 

[kingstrick]

Consider the function $g:[0,\frac{1}{2}] \rightarrow \mathbb{R}$ defined by $g(x)=f(x+\frac{1}{2})-f(x)$ and apply the Intermediate Value Theorem to $g$.

You usually talk about antipodal points on an $n$-sphere. In this case, you are talking about antipodal points on the circle $S^1$. If $x \in S^1$, then the antipodal point corresponding to $x$ is the point $-x$. So $x,-x$ form a pair of antipodal points. So for the second part, you are going to need to write your mapping as a function $S^1 \rightarrow \mathbb{R}$ and then apply the result above.

Does this follow then:

Let f be continuous on the interval [0,1] to ℝ such that f(0) = f(1). Consider the function $g:[0,\frac{1}{2}] \rightarrow \mathbb{R}$ defined by $g(x)=f(x+\frac{1}{2})-f(x)$. Now, 0 < 1/2 < 1. Since f is continuous from [0,1], f is also continuous from [0,1/2]. By the location of roots theorem there exists a c where 0 < c < 1/2 such that g(0) < 0 < g(1/2). Therefore f(1/2)-f(0) < 0 < f(1)-f(1/2) Thus 0 = g(c) = f(c+1/2) - f(c). So f(c) = f(c+1/2).

Am i allow to assume that 0 is between g(0) and g(1/2)?

 

[Robert1986]

Well, you would need to show that 0 is between g(0) and g(1/2). That is, 0 is between:

f(1/2) - f(0) and f(1) - f(1/2)

So, the thing to do is to show that either:

a)f(1/2) - f(0) < 0 and f(1) - f(1/2) > 0
or
b)f(1/2) - f(0) >0 and f(1) - f(1/2) < 0
or
c) I'll let you think about this one HINT: It's a special case that should be handled before the (a) and (b) are. Also, remember that f(0) = f(1).

Once you have done the above, then use IVT as jgens suggested.

 

[kingstrick]

Well, you would need to show that 0 is between g(0) and g(1/2). That is, 0 is between:

f(1/2) - f(0) and f(1) - f(1/2)

So, the thing to do is to show that either:

a)f(1/2) - f(0) < 0 and f(1) - f(1/2) > 0
or
b)f(1/2) - f(0) >0 and f(1) - f(1/2) < 0
or
c) I'll let you think about this one HINT: It's a special case that should be handled before the (a) and (b) are. Also, remember that f(0) = f(1).

Once you have done the above, then use IVT as jgens suggested.

Does this follow then:

Let f be continuous on the interval [0,1] to ℝ such that f(0) = f(1). Consider the function $g:[0,\frac{1}{2}] \rightarrow \mathbb{R}$ defined by $g(x)=f(x+\frac{1}{2})-f(x)$. Now, 0 < 1/2 < 1. Since f is continuous from [0,1], f is also continuous from [0,1/2]. By the location of roots theorem there exists a c where 0 < c < 1/2 such that g(0) < 0 < g(1/2). Therefore f(1/2)-f(0) < 0 < f(1)-f(1/2) Thus 0 = g(c) = f(c+1/2) - f(c). So f(c) = f(c+1/2).

Am i allow to assume that 0 is between g(0) and g(1/2)?

Modified:

Let f be continuous on the interval [0,1] to ℝ such that f(0) = f(1). Consider the function $g:[0,\frac{1}{2}] \rightarrow \mathbb{R}$ defined by $g(x)=f(x+\frac{1}{2})-f(x)$. Now, 0 < 1/2 < 1. Since f is continuous from [0,1], f is also continuous from [0,1/2]. g(0) = f(1/2) - f(1), g(1) = f(1) - f(1/2).

if f(1/2) < f (1) then g(1/2) < 0 < g(0) where there exists a c such that 0 = g(c) = f(c+1/2)-f(c). So f(c) = f(c+1/2).

if f(1) < f(1/2) then g(0) = 0 and g(1/2) =0 where there exist c such that 0 = g(c) = f(c+1/2) -f(c). so f(c) = f(c+1/2).

if f(1) = f(1/2) then g(0) = 0 then g(1/2) = 0 where there exists a c =0 such that 0 = g(0) =g(c)=f(c+1/2)-f(c) so then f(c) = f(c+1/2)

Thus f(c) = f(c+1/2)

***Now i didn't use the fact that f(0) = f(1) so I know I made a mistake. What am i still missing?

By the location of roots theorem there exists a c where 0 < c < 1/2 such that g(0) < 0 < g(1/2). Therefore f(1/2)-f(0) < 0 < f(1)-f(1/2) Thus 0 = g(c) = f(c+1/2) - f(c). So f(c) = f(c+1/2).

 

[jgens]

***Now i didn't use the fact that f(0) = f(1) so I know I made a mistake. What am i still missing?

There are three cases to consider. If $f(\frac{1}{2})<f(0)$, then $g(0) < 0$ and $g(\frac{1}{2}) > 0$. If $f(\frac{1}{2}) = f(0)$, then the proof is complete. If $f(\frac{1}{2}) > f(0)$, then $g(0) > 0$ and $g(\frac{1}{2}) < 0$. Apply the IVT to the relevant cases.

 

[kingstrick]

There are three cases to consider. If $f(\frac{1}{2})<f(0)$, then $g(0) < 0$ and $g(\frac{1}{2}) > 0$. If $f(\frac{1}{2}) = f(0)$, then the proof is complete. If $f(\frac{1}{2}) > f(0)$, then $g(0) > 0$ and $g(\frac{1}{2}) < 0$. Apply the IVT to the relevant cases.

So my first two cases were incomplete? Does this meen, I need to contract the interval some more for each case?

 

[jgens]

Does this meen, I need to contract the interval some more for each case?

Nope. The work I wrote out already allows you to apply the IVT directly. Notice that the inequalities involving $g$ explicitly utilized the fact that $f(0)=f(1)$.

 

[kingstrick]

Nope. The work I wrote out already allows you to apply the IVT directly. Notice that the inequalities involving $g$ explicitly utilized the fact that $f(0)=f(1)$.

But didn't my proof use the IVT?

 

[jgens]

Ah! I only read through the part where you noted that you didn't use f(0) = f(1). So you know, the following is more or less correct:

if f(1/2) < f (1) then g(1/2) < 0 < g(0) where there exists a c such that 0 = g(c) = f(c+1/2)-f(c). So f(c) = f(c+1/2).

if f(1) < f(1/2) then g(0) = 0 and g(1/2) =0 where there exist c such that 0 = g(c) = f(c+1/2) -f(c). so f(c) = f(c+1/2).

if f(1) = f(1/2) then g(0) = 0 then g(1/2) = 0 where there exists a c =0 such that 0 = g(0) =g(c)=f(c+1/2)-f(c) so then f(c) = f(c+1/2)

You need to go through and fix some "=" signs to "<" signs, but the idea here is right. You do realize though that any of the inequalities you have about g rely on the fact that f(0)=f(1), right?

 

[kingstrick]

yes, i didn't realize it earlier when i was working through g(0) and substituted f(1/2)+f(0) and changed it to f(1/2)+f(1) when I started working it out earlier on. Thank you for your help.

This stuff is very frustrating. I think i may need to change majors.

 

[jgens]

This stuff is very frustrating. I think i may need to change majors.

Obviously the choice of switching majors is up to you, but I would not give up on math just yet. It takes everyone some period of time to figure out how to write proofs. I know that for me, I spent about a year writing mostly incorrect proofs and getting discouraged before I figured things out. So my advice is stick to it a bit longer. Give yourself enough time to see if you can get the knack for these kinds of things.

 

[kingstrick]

Modified:

Let f be continuous on the interval [0,1] to ℝ such that f(0) = f(1). Consider the function $g:[0,\frac{1}{2}] \rightarrow \mathbb{R}$ defined by $g(x)=f(x+\frac{1}{2})-f(x)$. Now, 0 < 1/2 < 1. Since f is continuous from [0,1], f is also continuous from [0,1/2]. g(0) = f(1/2) - f(1), g(1) = f(1) - f(1/2).

if f(1/2) < f (1) then g(1/2) < 0 < g(0) where there exists a c such that 0 = g(c) = f(c+1/2)-f(c). So f(c) = f(c+1/2).

if f(1) < f(1/2) then g(0) = 0 and g(1/2) =0 where there exist c such that 0 = g(c) = f(c+1/2) -f(c). so f(c) = f(c+1/2).

if f(1) = f(1/2) then g(0) = 0 then g(1/2) = 0 where there exists a c =0 such that 0 = g(0) =g(c)=f(c+1/2)-f(c) so then f(c) = f(c+1/2)

Thus f(c) = f(c+1/2)

***Now i didn't use the fact that f(0) = f(1) so I know I made a mistake. What am i still missing?

By the location of roots theorem there exists a c where 0 < c < 1/2 such that g(0) < 0 < g(1/2). Therefore f(1/2)-f(0) < 0 < f(1)-f(1/2) Thus 0 = g(c) = f(c+1/2) - f(c). So f(c) = f(c+1/2).

Ah! I only read through the part where you noted that you didn't use f(0) = f(1). So you know, the following is more or less correct:

You need to go through and fix some "=" signs to "<" signs, but the idea here is right. You do realize though that any of the inequalities you have about g rely on the fact that f(0)=f(1), right?

What is wrong with the '="?

 

[kingstrick]

So, i am confused is my proof still incomplete?