REGULARITY OF OPTIMAL CONFIGURATIONS FOR A MODEL WITH
PERIMETER-TO-AREA RATIO PENALIZATION
XIN YANG LU
Abstract. In this paper we consider the functional
Z
Ep,λ (R) :=
distp (x, ∂R) dx + λ
R
H1 (∂R)
.
H2 (R)
Here p ≥ 1, λ > 0 are given parameters, the unknown R varies among compact, convex, Hausdorff
2
two-dimensional
R sets pof R , ∂R denotes the boundary of R, and dist(x, ∂R) := inf y∈∂R |x−y|. The
integral term R dist (x, ∂R) dx quantifies the “easiness” for points in R to reach the boundary,
1
while HH2(∂R)
is the perimeter-to-area ratio. The main aim is to prove existence and C 1,1 -regularity
(R)
of minimizers of Ep,λ .
Keywords. perimeter-to-area ratio, regularity
Classification. 49Q20, 49K10, 35B65
1. Introduction
The perimeter-to-area ratio (in 2D), or surface area-to-volume ratio (in 3D), plays a crucial
role in many processes. The size of prokaryote cells is limited by the efficiency of diffusion
processes, fundamental to transport nutrients across the cell, and strongly correlated with the
surface area-to-volume ratio. The large surface area-to-volume ratio also gives prokaryote cells a
high metabolic rate, fast growth, and short lifespan compared to eukaryote cells (see for instance
[9]).
The size of animals, for a given shape, is also limited by the efficiency of body cooling: beyond
a certain size, the surface area-to-volume ratio is too small to ensure effective body cooling
([16, 19]). This is one reason why animals of significantly different size (e.g., cats and elephants)
have very different shapes.
In chemistry, higher surface area-to-volume ratio increases the typical speed of chemical reactions. For instance, this leads to dust explosions, when dust particles of seemingly non-flammable
materials (e.g., aluminum, sugar, flour, etc.) can be ignited due to the significant surface areato-volume ratio ([10, 13]).
In this paper we will focus on the 2D case. In the above examples, another relevant quantity
is the “easiness” to access the boundary: a body whose shape approximates a Koch’s snowflake
(Figure 1) may have arbitrarily large perimeter-to-area ratio, but the “average-distance” of its
1
2
XIN YANG LU
Figure 1. This shape has very large perimeter-to-area ratio, but most of its points are
still “far away” from the boundary...
points to the boundary is also quite large. Hence such a shape would still exhibit drawbacks of
shapes with low perimeter-to-area ratio (e.g., inefficient waste heat disposal, difficult transport
of nutrients, etc.).
A term quantifying the “average distance” of points to the boundary is thus required. One
possible choice is
Z
Fp (R) :=
distp (x, ∂R) dx,
dist(x, ∂R) := inf |x − y|,
y∈∂R
R
where p ≥ 1 is a given parameter, and | · | denotes the Euclidean distance.
The main functional of this paper is
H1 (∂R)
Ep,λ (R) := Fp (R) + λ 2
=
H (R)
Z
distp (x, ∂R) dx + λ
R
H1 (∂R)
,
H2 (R)
(1)
where the argument R varies among compact, convex, Hausdorff two-dimensional sets of R2 ,
1
and ∂R denotes the boundary of R. The term HH2(∂R)
is thus the perimeter-to-area ratio. Note
(R)
1
2
that neither the perimeter H (∂R), nor the area H (R), is penalized, only their ratio is. This
makes compactness results quite challenging to prove, and several estimates (in Section 2) will be
required. The role of convexity is to ensure crucial compactness estimates (Lemmas 2.4 and 3.1).
Note that Ep,λ is invariant under rigid movements. Further details about the space of convex
sets, and its topology, will be discussed in Section 2. The main result is:
Theorem 1.1. For any λ > 0, p ≥ 1, the following assertions hold:
(1) Ep,λ admits a minimizer in X (defined in (2) below).
PERIMETER-AREA RATIO PENALIZATION
3
(2) All minimizers are compact, convex, C 1,1 -regular sets, with Hausdorff dimension equal to
2.
(3) The perimeter-to-area ratio of any minimizer R satisfies
p+2
H1 (∂R)
=
min Ep,λ .
2
H (R)
λ(p + 3) X
Here, and for future reference, the expression “R is C k -regular” means that its boundary ∂R
is C k -regular, i.e., ∂R admits a C k -regular parameterization.
Note that the functional Fp is only superficially similar to the average-distance functional
Σ 7→
Z
distp (x, Σ) dµ,
Ω
where Ω is a given domain, µ a given measure on Ω, and Σ varies among compact, path-wise
connected sets with Hausdorff dimension equal to 1. The average-distance functional has been
widely studied, and used in several modeling problems. For a (non exhaustive) list of references,
we cite the papers (and books) by Buttazzo and collaborators [1, 2, 3, 4, 5, 6, 7, 8]. Also related
are the papers by Paolini and Stepanov [14], Santambrogio and Tilli [15], Tilli [18], Lemenant
and Mainini [12], Slepčev [17], and the review paper by Lemenant [11].
A crucial difference is that the domain of integration of Fp is the unknown R itself, while, in the
case of average-distance functional, the domain was always given. This is a critical issue, since
key compactness results (used to prove the existence of minimizers) will require a completely
different approach.
This paper is structured as follows:
(1) in Section 2, we provide crucial estimates on the area (Lemma 2.2 and Corollary 2.3) and
perimeter (Lemma 2.4) of elements of minimizing sequences.
(2) In Section 3, we prove Theorem 1.1.
2. Preliminary results
In this section we present the main setting of the problem, and prove some auxiliary results.
Set
X := {R : R is convex, compact, Hausdorff 2-dimensional set of R2 },
(2)
and endow X with the distance
d(R1 , R2 ) := H2 (R1 4R2 ),
(3)
where 4 denotes the symmetric difference. Since X is clearly not complete in the metric d (e.g.,
a sequence of rectangles may shrink to a point), we set
X̄ := completion of X with respect to d.
It is straightforward to check that, given a sequence Rn ⊆ X converging to R ∈ X̄ in the metric
d, it holds:
4
XIN YANG LU
(1) R is convex,
(2) R can be unbounded,
(3) the Hausdorff dimension of R can be strictly less than 2.
Thus all elements of X̄\X are either unbounded, or have Hausdorff dimension strictly less than
H1 (∂R)
2. In particular, we define the ratio of
of an element R ∈ X̄\X as follows:
H2 (R)
H1 (∂R)
:= +∞. This is because iso-perimetric inequality gives that
H2 (R)
1
among all convex sets with given area, the ratio HH2(∂R)
is minimum when R is a circle,
(R)
(1) if H2 (R) = 0, then
√
π
.
H2 (R)
where it attains the value √2
Thus
√
H1 (∂R)
2 π
p
≥
H2 (R)
H2 (R)
H2 (R)→0
−→
+∞.
H1 (∂R)
:= +∞.
H2 (R)
H1 (∂R)
is already well-defined.
(3) If H1 (∂R) < +∞ and 0 < H2 (R) < +∞, then the ratio
H2 (R)
(4) Undetermined if H1 (∂R) = H2 (R) = +∞. However, this will not be an issue, since (see
Lemma 2.4 below) we can consider only sets R with H1 (∂R) < +∞.
(5) Case H1 (∂R) < +∞, H2 (R) = +∞ is impossible due to the iso-perimetric inequality.
(2) If H1 (∂R) = +∞ and H2 (R) < +∞, then
We will prove the existence of minimizers of Ep,λ in X (instead of just X̄). This is one of the
key difficulties, since X is not closed with respect to d. To achieve this, the key point is to prove
an uniform a priori upper bound on the perimeter (Lemma 2.4), and lower bound on the area
(Lemma 2.2 and Corollary 2.3).
Lemma 2.1. Given p ≥ 1, λ > 0, and a sequence Rn ⊆ X converging to R ∈ X̄ with respect to
d, such that
R∪
[
Rn ⊆ K
n
for some compact set K, then it holds:
H2 (R) = lim H2 (Rn ),
(4)
H1 (∂R) ≤ lim inf H1 (∂Rn ),
(5)
n→+∞
n→+∞
Z
R
distp (x, ∂R) dx = lim
Z
n→+∞ Rn
distp (x, ∂Rn ) dx.
Proof. Estimate (4) follows straightforwardly from the definition of the metric d.
(6)
PERIMETER-AREA RATIO PENALIZATION
5
To prove (5), recall that the perimeter H1 (∂Rn ) is the total variation of the characteristic
function of Rn . Convergence Rn → R with respect to d implies
1Rn → 1R strongly in L1 (R2 ),
with “1” denoting the characteristic function of the subscribed set. Thus (5) follows from the
lower-semicontinuity of the total variation semi-norm.
To prove (6), note that
Z
distp (x, ∂Rn ) dx =
Rn
Z
Z
Rn \R
Z
p
R
Rn ∩R
Z
p
dist (x, ∂R) dx =
Z
distp (x, ∂Rn ) dx +
distp (x, ∂Rn ) dx
distp (x, ∂R) dx,
dist (x, ∂R) dx +
Rn ∩R
R\Rn
hence
Z
distp (x, ∂Rn ) dx −
Rn
Z
R
≤
Z
distp (x, ∂R) dx
Z
p
Rn \R
Z
+
Rn ∩R
dist (x, ∂Rn ) dx +
distp (x, ∂R) dx
(7)
R\Rn
| distp (x, ∂Rn ) − distp (x, ∂R)| dx.
(8)
For arbitrary set H, denote by
diam H := sup |x − y|,
x,y∈H
and observe that
Z
Rn \R
distp (x, ∂Rn ) dx ≤ H2 (Rn \R) diam Rn ≤ H2 (Rn \R) diam K → 0,
Z
distp (x, ∂R) dx ≤ H2 (R\Rn ) diam R ≤ H2 (R\Rn ) diam K → 0,
R\Rn
hence the sum in (7) goes to zero. To estimate (8), denote by dH the Hausdorff distance, and
note that, by the Mean Value theorem, it holds
Z
Rn ∩R
| distp (x, ∂Rn ) − distp (x, ∂R)| dx
≤
Z
Rn ∩R
| dist(x, ∂Rn ) − dist(x, ∂R)|
· p sup
max{dist(x, ∂Rn ), dist(x, ∂R)}
p−1
dx
x∈Rn ∩R
p−1
≤ H2 (Rn ∩ R)dH (∂Rn , ∂R) · p max{diam Rn , diam R}
≤ H2 (K)dH (∂Rn , ∂R) · p(diam K)p−1 → 0.
Thus the term in (8) goes to zero too, and (6) is proven.
6
XIN YANG LU
Lemma 2.2 and Corollary 2.3 will give a lower bound on the H2 -measure of elements of minimizing sequences.
Lemma 2.2. Given p ≥ 1 and λ > 0, for any R ∈ X it holds
H2 (R) ≥
4πλ2
.
Ep,λ (R)2
Proof. Consider an arbitrary R ∈ X. By the iso-perimetric inequality, among all convex sets
with area
H2 (R), the perimeter-to-area ratio is minimum for a circle, where it attains the value
√ p 2
2 π/ H (R). Hence
√
2λ π
H1 (∂R)
p
≤
λ
≤ Ep,λ (R),
H2 (R)
H2 (R)
concluding the proof.
Corollary 2.3. Given p ≥ 1, λ > 0, any minimizing sequence Rn ⊆ X satisfies
H2 (Rn ) ≥ 4πλ2
2π
+ 2λ + 1
2
p + 3p + 2
−2
=: Ap,λ ,
H1 (∂Rn )
1
2π
≤
+ 2λ + 1 =: %p,λ ,
2
2
H (Rn )
λ p + 3p + 2
(9)
(10)
for any sufficiently large n.
Proof. First we prove inf X Ep,λ < +∞. Let B1 ∈ X be a circle of radius 1. Direct computation
gives
H1 (∂B1 )
H2 (B1 )
B1
Z 1
2π
(1 − r)p r dr + 2λ = 2
= 2π
+ 2λ < +∞.
p + 3p + 2
0
inf Ep,λ ≤ Ep,λ (B1 ) =
X
Z
distp (x, ∂B1 ) dx + λ
(11)
Thus, given a minimizing sequence Rn ⊆ X, there exists n0 such that for any n ≥ n0 it holds
Ep,λ (Rn ) ≤
p2
2π
+ 2λ + 1,
+ 3p + 2
(12)
and Lemma 2.2 gives
inf H2 (Rn ) ≥ 4πλ2
n≥n0
2π
+ 2λ + 1
2
p + 3p + 2
−2
,
hence (9). To prove (10), note that (12) forces
H1 (∂Rn )
2π
+
2λ
+
1
≥
E
(R
)
≥
λ
,
n
p,λ
p2 + 3p + 2
H2 (Rn )
concluding the proof.
PERIMETER-AREA RATIO PENALIZATION
7
Remark. Corollary 2.3 ensures that any minimizing sequence is definitely composed of sets
whose areas are uniformly bounded away from zero, thus any (potential) minimizer of Ep,λ has
Hausdorff dimension equal to 2.
Lemma 2.4. Given p ≥ 1 and λ > 0, for any minimizing sequence Rn ⊆ X, it holds
1
H (∂Rn ) ≤ (p +
1)%p+1
p,λ
2π
+ 2λ + 1 =: Pp,λ
2
p + 3p + 2
(13)
for any sufficiently large n, with %p,λ defined in (10).
Proof. We first claim:
• for any R ∈ X it holds
Z
1 H2 (R)p+1
.
p + 1 H1 (∂R)p
distp (x, ∂R) dx ≥
R
(14)
Consider an arbitrary R ∈ X. For any z ∈ ∂R, define
`z := {x ∈ R : dist(x, ∂R) = |x − z|} ⊆ R.
Since R is convex, it is straightforward to check that `z is a segment with an endpoint in
z. Moreover, given z1 , z2 ∈ ∂R, z1 6= z2 , the intersection `z1 ∩ `z2 is at most a singleton.
Thus the area H2 (R) can be written as “the area of the union of all segments `z ”. That
is,
H2 (R) =
Z
H1 (`z ) d`z .
(15)
z∈∂R
Let γ : [0, H1 (`z )] −→ `z be an arc-length parameterization, with γ(0) = z, and note that
Z
`z
1
|x − z|p dHx`
(x) =
z
Z H1 (`z )
sp ds =
0
H1 (`z )p+1
.
p+1
By Hölder’s inequality it follows
Z
distp (x, ∂R) dx =
Z
R
Z
z∈∂R `z
1
=
p+1
Z
1
≥
p+1
Z
1
|x − z|p dHx`
(x) d`z
z
H1 (`z )p+1 d`z
z∈∂R
H (`z ) d`z
This proves (14).
H1 (∂R)−p
z∈∂R
1
H2 (R)
=
p + 1 H1 (∂R)
(15)
p+1
1
p+1
H1 (∂R).
(16)
8
XIN YANG LU
Now consider an arbitrary minimizing sequence Rn ⊆ X. Estimate (11) thus gives
2π
+ 2λ + 1
Ep,λ (Rn ) ≤ 2
p + 3p + 2
for all sufficiently large n. Corollary 2.3 gives also
(17)
H1 (∂Rn )
H2 (Rn )
≤
%
<
+∞
=⇒
≥ %−1
p,λ
p,λ > 0,
H2 (Rn )
H1 (∂Rn )
hence (16) gives
%−p−1
p,λ
H2 (Rn )
1
H (∂Rn ) ≤
p+1
p + 1 H1 (∂Rn )
1
(17)
≤ Ep,λ (Rn ) ≤
p2
p+1
1
H (∂Rn ) ≤
Z
distp (x, ∂Rn ) dx
Rn
2π
+ 2λ + 1,
+ 3p + 2
concluding the proof.
3. Proof of Theorem 1.1
Now we are ready to prove point (1) of Theorem 1.1, i.e., the existence of minimizers in X
(instead of just X̄).
Lemma 3.1. For any p ≥ 1, λ > 0, the functional Ep,λ admits a minimizer R ∈ X, which
satisfies:
H2 (R) ≥ Ap,λ ,
H1 (∂R) ≤ Pp,λ ,
with Ap,λ (resp. Pp,λ ) defined in (9) (resp. (13)).
Proof. Consider a minimizing sequence Rn ⊆ X (since X is dense in X̄, this is not restrictive),
i.e., Ep,λ (Rn ) & inf X̄ E. Since Ep,λ is invariant under rigid movements, upon translation, we can
assume Rn 3 (0, 0) for any n. Estimate (11) allows to assume, without loss of generality,
2π
+ 2λ + 1.
(∀n)
Ep,λ (Rn ) ≤ 2
p + 3p + 2
Lemma 2.4 gives H1 (∂Rn ) ≤ Pp,λ . Note also that
diam Rn ≤ H1 (∂Rn ).
By our assumption (without loss of generality) that Rn 3 (0, 0) for any n, we conclude
(∀n)
Rn ⊆ {z ∈ R2 : kzk ≤ Pp,λ },
(18)
i.e., Rn are all contained in the ball of center (0, 0) and radius Pp,λ . Thus there exists (upon
subsequence, which we do not relabel) R ∈ X̄ such that Rn → R in (X̄, d). Note that:
(1) Corollary 2.3 gives H2 (Rn ) ≥ Ap,λ for any n, and Lemma 2.1 gives
H2 (R) = lim H2 (Rn ) ≥ Ap,λ .
n→+∞
(19)
PERIMETER-AREA RATIO PENALIZATION
9
(2) Since Rn ⊆ {z ∈ R2 : kzk ≤ Pp,λ }, we infer also
2
H2 (R) = lim H2 (Rn ) ≤ πPp,λ
.
n→+∞
(20)
(3) Lemmas 2.1 and 2.4 give
2
H1 (∂R) ≤ lim inf H1 (∂Rn ) ≤ Pp,λ
.
(21)
H1 (∂Rn )
H1 (∂R)
≤
lim
inf
.
n→+∞ H2 (Rn )
H2 (R)
(22)
n→+∞
Combining (19) and (21) gives
Lemma 2.1 gives
Z
Z
p
dist (x, ∂R) dx = lim
n→+∞ Rn
R
distp (x, ∂Rn ) dx.
(23)
Combining (22) and (23) gives
Z
Ep,λ (R) =
distp (x, ∂R) dx + λ
R
≤ lim
Z
n→+∞ Rn
H1 (∂R)
H2 (R)
distp (x, ∂Rn ) dx + lim inf λ
n→+∞
H1 (∂Rn )
≤ lim inf Ep,λ (Rn ) = inf Ep,λ ,
n→+∞
H2 (Rn )
X̄
hence R is effectively a minimizer of Ep,λ . Since R is limit of Rn ⊆ X, it is also convex, and
compact in view of (18), while area estimate (19) gives that the Hausdorff dimension of R is 2.
Thus R ∈ X.
Consequently, we have:
Corollary 3.2. For any p ≥ 1, λ > 0, any minimizer of Ep,λ belongs to X.
Proof. Consider and arbitrary minimizer R ∈ X̄. Choose an arbitrary minimizing sequence
Rn ⊆ X, such that Rn → R with respect to d. Clearly, R is convex. Without loss of generality,
assume also
sup Ep,λ (Rn ) ≤ min Ep,λ + 1.
n
X̄
Lemma 2.4 gives H1 (∂Rn ) ≤ Pp,λ , hence
R, Rn ⊆ {z ∈ R2 : kzk ≤ Pp,λ },
n = 1, 2, · · · .
Thus R is compact, and satisfies
2
H2 (R) ≤ πPp,λ
.
(24)
H2 (R) = lim H2 (Rn ) ≥ Ap,λ ,
(25)
Conversely, Lemma 2.2 gives
n→+∞
hence R has Hausdorff dimension equal to 2. Since we proved that R is also compact and convex,
it follows R ∈ X.
10
XIN YANG LU
Now we prove point (2) of Theorem 1.1. The proof will be split over Lemmas 3.3 and 3.5.
Lemma 3.3. (C 1 -regularity) For any p ≥ 1, λ > 0, any minimizer of Ep,λ is C 1 -regular.
The proof uses the following (simple) observation.
Lemma 3.4. Let S be a compact, convex set, with Hausdorff dimension equal to 2. Let w1 ,
w2 ∈ ∂S be arbitrary distinct points, and let σ be the segment with endpoints w1 and w2 . Denoting
by S1 and S2 the two connected components of S\σ, then both S1 , S2 are convex.
Proof. Endow R2 with a Cartesian coordinate system. Upon rotation and reflection, assume that
σ lies in the x-axis, and S1 ⊆ {x > 0}, S2 ⊆ {x < 0}. Clearly, given points u, v ∈ S1 , the segment
ξ between u and v lies entirely in S ∩ {x > 0} = S1 , hence S1 is compact. The proof for S2 is
analogous.
Proof. (of Lemma 3.3) Consider an arbitrary minimizer R ∈ X. Endow R2 with a polar
coordinate system. Similarly to the proof of Lemma 2.4, we parameterize ∂R by a closed curve
γ : [0, 2π] −→ ∂R,
γ(θ) := {`θ ∩ ∂R},
where `θ denotes the half-line starting in (0, 0) with anomaly θ. The proof is achieved by a
contradiction argument. Assume that R is not C 1 -regular. That is, γ is not C 1 -regular at some
point θ0 . Upon rotating the coordinates, we can also assume θ0 ∈ (0, 2π). Since R is convex,
both one-sided derivatives
l− := lim γ 0 (θ),
l+ := lim γ 0 (θ)
θ→θ0−
θ→θ0+
are well-defined. Denote by α the angle between l− and l+ . Clearly, α 6= π.
γ(t0 )
∂R
γ(t1 )
α α
2 2
σ
γ(t2 )
R
Figure 2. A schematic representation (near γ(t0 ), in first order approximation in ε) of
the construction of Rε0 .
Figure 2 is a representation (in first order approximation) of ∂R near γ(t0 ). For small parameters 0 < ε 1, construct the competitor Rε0 as follows:
PERIMETER-AREA RATIO PENALIZATION
11
(1) choose t1 < t0 < t2 such that (in first order approximation in ε)
H1 (γ([t1 , t0 ])) = H1 (γ([t0 , t2 ])) = ε + O(ε2 ).
(2) Denote by
σ := {(1 − s)γ(t1 ) + sγ(t2 ) : s ∈ [0, 1]}
the line segment between γ(t1 ) and γ(t2 ), and set
L := ∂R\γ([t1 , t2 ]) ∪ σ.
(26)
Note that such L is a convex Jordan curve, and denote by Rε0 the bounded region delimited
by L.
By construction, in first order approximation in ε, it holds
H1 (∂Rε0 ) = H1 (∂R) − 2ε(1 − sin(α/2)) + O(ε2 ),
H2 (Rε0 ) = H2 (R) −
ε2 sin α
2
+ o(ε2 ) = H2 (R) + O(ε2 ).
(27)
(28)
Moreover, we claim
Z
Rε0
dist
p
(x, ∂Rε0 ) dx
≤
Z
distp (x, ∂R) dx.
(29)
R
This holds because:
(1) for any x ∈ R such that dist(x, ∂R) = |x − y| for some y ∈
/ γ([t1 , t2 ]), it holds
dist(x, ∂Rε0 ) ≤ |x − y| = dist(x, ∂R)
since y ∈ ∂Rε0 ,
(2) for any x ∈ R such that dist(x, ∂R) = |x − y| for some y ∈ γ([t1 , t2 ]), due to the convexity
of R, the segment {(1 − s)x + sy : s ∈ [0, 1]} must intersect the segment σ. Denoting by
y 0 := {(1 − s)x + sy : s ∈ [0, 1]} ∩ σ,
it follows
dist(x, ∂Rε0 ) ≤ |x − y 0 | ≤ |x − y| = dist(x, ∂R).
Thus in both cases we have dist(x, ∂Rε0 ) ≤ dist(x, ∂R), and (29) is proven.
12
XIN YANG LU
Recalling that H2 (R) > 0 (since R is a minimizer), combining (27), (28) and (29) gives (in
first order approximation in ε)
Ep,λ (Rε0 )
Z
=
≤
Rε0
Z
distp (x, ∂Rε0 ) dx + λ
distp (x, ∂R) dx + λ
R
H1 (∂Rε0 )
H2 (Rε0 )
H1 (∂R) − 2ε(1 − sin(α/2)) + O(ε2 )
H2 (R) + O(ε2 )
H1 (∂R) 2λε(1 − sin(α/2))
−
+ O(ε2 )
2 (R)
2 (R)
H
H
R
2λε(1 − sin(α/2))
= Ep,λ (R) −
+ O(ε2 )
H2 (R)
2λε(1 − sin(α/2))
+ O(ε2 ),
= min Ep,λ −
X
H2 (R)
Z
=
distp (x, ∂R) dx + λ
which is a contradiction for sufficiently small ε. Thus R must be C 1 -regular.
Lemma 3.5. Given p ≥ 1, λ > 0, a minimizer R of Ep,λ , let γ : [0, H1 (∂R)] −→ ∂R be an
arc-length parameterization. Then it holds
lim sup
h→0
|γ(t + 2h) − 2γ(t) + γ(t − 2h)|
≤ 4C(λ, p),
h2
(30)
for any t, where C(λ, p) is some constant depending only on λ and p (and independent of R).
We remark that (30) implies C 1,1 -regularity of ∂R.
Proof. Consider an arbitrary point p0 ∈ ∂R. Since we proved that R is C 1 -regular, consider a
(local) orthogonal coordinate system with origin in p0 , and x-axis oriented along the tangent
derivative (at p0 ), such that R is entirely contained in the half-plane {y ≥ 0}. The boundary ∂R
is thus (locally) the graph of some nonnegative function f . Clearly, such f satisfies f (0) = 0.
Choose an arbitrary 0 < ε 1. Denote by
σε := {(x, y) : 0 ≤ x ≤ ε, y = x · f (ε)/ε}
the segment between the origin and (ε, f (ε)). Let Lε be the curve obtained by replacing f ([0, ε])
with σε . That is,
Lε := (∂R\f ([0, ε])) ∪ σε .
By construction (see Lemma 3.4) Lε is a convex Jordan curve, and let Rε be the bounded
connected component of R2 \Lε . Note that:
(1) using the same arguments from the proof of (29), we infer
Z
Rε
distp (x, ∂Rε ) dx ≤
Z
R
distp (x, ∂R) dx.
(31)
PERIMETER-AREA RATIO PENALIZATION
13
y
(ε, f (ε))
R
σε
f
ε
0
x
Figure 3. A schematic representation of the construction near p0 = (0, 0).
(2) For areas, since by construction it holds Rε ⊆ R, we have
H2 (R) − H2 (Rε ) = H2 (R\Rε )
= H2 ({(x, y) : 0 ≤ x ≤ ε, f (x) ≤ y ≤ x · f (ε)/ε})
Z ε
=
0
f (ε)ε
−
[xf (ε)/ε − f (x)] dx =
2
Z ε
f (x) dx.
(32)
0
(3) For perimeters, note that f 0 (0) = 0, so |f 0 | is small near 0. In particular, by choosing
sufficiently small ε, we can ensure that
q
1 + |f 0 (x)|2 ≤ 2
for all x ∈ (0, ε).
Thus
H1 (∂R) − H1 (∂Rε ) =
Z ε q
1 + |f 0 (x)|2 − 1 dx
0
Z ε
=
0
1
|f 0 (x)|2
dx ≥
0
2
3
1 + |f (x)| + 1
p
Z ε
|f 0 (x)|2 dx.
(33)
0
Combining (31), (32) and (33) gives
Z
Ep,λ (Rε ) =
≤
distp (x, ∂Rε ) dx + λ
Rε
Z
p
dist (x, ∂R) dx + λ
R
H1 (∂Rε )
H2 (Rε )
1 Rε 0
2
3 0 |f (x)| dx
.
Rε
( f (ε)ε
−
f
(x)
dx)
0
2
H1 (∂R) −
H2 (R) −
(34)
14
XIN YANG LU
Since
H1 (∂R)
H2 (R) − ( f (ε)ε
2 −
Rε
0
=
f (x) dx)
H1 (∂R)
H2 (R)
·
R
H2 (R) H2 (R) − ( f (ε)ε − ε f (x) dx)
0
2
f (ε)ε
ε
H1 (∂R)
2 − 0 f (x) dx
=
,
·
1
+
Rε
H2 (R)
H2 (R) − ( f (ε)ε
2 − 0 f (x) dx)
R
estimate (34) reads
1 Rε 0
2
3 0 |f (x)| dx
Ep,λ (Rε ) ≤
dist (x, ∂R) dx − λ
R
ε
R
H2 (R) − ( f (ε)ε
2 − 0 f (x) dx)
Rε
f (ε)ε
H1 (∂R)
2 − 0 f (x) dx
+λ 2
· 1+
Rε
H (R)
H2 (R) − ( f (ε)ε
2 − 0 f (x) dx)
R
R
H1 (∂R) f (ε)ε
( 2 − 0ε f (x) dx) − 13 0ε |f 0 (x)|2 dx
H2 (R)
.
= Ep,λ (R) + λ
Rε
−
f
(x)
dx)
H2 (R) − ( f (ε)ε
0
2
2
minimizer, Lemma 3.1 gives H (R) > 0, and note that
Z
Since R is a
p
(35)
ε
f (ε)ε
H2 (R)
f (x) dx ≤
−
2
2
0
for all sufficiently small ε, hence the denominator in (35) is positive. Thus the minimality of R
forces the numerator in (35) to be nonnegative, i.e.,
Z
H1 (∂R) f (ε)ε
3 2
−
H (R)
2
Z ε
f (x) dx −
Z ε
|f 0 (x)|2 dx ≥ 0.
(36)
0
0
Lemma 3.1 gives
H2 (R) ≥ Ap,λ ,
hence
3
H1 (∂R) ≤ Pp,λ ,
3Pp,λ
H1 (∂R)
≤
=: C(λ, p),
H2 (R)
Ap,λ
and (36) forces
ε
ε
f (ε)ε
−
f (x) dx ≥
|f 0 (x)|2 dx.
(37)
2
0
0
Since R is convex, and we assumed (at the beginning of this proof) that R ⊆ {y ≥ 0}, f is
nonnegative, hence (37) forces
Z
Z
C(λ, p)
C(λ, p)
f (ε)ε ≥
2
Note that, since f (0) = 0, it follows
Z ε
f (ε) =
0
0
Z ε
|f 0 (x)|2 dx.
(38)
0
f (x) dx ≤
Z ε
0
|f 0 (x)| dx.
(39)
PERIMETER-AREA RATIO PENALIZATION
15
By Hölder’s inequality,
Z ε
|f 0 (x)|2 dx ≥
0
1
ε
Z ε
|f 0 (x)| dx
2
,
(40)
0
hence
2
ε
ε
(40) 1
(38)
C(λ, p)
|f 0 (x)| dx
|f (x)| dx ≥
|f 0 (x)|2 dx ≥
f (ε)ε ≥
2
ε 0
0
0
Z ε
Z ε
C(λ, p) 2
=⇒
f 0 (x) dx = f (ε).
|f 0 (x)| dx ≥
ε ≥
2
0
0
The above arguments can be repeated for ε < 0, |ε| 1 (or equivalently, when the orientation
of x-axis is inverted). The arbitrariness of ε then gives
C(λ, p)
ε
2
Z ε
(39)
0
lim sup
ε→0
Z
Z
|f (ε) − 2f (0) + f (−ε)|
≤ 4C(λ, p),
(ε/2)2
concluding the proof.
Now we prove point (3) of Theorem 1.1.
Lemma 3.6. Given p ≥ 1, λ > 0, any minimizer R of Ep,λ satisfies
H1 (∂R)
p+2
=
min Ep,λ .
2
H (R)
λ(p + 3) X
Proof. Let R be an arbitrary minimizer. Endow R2 with a Cartesian coordinate system, and
assume without loss of generality that (0, 0) is in the interior part of R. For any r > 0, denote by
Tr : R2 −→ R2 ,
Tr (x) := rx
the homothety of center (0, 0) and ratio r. Note that Tr (R) ∈ X for any r > 0, and the scalings
are
Z
distp (x, ∂Tr (R)) dx = rp+2
Z
distp (x, ∂R) dx,
R
1
H (∂R)
Tr (R)
H1 (∂Tr (R))
1
= · 2
.
2
H (Tr (R))
r H (R)
Define the function
f (r) := Ep,λ (Tr (R)) = rp+2
f : (0, +∞) −→ (0, +∞),
Z
distp (x, ∂R) dx +
R
Since f is smooth, and attains a global minimum at r = 1, it follows
f 0 (1) = (p + 2)
Z
distp (x, ∂R) dx − λ ·
R
Z
=⇒
R
distp (x, ∂R) dx =
H1 (∂R)
=0
H2 (R)
λ
H1 (∂R)
· 2
,
p + 2 H (R)
λ H1 (∂R)
·
.
r H2 (R)
16
XIN YANG LU
hence
Ep,λ (R) =
λ(p + 3) H1 (∂R)
· 2
= min Ep,λ ,
X
p+2
H (R)
and the proof is complete.
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