Pu's original proof relies on the uniformization theorem and employs an averaging argument, as follows. By uniformization, the Riemannian surface is conformally diffeomorphic to a round projective plane. This means that we may assume that the surface is obtained from the Euclidean unit sphere by identifying antipodal points, and the Riemannian length element at each point is where is the Euclidean length element and the function, called the conformal factor, satisfies. More precisely, the universal cover of is, a loop is noncontractible if and only if its lift goes from one point to its opposite, and the length of each curve is Subject to the restriction that each of these lengths is at least, we want to find an that minimizes the where is the upper half of the sphere. A key observation is that if we average several different that satisfy the length restriction and have the same area, then we obtain a better conformal factor, that also satisfies the length restriction and has and the inequality is strict unless the functions are equal. A way to improve any non-constant is to obtain the different functions from using rotations of the sphere, defining. If we average over all possible rotations, then we get an that is constant over all the sphere. We can further reduce this constant to minimum value allowed by the length restriction. Then we obtain the obtain the unique metric that attains the minimum area.
Reformulation
Alternatively, every metric on the sphere invariant under the antipodal map admits a pair of opposite points at Riemannian distance satisfying A more detailed explanation of this viewpoint may be found at the page Introduction to systolic geometry.
Filling area conjecture
An alternative formulation of Pu's inequality is the following. Of all possible fillings of the Riemannian circle of length by a -dimensional disk with the strongly isometric property, the round hemisphere has the least area. To explain this formulation, we start with the observation that the equatorial circle of the unit -sphere is a Riemannian circle of length. More precisely, the Riemannian distance function of is induced from the ambient Riemannian distance on the sphere. Note that this property is not satisfied by the standard imbedding of the unit circle in the Euclidean plane. Indeed, the Euclidean distance between a pair of opposite points of the circle is only, whereas in the Riemannian circle it is. We consider all fillings of by a -dimensional disk, such that the metric induced by the inclusion of the circle as the boundary of the disk is the Riemannian metric of a circle of length. The inclusion of the circle as the boundary is then called a strongly isometric imbedding of the circle. Gromov conjectured that the round hemisphere gives the "best" way of filling the circle even when the filling surface is allowed to have positive genus.
Isoperimetric inequality
Pu's inequality bears a curious resemblance to the classical isoperimetric inequality for Jordan curves in the plane, where is the length of the curve while is the area of the region it bounds. Namely, in both cases a 2-dimensional quantity is bounded by a 1-dimensional quantity. However, the inequality goes in the opposite direction. Thus, Pu's inequality can be thought of as an "opposite" isoperimetric inequality.
