Comparing Opposite Points on the Circle
A point on the circle has a natural partner: the point directly opposite it. Given a real-valued function on the circle, we can compare its values at each such pair. The resulting difference has a symmetry that forces a sign relation between the ends of any semicircle. Continuity then prevents that difference from staying strictly positive or strictly negative all the way across.
We use \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) with the subspace topology inherited from the standard topology on \(\mathbb{R}^2\), and \(\mathbb{R}\) with its standard topology. The antipode of \(p\in S^1\) is \(-p\). Let \(f:S^1\to\mathbb{R}\) be continuous.
The definition compares ordered values: the value at \(p\) comes first and the value at \(-p\) is subtracted. Reversing the point reverses that order. The symmetry is the key fact, and it does not depend on a special formula for \(f\).
Proof. The map \(a:S^1\to S^1\) given by \(a(p)=-p\) is continuous, since it is the restriction of the continuous linear map \((x,y)\mapsto(-x,-y)\) on \(\mathbb{R}^2\). Thus \(f\circ a\) is continuous. The difference of two continuous real-valued functions is continuous, so \(D_f=f-f\circ a\) is continuous. For the symmetry, use the definition at \(-p\): \(D_f(-p)=f(-p)-f(-(-p))=f(-p)-f(p)=-\bigl(f(p)-f(-p)\bigr)=-D_f(p)\). This holds for every \(p\), including points where the difference is zero. \(\square\)
What Happens Along a Semicircle
A semicircle can be described by choosing a starting point \(p=(p_1,p_2)\in S^1\) and moving along the circle to \(-p\). The vector \(q=(-p_2,p_1)\) is a unit vector perpendicular to \(p\), so one such parametrization is \(\gamma(t)=(\cos t)p+(\sin t)q\), for \(0\le t\le\pi\). In particular, \(\gamma(0)=p\), \(\gamma(\pi)=-p\), and \(\gamma(t)\in S^1\) throughout: its squared norm is \(\cos^2t+\sin^2t=1\).
Proof. The endpoint identity is the proposition’s antipodal symmetry. The map \(\gamma:[0,\pi]\to S^1\) is continuous, and \(D_f\) is continuous, so \(D_f\circ\gamma:[0,\pi]\to\mathbb{R}\) is continuous. If \(D_f(p)>0\), then \(D_f(\gamma(0))>0\) and \(D_f(\gamma(\pi))=-D_f(p)<0\). If \(D_f(p)<0\), the first endpoint is negative and the second is positive. In either case the intermediate value theorem for continuous real-valued functions on an interval gives \(t_0\in(0,\pi)\) with \(D_f(\gamma(t_0))=0\). If instead \(D_f(p)=0\), then the symmetry gives \(D_f(-p)=0\) as well, so both endpoints are zeros. Thus in every case there is a zero on the closed semicircle; strict opposite signs at the endpoints are guaranteed when the endpoint difference is nonzero. \(\square\)
The zero has a direct interpretation. If \(D_f(z)=0\), then the definition gives \(f(z)-f(-z)=0\), hence \(f(z)=f(-z)\). This conclusion only says that the difference is zero somewhere on the semicircle; it does not say that there is exactly one zero, or that the difference has opposite strict signs at the endpoints when those values are already zero.
Worked Example: A Linear Function on the Circle
Let \(f(x,y)=x+2y\) on \(S^1\). This is continuous as the restriction of a linear function on \(\mathbb{R}^2\). Compute its antipodal difference and locate a zero on the semicircle parametrized by \(\gamma(t)=(\cos t,\sin t)\), \(0\le t\le\pi\).
For \((x,y)\in S^1\), the antipode is \((-x,-y)\). Substitution gives \(f(-x,-y)=-x-2y\), so \(D_f(x,y)=(x+2y)-(-x-2y)=2x+4y\). Along the chosen semicircle, this becomes \(D_f(\gamma(t))=2\cos t+4\sin t\). At \(t=0\) its value is \(2\cos 0+4\sin 0=2(1)+4(0)=2\). At \(t=\pi\) it is \(2\cos\pi+4\sin\pi=2(-1)+4(0)=-2\). The endpoint values are strictly opposite, as the theorem predicts.
To find a zero, solve \(2\cos t+4\sin t=0\), or \(\cos t+2\sin t=0\). At \(t=\pi-\arctan(1/2)\), the angle is in quadrant II and its trigonometric values are \(\cos t=-2/\sqrt{5}\) and \(\sin t=1/\sqrt{5}\). Substitution gives \(2(-2/\sqrt{5})+4(1/\sqrt{5})=0\). The corresponding point is \((-2/\sqrt{5},1/\sqrt{5})\), which lies on \(S^1\) because \(4/5+1/5=1\).
Check: At this point, \(f(-2/\sqrt{5},1/\sqrt{5})=-2/\sqrt{5}+2/\sqrt{5}=0\). Its antipode is \((2/\sqrt{5},-1/\sqrt{5})\), where \(f(2/\sqrt{5},-1/\sqrt{5})=2/\sqrt{5}-2/\sqrt{5}=0\). The two values agree, so their difference is indeed zero.
Worked Example: Zero Endpoints but Positive Interior Values
Let \(f(x,y)=x^2+2y\), again restricted to \(S^1\), and use the right-hand semicircle \(\gamma(t)=(\cos t,\sin t)\), \(0\le t\le\pi\). The function is continuous because it is the restriction of a polynomial.
At the antipode, \(f(-x,-y)=(-x)^2+2(-y)=x^2-2y\). Therefore \(D_f(x,y)=(x^2+2y)-(x^2-2y)=4y\). On the semicircle this is \(D_f(\gamma(t))=4\sin t\). At the endpoints the values are \(4\sin 0=0\) and \(4\sin\pi=0\). For every \(t\) with \(0<t<\pi\), \(\sin t>0\), so the difference is strictly positive at every interior point.
This does not contradict the theorem: the endpoint values are negatives of each other because each is zero. It also shows why the strongest correct statement distinguishes the zero-endpoint case from strict sign reversal. For instance, at \(t=\pi/2\), the point is \((0,1)\) and the difference is \(4\sin(\pi/2)=4\). Directly, \(f(0,1)=2\) and \(f(0,-1)=-2\), so the difference is \(2-(-2)=4\).
Check: The direct difference at the starting endpoint \((1,0)\) is \(f(1,0)-f(-1,0)=1-1=0\); at the other endpoint \((-1,0)\) it is also \(1-1=0\). These values match the formula \(4y\), since both endpoints have \(y=0\).
Worked Example: A Difference That Vanishes Everywhere
Let \(f(x,y)=x^2+y^2\) on \(S^1\). This is continuous. Since every point of \(S^1\) satisfies \(x^2+y^2=1\), the function has value \(1\) at every point. In particular, \(f(-x,-y)=(-x)^2+(-y)^2=x^2+y^2\), and thus \(D_f(x,y)=1-1=0\) for every \((x,y)\in S^1\).
Along any semicircle the antipodal difference is therefore zero at every parameter value, not merely at some point. For a numerical check, at \(p=(1,0)\) we have \(f(1,0)=1^2+0^2=1\) and \(f(-1,0)=(-1)^2+0^2=1\), giving difference \(1-1=0\). At \(p=(0,1)\), the values are also \(0^2+1^2=1\) and \(0^2+(-1)^2=1\), again giving difference \(1-1=0\).
A Common Sign Mistake
A frequent error is to claim that \(D_f(-p)=D_f(p)\), on the grounds that the same two function values appear in both differences. The order matters: at \(-p\), the value at \(-p\) comes first, not second. The correct computation is \(D_f(-p)=f(-p)-f(p)=-D_f(p)\).
For a concrete check, take \(f(x,y)=x\) and \(p=(1,0)\). Then \(D_f(1,0)=1-(-1)=2\), while \(D_f(-1,0)=-1-1=-2\), not \(2\). Thus the values are negatives, as the proposition states. Another overstatement is that the difference must strictly change sign on every semicircle. The example \(f(x,y)=x^2+y^2\) disproves that: its difference is zero everywhere. The precise conclusion is that opposite endpoint values are negatives; they are strictly opposite only when they are nonzero.
The argument is useful because it combines an algebraic symmetry with continuity along an interval. The symmetry determines the endpoint values, while the intermediate value theorem supplies a zero between strict opposite signs. For antipodal differences, the calculation is always the same; the shape of the original function affects where the zero occurs and whether there are additional zeros.
Check Your Understanding
Use the definition and the sign relation to answer each question. When computing a difference, keep the order \(f(p)-f(-p)\).
- For \(f(x,y)=3x-4y\) on \(S^1\), compute \(D_f(x,y)\) and its values at \((1,0)\) and \((-1,0)\).
- If \(D_f(p)=5\), what is \(D_f(-p)\), and what does the theorem guarantee along a semicircle from \(p\) to \(-p\)?
- For \(f(x,y)=y^2\) on \(S^1\), compute the antipodal difference. Does it have strict opposite signs at the endpoints of any semicircle?
- Explain why \(D_f(p)=0\) is equivalent to \(f(p)=f(-p)\).
- Can the difference be positive at every interior point of a semicircle whose endpoint values are both zero? Give an example from the tutorial and verify it from its formula.