I’m having trouble understanding why we can’t just fix n=2 in your proof. Then at each iteration we bisect the interval, so we wouldn’t be using the “full power” of the 1-D Sperner’s lemma (we would just be using something close to the base case).
Also if we are only given that f is continuous, does it make sense to talk about the gradient?
“I’m having trouble understanding why we can’t just fix n=2 in your proof. Then at each iteration we bisect the interval, so we wouldn’t be using the “full power” of the 1-D Sperner’s lemma (we would just be using something close to the base case).”—You’re right, you can prove this without using the full power of Sperner’s lemma. I think it becomes more useful for the multi-dimensional case.
Yeah agreed, in fact I don’t think you even need to continually bisect, you can just increase n indefinitely. Iterating becomes more dangerous as you move to higher dimensions because an n dimensional simplex with n+1 colours that has been coloured according to analogous rules doesn’t necessarily contain the point that maps to zero.
On the second point, yes I’d been assuming that a bounded function had a bounded gradient, which certainly isn’t true for say sin(x^2), the final step needs more work, I like the way you did it in the proof below.
I hit that stumbling block as well. I handwaved it by saying “continue iterating until you have x(k,B) and x(k,G) such that f(xk,B)<0, f(xk,G)≥0, and f has no local maxima or local minima on the open interval (xk,B,xk,G)”, but that doesn’t work for the Weierstrass function, which will (I believe) never meet that criterion.
I’m having trouble understanding why we can’t just fix n=2 in your proof. Then at each iteration we bisect the interval, so we wouldn’t be using the “full power” of the 1-D Sperner’s lemma (we would just be using something close to the base case).
Also if we are only given that f is continuous, does it make sense to talk about the gradient?
“I’m having trouble understanding why we can’t just fix n=2 in your proof. Then at each iteration we bisect the interval, so we wouldn’t be using the “full power” of the 1-D Sperner’s lemma (we would just be using something close to the base case).”—You’re right, you can prove this without using the full power of Sperner’s lemma. I think it becomes more useful for the multi-dimensional case.
Yeah agreed, in fact I don’t think you even need to continually bisect, you can just increase n indefinitely. Iterating becomes more dangerous as you move to higher dimensions because an n dimensional simplex with n+1 colours that has been coloured according to analogous rules doesn’t necessarily contain the point that maps to zero.
On the second point, yes I’d been assuming that a bounded function had a bounded gradient, which certainly isn’t true for say sin(x^2), the final step needs more work, I like the way you did it in the proof below.
I hit that stumbling block as well. I handwaved it by saying “continue iterating until you have x(k,B) and x(k,G) such that f(xk,B)<0, f(xk,G)≥0, and f has no local maxima or local minima on the open interval (xk,B,xk,G)”, but that doesn’t work for the Weierstrass function, which will (I believe) never meet that criterion.