Assume a minimal surface $\Sigma$ has boundary on the unit sphere in the Euclidean space and $r$ is the distance from $\Sigma$ to the center of the ball. Is it true that $$\mathop{\rm area} \Sigma\ge \pi\cdot(1-r^2).$$


  • The problem is solved in all dimensions and codimension, see "Area bounds for minimal..." by Brendle and Hung in 2016. (Thanks Rbega for the ref.)

  • If $r=0$, the statement follows directly from the monotonicity formula.

  • If $\Sigma$ is topological disc the answer is YES, see answer of Oleg Eroshkin below.
  • The general question is formulated as a conjecture in 1975 --- see comment of Ian Agol.
  • There is an analog in all dimension and codimension for area minimizing surfaces, see Alexander, H.; Hoffman, D.; Osserman, R. Area estimates for submanifolds of Euclidean space. 1974.
  • 2
    $\begingroup$ A little more background would be helpful. Why do you want to know this? $\endgroup$ – Scott Morrison Nov 3 '09 at 18:02
  • 6
    $\begingroup$ This question is Conjecture 2 of: books.google.com/… $\endgroup$ – Ian Agol Nov 22 '09 at 16:26

This has just been solved (in full generality) by Brendle and Hung using the first variation formula together with a clever (if mysterious) choice of vector field.


This result (and several similar) proved in a nice paper Alexander, H.; Osserman, R. "Area bounds for various classes of surfaces." Amer. J. Math. 97 (1975), no. 3, 753--769.

  • $\begingroup$ Only for disc... but that is already a lot. !!!THANK YOU VERY MUCH!!! $\endgroup$ – Anton Petrunin Nov 15 '09 at 4:56
  • $\begingroup$ The problem was solved --- so I marked the latter answer as accepted. $\endgroup$ – Anton Petrunin Jul 20 '16 at 10:37

One obvious observation (of which you are probably already aware) is that if the boundary of the surface is connected, it must have length at least $2\pi\sqrt{1-r^2}$, or else it is contained in a lune whose convex hull does not contain a point at distance $r$ from the center. In the very special case that your surface is a topological disk transverse to a foliation of the ball by concentric spheres, the coarea formula (obtained by integrating the lengths of the intersection of your surface with concentric spheres, and using this observation) gives an estimate for the area, but a quick calculation shows that it is not good enough to prove what you want.

  • $\begingroup$ It seems you only use that $\Sigma$ is saddle. This way you will only get that length $\ge 4\cdot\arccos r$ and you can not make it better. $\endgroup$ – Anton Petrunin Nov 13 '09 at 13:30

I think there is such a minimal surface with area less than pi(1-r^2). Look at the catenoid formed by rotation of y a(cosh a) about the x axis where a is small the the distance from the origin will be 2a or more with the minimum at x=0. Now it will intersect the circle at a value of x less than (ln (a^-1))^2 actually well before that but clearly e^((log a)^2) is a^(ln a) (note this is a small number to a large negative power which gives a large number which is bigger than 2/a which is all I need (note I am also overestimating the length by assuming the radius where the catenoid intersects the sphere is one) I estimate the area with two cones using the formula 2(pi)rs which will have a greater area than the minimal surface and get an upper bound of 2(ln a)^2 - a small factor due to the circle at the origin which I ignore since the number is small enough already not to mention I am already grossly overestimating the length. So I get an upper bound of the area of 2(ln a)^2 which is less than pi(1-4a^2).

  • $\begingroup$ (1) Did you meant that you have example for all small $a$? - (2) guess you mean rotation of $y=a\cosh ax$ around $x$-axis? Then $r=a$, but not $2a$(?) (3) In the very end, $2(\ln a)^2$ is huge number while $\pi(1-4a^2)$ is small --- not other way arround (?) (4) I see that as $a\to0$ the we have that $area-\pi\cdot(1-a^2)$ is very small, but it seems that it is still positive... $\endgroup$ – Anton Petrunin Nov 12 '09 at 4:27
  • $\begingroup$ As it is written it does not work. I am still looking for ways to fix it by rotating y=acosh(ax) around x-axis and then translating vertically and horizontally by various amounts. It is still unclear to me if this works and the translations introduce difficulties in computation. $\endgroup$ – Kristal Cantwell Nov 12 '09 at 18:11

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