Let f be a continuous real-valued function on R3. Suppose that for every sphere S of radius 1, the integral of f(x, y, z) over the surface of S equals 0. Must f(x, y, z) = 0 for all points (x, y, z)?
Let f be a continuous real-valued function on R3. Suppose that for every sphere S of radius 1, the integral of f(x, y, z) over the surface of S equals 0. Must f(x, y, z) = 0 for all points (x, y, z)?
Maybe my math is not the best, but wouldn't a simple function like f = cos(pi * z) work?
Say the sphere is centred at the origin. If I make this an integral in z, the area element is dA = 2*pi dz (the curvature of the sphere exactly compensates for the chord becoming smaller as we go from z=0 to z=1, so the area element is dA = 2*pi*cos(theta) / cos(theta), where cos(theta)=sqrt(1-z^2) and the first part is the radius of the circle defined by intersecting the sphere with a plane of constant z, and the second part is to take into account the slantedness of the area element relative to z)
Say the sphere center has coordinate z=alpha (the other coordinates don't matter). Now the integral becomes: integral from -1+alpha to 1+alpha of 2*pi*f(z). or, equivalently: integral from -1 to 1 of 2*pi*f(z - alpha).
And now it's clear that whatever the center of the sphere, this integral is zero for our function f = cos(pi*z) because of its periodicity.
This answer seems suspiciously simple, so I'm ready to be corrected here
This is correct
Yay! What do I get?
It's a question from a math competition, so if you'd answered it then you would get something