I remember seeing somewhere that for every connected compact set $\Omega$ in $\mathbb{R}^2$ with piecewise $C^1$ boundary we have $$A(\Omega_r)\leq A(\Omega)+L(\partial \Omega)r+ \pi r^2,$$ where $$\Omega_r=\{x\in \mathbb{R}^2: d(x,\Omega)\leq r\},$$ $A$ denotes the area and $L$ the length. I tried to find a reference for this inequality, but I only found Steiner's formula which states that equality holds when $\Omega$ is convex. Can someone please give me a reference for that inequality?


4 Answers 4


In the first note to section 4.2 of

Schneider, Rolf, Convex bodies: the Brunn-Minkowski theory, Encyclopedia of Mathematics and Its Applications. 44. Cambridge: Cambridge University Press. xiii, 490 p. (1993). ZBL0798.52001.

Rolf Schneider cites several articles with generalizations of the Steiner formula. Of these, the following two:

Sz.-Nagy, Béla, Über Parallelmengen nichtkonvexer ebener Bereiche, Acta Sci. Math. 20, 36-47 (1959). ZBL0101.14701.

Makai, E, Steiner type inequalities in plane geometry, Period. Polytech. Elec. Engrg. 3, 345-355 (1959). link to pdf

prove your inequality for simply-connected domains. For non simply-connected it follows for example by cutting the domain into pieces.

Makai also cites

Hadwiger, Hugo, Die erweiterten Steinerschen Formeln für ebene und sphärische Bereiche, Comment. Math. Helv. 18, 59-72 (1945). ZBL0063.01850.

(and a four-page article in Revista Hispano-Americana, of which I found no pdf, but according to MathReviews it contains the same result as the previous article of Hadwiger) as the first reference to this inequality. However, I have not found this result there. It speaks only about the Steiner formula for inner and outer bodies for $r$ within the injectivity radius.


Theorem 1. If $\Omega\subset\mathbb{R}^2$ is a bounded $C^2$ domain, $\Omega_r$ is defined as in the question, then there is $r_0>0$ such that $$ \operatorname{A}(\Omega_r)= \operatorname{A}(\Omega)+L(\partial\Omega)r+\pi r^2\chi(\Omega), \quad \text{for $0\leq r\leq r_0$} $$ where $\chi(\Omega)$ is the Euler characteristic of $\Omega$.

Remark. If $\Omega$ has no holes, then $\chi(\Omega)=1.$ For example if $\Omega=B(R)$ is a disc of radius $R$, then $\chi(B)=1$ and the above equality reads as: $$ \pi(R+r)^2=\pi R^2+2\pi Rr+\pi r^2 $$ which is trivially true.

Proof. The proof will be in the case in which $\Omega$ is simply connected i.e. it has no holes so $\chi(\Omega)=1$. If $\Omega$ has a holes (finitely many because of the regularity of the boundary), then one can apply the formula to each of the holes (which is a simply connected domain) and the general case will follow.

Let $\alpha:[0,L]\to\partial\Omega$, $L=L(\partial\Omega)$ be a counter-clockwise parametrization. Then $\alpha'(s)=T(s)$ is the unit tangent vector, Let $N(s)$ be the unit normal vector so that $\langle T(s),N(s)\rangle$ is a positively oriented basis of $\mathbb{R}^2$. It is well known (Frenet formulas) that $$ T'(s)=\kappa(s)N(s), \quad N'(s)=-\kappa(s)T(s), $$ where $\kappa(s)$ is the oriented curvature of $\alpha$ at $s$.

Consider the mapping $$ \Phi:[0,L]\times [0,r]\to\mathbb{R}^2, \quad \Phi(s,t)=\alpha(s)-tN(s). $$ Note that due to the choice of the orientation $N(s)$ is the inner normal and $-N(s)$ is the outer normal so for small values of $r$, $\Phi$ will map $[0,L]\times[0,r]$ outside of $\Omega$.

We have $\Phi_s=\alpha'(s)-tN'(s)=T(s)+t\kappa(s)T(s)$, $\Phi_t=N(s)$. Thus the Jacobian of $\Phi$ equals $$ J\Phi=\det D\Phi=1+t\kappa(s) $$ as it is equal to the oriented area of the rectangle with sides $\Phi_s$ and $\Phi_t$. If $r$ is small, $J\Phi>0$ and it follows that there is $r_0>0$ such that $\Phi$ is a diffeomorphism when $0<r\leq r_0$. Clearly, for $0<r\leq r_0$, $\Phi$ maps $[0,L]\times [0,r]$ onto $\Omega_r\setminus\Omega$ (because the distance of points in $\Omega_r\setminus\Omega$ to $\Omega$ is along the normal vector). Hence the change of variables formula shows that the area of $\Omega_r\setminus\Omega$ equals $$ A(\Omega_r\setminus\Omega)=\int_0^L\int_0^r J\Phi= Lr+\frac{r^2}{2}\int_0^L\kappa(s)\, ds $$ and the result readily follows from the theorem on turning tangents according to which $$ \int_0^L\kappa(s)\, ds=2\pi. $$ The theorem on turning tangents is a special case of the Gauss-Bonnet theorem. $\Box$

Theorem 2. If $\Omega$ is bounded and convex, then $$ \operatorname{A}(\Omega_r)= \operatorname{A}(\Omega)+L(\partial\Omega)r+\pi r^2, \quad \text{for all $r>0$.} $$

Proof. If $\Omega$ is convex and $C^2$, then the mapping $\Phi$ is a diffeomorphism for all $r>0$ and we do not need to restrict it by $r_0$. If seems that for general convex domains we can approximate them by $C^2$ convex so that we approximate both the area and the length of the boundary.

The above result (Theorem 1) is in [3] pp. 10-11. I haven't read the proof from [3], but it seems similar. The book [3] contains many generalizations of the formula, including higher dimensional ones. Another reference is [2]. See also the introduction in [1]. The paper [1] is available on arXiv and if you know how to look for books in the Internet, you can find the books [2,3] online too.

[1] Z. M. Balogh, F. Ferrari, B. Franchi, E. Vecchi, K. Wildrick, Steiner's formula in the Heisenberg group. Nonlinear Anal. 126 (2015), 201–217.

[2] Yu. D. Burago, V. A. Zalgaller, Geometric inequalities. Grundlehren der Mathematischen Wissenschaften, 285. Springer Series in Soviet Mathematics. Springer-Verlag, Berlin, 1988.

[3] A. Gray, Tubes. Second edition. Progress in Mathematics, 221. Birkhäuser Verlag, Basel, 2004.

  • 2
    $\begingroup$ Declaring the identity for all finite positive $r$ is a bit too much: once we reach the minimal outer radius of curvature for the boundary, the regularity of the equidistant is lost and overlaps begin. $\endgroup$
    – fedja
    Jun 18, 2018 at 2:59
  • $\begingroup$ @PiotrHajlasz Is the inequality I asked true in general? $\endgroup$
    – Michael
    Jun 18, 2018 at 12:24
  • 1
    $\begingroup$ To comment on the comment of fedja, the injectivity radius can also be zero. Intuitively, the $\epsilon$-neighborhood of the set starts to overlap, which is only good for our inequality, but I don't see if this can easily be formalized. By lifting the neighborhood to the tangent bundle of the Euclidean space, where it is non self-intersecting, computing the volume there, and projecting back? $\endgroup$ Jun 18, 2018 at 15:14
  • 1
    $\begingroup$ @fedja Are you satisfied with my answer now? $\endgroup$ Jun 18, 2018 at 17:40
  • 2
    $\begingroup$ @PiotrHajlasz Sure, I am :-) $\endgroup$
    – fedja
    Jun 18, 2018 at 17:50

Rather than editing my previous answer I prefer to write it as a new answer since it is very different from the previous one.

Theorem (Steiner's inequality). Let $\Omega\subset\mathbb{R}^2$ be a closed and bounded set with piecewise $C^1$ boundary. Then for all $r>0$ we have $$ A(\Omega_r)\leq A(\Omega)+L(\partial\Omega)r+\pi r^2, \quad L(\partial\Omega_r)\leq L(\partial\Omega)+2\pi r. $$ If $\Omega$ is convex, then the inequalities are equalities.

You can find a proof in: A. Treibergs, Inequalities that Imply the Isoperimetric Inequality. On page 14 it is proved for general domains by approximating them by polygons. The case of polygons is proved later, at the end of the paper on page 29.


One more reference (making no smoothness assumptions on $\partial \Omega$) is

G. Fast. Area of a generalized circle as a function of its radius. I, II (in Russian). Fund. Math. 46:137--163, 1959.

I tried to find more modern results than this one and those cited by Ivan Izmestiev. I expected to get an answer from

D. Hug, G. Last and W. Weil (2004) A local Steiner-type formula for general closed sets and applications. Mathematische Zeitschrift, 246, 237-272.

However, after discussing with the authors, we became convinced that Steiner's inequality does NOT follow easily from the results of this paper.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.