## Proof of the isoperimetric inequality 2

I will continue the series of proofs for the isoperimetric inequality in the two dimensional case, i.e. if a simple closed curve (which we suppose for simplicity that it consists of piecewise curves) of length encloses an area then and the equality is attained only if is the boundary of a circle.

The proof I give here assumes again the existence of the optimal shape, a non-trivial fact proved here. Let’s make a few remarks about the optimal shape. Like in the proof presented here, it is reasonable to assume that the plane region enclosed by denoted by is convex.

One of the basic applications of continuity to plane geometry to see that for any direction in the plane there exists a line parallel to this direction which splits in two region of equal areas. To do this, take a ‘sweep’ of the plane by a line parallel to , i.e. choose a coordinate system in the plane such that is included in the halfplane , and denote , where is the Lebesgue measure. It can be proved that is in fact Lipschitz continuous (note that is considered to be bounded). Since and , the continuity of implies that there exists such that , and the line has the desired property.

Using the fact presented above and another continuity argument we can prove that there exists a line which splits into two regions which have the same area and the same perimeter. For every direction which we can characterize by the angle which this direction makes with the positive side of the axis there exists a line which splits in two regions of equal areas. For such a line we denote by and the perimeter of on one side and the other of . We choose the and signs such that the orientation is consistent with the case . Then we can define by

a function which is again continuous. We note that since after a rotation by the line support of the direction is the same, the direction points the other way. Therefore we can find a such that and we are done.

Using this trick we can choose this line which splits into two parts with equal areas and equal perimeters and reduce the problem to proving that half the optimal shape is half a circle. For this denote the intersection of with and pick one half of . Suppose that this half is not a half-circle. Then we can find a point on the boundary (not on ) such that the angle is less or more than . The area of is given by the formula

so if we ‘glue’ the portions of to the sides , and consider a hinge in such that these regions are mobile (see the figure). If we move the considered regions until then we have increased the area preserving the perimeter, and by the symmetry by we get a contradiction to the fact that is optimal.

This contradiction shows that half of is half a circle, and therefore itself is a circle.