Equality cases for condenser capacity inequalities under symmetrization

It is well known that certain transformations decrease the capacity of a condenser. We prove equality statements for the condenser capacity inequalities under symmetrization and polarization without connectivity restrictions on the condenser and without regularity assumptions on the boundary of the condenser.


Introduction.
A condenser in R n , n ≥ 2, is a pair (D, K), where D is an open subset of R n and K is a nonempty compact subset of D. Let ACL 2 (G) be the set of continuous functions φ : G → R on the open set G ⊂ R n , which are absolutely continuous on lines and their partial derivatives are in L 2 loc (G) (see e.g.[23, pp. 88-89]).The (Newtonian) capacity of (D, K) is where the infimum is taken over all functions u ∈ ACL 2 (D \K) with boundary limits 0 on ∂D and 1 on ∂K.The boundary of an open set in R n is taken in the topology of the one-point compactification R n ∪ {∞} of R n .Let T : D → R n be a geometric transformation such that (T (D), T (K)) is a condenser.It is well known that certain transformations decrease the capacity of a condenser, that is (1.1)Cap(T (D), T (K)) ≤ Cap(D, K).
Some examples of transformations such that (1.1) is valid are Steiner and Spherical symmetrizations and polarization; see [21,10,12].Inequality (1.1) for various types of symmetrization, in its most general form, was proved by Sarvas in [21], where one can find references to earlier work of Hayman, Gehring, Mostow, Anderson and Pfaltzgraff.Inequalities of the type (1.1) have important applications in potential theory, complex analysis, and mathematical physics; see [2,12,13,20].
In the present article, we give necessary and sufficient conditions on (D, K) such that equality occurs in (1.1).We will see that, roughly speaking, equality occurs if and only if the original condenser is already symmetric in the sense that T (D) ≈ D and T (K) ≈ K. Equality statements have been investigated by several authors under connectivity assumptions and certain regularity conditions on the boundary of the condenser.A condenser (D, K) is called admissible if D \ K is regular for the Dirichlet problem and (D, K) is called connected if D \ K is connected.J. A. Jenkins [15,16] proved that, if (D, K) is an admissible and connected condenser in R 2 , then equality in (1.1) for circular symmetrization with respect to a ray l emanating from the origin is attained if and only if D \ K is circularly symmetric with respect to some ray l emanating from the origin; see also [19, p. 177].V. A. Shlyk [22] proved an equality statement for circular symmetrization of connected condensers, without the admissibility condition.Equality statements for Schwarz symmetrization have been proved under regularity (or smoothness) conditions on the boundary of the condenser; see [2, p. 57], [12, p. 17], [17, pp. 71-72].V. N. Dubinin [12] proved that, if (D, K) is an admissible and connected condenser in R 2 , then equality in (1.1) for polarization with respect to a straight line α is attained if and only if the polarization of (D, K) coincides with (D, K) or it is symmetric to the latter with respect to α.Also, in [9], he proved an equality statement for polarization without the connectivity assumption.D. Betsakos [3,4] proved equality statements for symmetrization and polarization inequalities for Green functions, Brownian transition functions, Dirichlet heat kernels and harmonic measure.A. Cianchi and N. Fusco [8] proved equality statements in Steiner symmetrization inequalities for Dirichlet-type integrals under connectedness, boundedness and boundary conditions.Our purpose is to prove equality statements for condensers in R n under Steiner and Schwarz symmetrizations and polarization, without any connectivity or regularity assumptions.
We note that inequality (1.1) is true when Newtonian capacity is replaced by several other capacities, for example the well-known p-capacity, under polarization and symmetrization; see V. N. Dubinin [11].The characterization of the equality cases for the general capacities is an open problem.
For the sake of concreteness, we will state and prove symmetrization results only for Schwarz and 1-dimensional Steiner symmetrization but similar results hold for other kinds of symmetrization.We give here the definition of 1-dimensional Steiner symmetrization.Let H be an (n − 1)-dimensional hyperplane in R n .We define the symmetrization S H (A) of an open or compact set A ⊂ R n by determining its intersections with every line perpendicular to H. Let Σ(x) be the line which is perpendicular to H and passes through the point x ∈ H. Let 2r x be the 1-dimensional Lebesgue measure of the set Σ(x) ∩ A. Let (−r x , r x ) be the open linear segment on Σ(x) centered at x with length 2r x and let [−r x , r x ] be the corresponding closed segment.Then (see Figure 1) The Steiner symmetrization of a condenser (D, K) with respect to H is the condenser (S H (D), S H (K)).We refer to [7,12,13,17,21] for more information about symmetrization.We need to introduce some terminology and notation.Every (n − 1)dimensional hyperplane in R n will be simply called plane.If H is a plane, we denote by Π H (A) the orthogonal projection of a set A ⊂ R n on H.We denote by C 2 (E) the logarithmic (n = 2) or Newtonian (n ≥ 3) capacity of the Borel set E ⊂ R n (see e.g.[1,14,18]).If two Borel sets A, B ⊂ R n differ only on a set of zero capacity (namely, C 2 (A \ B) = C 2 (B \ A) = 0), then we say that A, B are nearly everywhere equal and write A n.e.= B.By convention, C 2 ({∞}) = 0 if n = 2 and C 2 ({∞}) > 0 if n ≥ 3 in the sense that {∞} is polar when n = 2 and is non-polar when n ≥ 3; see [14, p. 206].
A condenser (D, K) will be called normal if for every connected component Every normal condenser has positive capacity and every connected condenser with positive capacity is normal.The normality condition for a condenser is natural.If a connected component of D \ K does not satisfy the above inequalities, then it does not contribute to the capacity; in particular it can be viewed as a part of K or it can be removed from D, according to which one of the above inequalities fails, without affecting the capacity of the condenser.Therefore, normality is a necessary condition for equality statements.
Our main result is the following equality statement for Steiner symmetrization.We do not suppose that D \ K or D is connected and we do not suppose that the condenser is admissible.
Let A ⊂ R n be an open (or compact) set and let m n (A) be the ndimensional Lebesgue measure of A. The Schwarz symmetrization of A, denoted by A * , with respect to a point z 0 ∈ R n , is the open (or closed) ball with center at z 0 such that m n (A) = m n (A * ).The Schwarz symmetrization of a condenser (D, K) with respect to z 0 is the condenser (D * , K * ).We shall prove that we have equality in (1.1) for Schwarz symmetrization if and only if there exists a translation F such that D * n.e.
= F (D) and K * n.e.= F (K).The proof of Theorem 1 is based on the approach to symmetrization via polarization.In Section 2 we describe polarization and we prove an equality statement in inequality (1.1) for polarization.Also we describe the extended Dirichlet principle which we use to deal with the irregular boundary points.Theorem 1 is proved in Section 3. In Section 4 we prove the equality statement for Schwarz symmetrization based on the corresponding statement for Steiner symmetrization.

Equality statement for polarization.
2.1.The extended Dirichlet Principle.First we introduce some more terminology.If K is a compact subset of R n , n ≥ 2, the reduced kernel of K is the compact set It is known that C 2 (K \ Ǩ) = 0; see [18, p. 164 = Ǧ.Let φ be a continuous real valued function on ∂G.We consider the set According to the extended Dirichlet principle, we can enlarge the class of functions in the definition of the condenser capacity.In particular, (2.1) where The function h will be called the equilibrium potential of the condenser (D, K).We shall denote D(D \ K, ϕ) by D(D, K).

Remark 2.2. A condenser (D, K) is normal if and only if its equilibrium potential is non-constant on each connected component of D \ K.
We state here some well-known properties of condenser capacity.

Definition of polarization.
Let H be an oriented plane in R n .Let H + and H − be the closed half-spaces into which H divides R n , with respect to the given orientation.We denote by R H (•) the reflection of a point or a subset of R n in H.
We proceed to define polarization with respect to H. Let E be any set in R n .We divide E into three disjoint sets as follows: The symmetric part of E is the set and the lower non-symmetric part of E is the set The polarization P H (E) of E with respect to H is the set (see Figure 2)

The polarization of an open (or closed) set is open (or closed)
. The polarization of a condenser (D, K) with respect to the plane H is the condenser (P H (D), P H (K)).For more information about polarization, see [3,7,12].
A set E and its polarization P H (E).

Polarization and capacity.
We will use the fact that a closed set with zero capacity cannot disconnect a domain: We shall also need the following lemma for the identification of two condensers.
Lemma 2.7.Let (D 1 , K 1 ) and (D 2 , K 2 ) be two normal condensers and let h 1 and h 2 be their equilibrium potentials, respectively.Let A be a connected component of the set Proof.From the identity principle of harmonic functions we obtain that Consider the decomposition Since ∂A ⊂ (∂Ω 1 ∪ ∂Ω 2 ), the set Ω 1 ∩ ∂A is a subset of ∂Ω 2 with positive capacity.From the boundary behavior of h 2 we obtain that there exists Therefore, since h 1 = h 2 on A, we obtain that From the extended maximum principle, in either case, we obtain that h 1 is constant on A. So, by the identity principle, h 1 is constant on Ω 1 which contradicts the fact that = A. In a similar way we show that Let (D, K) be a condenser and let H be a plane.Inequality (1.1) for polarization is (see [10] and Figure 3) . The domain D is the interior of the polygon and the compact set K is the union of the three disks.The condenser (P H (D), P H (K)) is the polarization of the condenser (D, K).
We proceed to state and prove the equality statement for polarization.For the proof we use polarization of functions which was first introduced by V. N. Dubinin.Dubinin [9] proved a similar statement with the additional assumption that the condenser is admissible.Our proof follows the general scheme of Dubinin's proof with several differences in technical details arising from the presence of irregular boundary points.= R H (∂Ω ∩ ∂D).
Proof.Suppose that equality (2.4) holds.Let h be the equilibrium potential of (D, K).We extend the function h on R n by setting h = 1 on K and h = 0 on R n \ D. We note that the function The polarization of h is the function It is well known (see e.g.[7, pp. 1768-1769]) that P h ∈ D(P H (D), P H (K)) and and therefore From the extended Dirichlet principle we obtain that P h is the equilibrium potential of (P H (D), P H (K)). Since (D, K) is a normal condenser, h is not constant on any connected component of D \ K. Therefore, from the definition of P h and Remark 2.2, we obtain that P h is not constant on any connected component of P H (D) \ P H (K) and hence (P H (D), P H (K)) is a normal condenser.We shall show that every connected component of D \ K contains a ball B such that P h = h or P h = Rh on B and every connected component of P H (D) \ P H (K) contains a ball B such that P h = h or P h = Rh on B .Let Ω be a connected component of D \ K and let J be the union of the boundaries of the condensers (D, K), (R H (D), R H (K)) and (P H (D), P H (K)).The functions h, Rh and P h are continuous on the open set Ω \ J.If h = Rh = P h on Ω \ J, the assertion for Ω follows.Suppose that P h(z) < h(z) for a point z ∈ Ω \ J.Then, by the continuity of h, P h, we obtain that P h < h on a ball B(z, ) ⊂ Ω \ J. Therefore P h = Rh on B(z, ) and the assertion is proved.In a similar way we treat the other possible cases.
Let Ω be a connected component of P H (D) \ P H (K). Suppose that = R H (∂G∩ ∂D).Therefore, the correspondence between the connected components of P H (D) \ P H (K) and the connected components of D \ K is one-to-one and onto.
Conversely, let Λ 1 denote the union of the connected components Ω of P H (D) \ P H (K) such that there exists a connected component = ∂Ω ∩ ∂D.Also, let Λ 2 denote the union of the remaining connected components of P H (D) \ P H (K). Then the equilibrium potential of (P H (D), P H (K)) is the function and Corollary 2.9.Let (D, K) be a connected condenser with positive capacity.Then equality in (2.4) The open set D is the interior of the large square minus the four horizontal segments.The compact set K is the union of the boundaries of the two small rectangles.
In the terminology of the proof of Theorem 2.8,

Equality statement for Steiner symmetrization.
3.1.Green capacity.Let (D, K) be a condenser.We assume that D is a Greenian open subset of R n and we denote by G D (x, y) the Green function of D. The Green equilibrium energy of K relative to D is defined by where the infimum is taken over all unit Borel measures μ supported on K.
The Green capacity of K relative to D is the number .
When I(K, D) < +∞, the unique unit Borel measure for which the above infimum is attained is the Green equilibrium measure and, since G D is infinite on the diagonal {(x, x) : x ∈ K}, See e.g.[18, p. 174] or [1, p. 134].
It is well known that condenser capacity is proportional to Green capacity by a positive constant which depends only on the dimension; see e.g.[18].
It follows from the boundary behavior of the Green function that the following lemma holds.

Steiner symmetrization and capacity.
Let (D, K) be a condenser and let H be a plane.Inequality (1.1) for Steiner symmetrization is (see [21]) The proof of Theorem 1 is based on the equality statement for polarization.We note that for any open or compact set A, for any plane H and for any oriented plane Y parallel to H, We shall need several auxiliary lemmas: some strict inequalities for condenser capacity and a characterization of Steiner symmetric sets by appropriate polarizations.
Let μ and μ be the Green equilibrium measures of K relative to D and D , respectively.Then and therefore For the next lemma we need some definitions.For every c ∈ R, we denote by H c the horizontal plane {(x 1 , x 2 , . . ., x n ) : x n = c}.A line will be called vertical if it is perpendicular to H 0 .A set A will be called striplike if for every vertical line Σ that intersects A, we have Σ ∩ A = Σ.A set B will be called essentially striplike, if B n.e.
= A for some striplike set A. Let Ω be an open set in R n .We say that Ω ∈ A 1 if there exists a horizontal plane H such that for every vertical line Σ that intersects Ω, the set Σ ∩ (R n \ Ω) is either empty or a nonempty, bounded, vertical segment, symmetric with respect to H. We say that Ω ∈ A 2 if for every vertical line Σ that intersects Ω, the set Σ ∩ Ω is either the whole line Σ or an upward half-line.We say that Ω ∈ A 3 if for every vertical line Σ that intersects Ω, the set Σ ∩ Ω is either the whole line Σ or a downward half-line.We say that D ∈ G i , i = 1, 2, 3 if D is not essentially striplike and D n.e.
= Ω, for some Ω ∈ A i , i = 1, 2, 3 respectively.Let A be a compact subset of R n .Recall that the reduced kernel Ǎ has been defined in Subsection 2.1.We consider the numbers and a = m+M 2 .Suppose that (3.3) holds for every oriented horizontal plane H c , c ∈ R. We denote by H + c and H − c the closed half-spaces {x n ≥ c} and {x n ≤ c}, respectively and we assume that H c is upward oriented.From the hypothesis and the definition of a we obtain that P Hc (A) n.e.
= A, for every c < a.We will show that Ǎ is vertically convex.Let Applying polarization with respect to H d , we obtain that

But P H d (A)
n.e.
Next we will show that Ǎ is Steiner symmetric with respect to H a .Suppose that there exists a vertical line Σ such that is a segment which is not symmetric with respect to H a .Let s = b 1 +b 2 2 and suppose that s < a.Let s ∈ (s, a) and suppose that H s is upward oriented.Then, if B is a ball with center (ζ 1 , . . ., ζ n−1 , b 1 ) and with sufficiently small radius, = A, and we derived a contradiction.In a similar way we can treat the case s > a.We conclude that, if Σ is a vertical line that intersects Ǎ, Σ ∩ Ǎ is either a singleton {x} for some x ∈ H a or it is a segment symmetric with respect to H a .So S Ha ( Ǎ) = Ǎ.Therefore, since A n.e.

= A.
The opposite direction is obvious.
We now prove two versions of Theorem 1 under connectivity restrictions on the condenser.We shall use these results for the proof of Theorem 1. = K.
Proof.Suppose that (3.4) is valid.Without loss of generality we may assume that H is a horizontal plane.Let Y be an oriented horizontal plane.Then, by (2.3), (3.1) and (3.2), By Corollary 2.9, the relations (2.5) or the relations (2.6) are valid for every oriented horizontal plane Y .By Lemma 3.3, there exists a horizontal plane Π such that S Π (K) n.e.
= K.We need to show that S Π (D) Since S H (D) is striplike and K is Steiner symmetric with respect to the horizontal plane Π, By Lemma 3.2, Proposition 2.3 and (3.5) which is a contradiction.Therefore D ∈ G 1 ∪ G 2 ∪ G 3 and by Lemma 3.3 we obtain that there exists a horizontal plane Π such that S Π (D) n.e.
= D. Also, since the relations (2.5) or the relations (2.6) hold for Π and Π , we obtain that Π = Π and the one direction is proved.The other direction is obvious.
Consider a normal condenser (D, K) such that the open set D is connected.In general, D \ K is not connected.Every connected component of D \K may be viewed as a "connected subcondenser" of (D, K).We proceed to give the precise definition.Let Ω 0 be the connected component of D \ K that intersects the unbounded connected component of R n \ K. Let Ω i , i ≥ 1 (the set of indices i at most countable) be the remaining connected components of D \ K.The subcondensers corresponding to Ω 0 and Ω i will be defined separately.Let Q j , j ∈ J (J a set of indices) be the connected components of R n \ D. Let and let J 0 = ∪ i J i (see Figure 5).We consider the sets . The domain D is the interior of the large rectangle minus the two small rectangles Q 1 and Q 2 .The compact set K is the union of the boundaries of the two remaining rectangles.Ω 0 , Ω 1 and Ω 2 are the connected components of D \ K. Lemma 3.5.Let (D, K) be a normal condenser such that D is connected and let (D 0 , K 0 ) and (D i , K i ) be the subcondensers of (D, K).Then Proof.If h is the equilibrium potential of (D, K), then the equilibrium potential h 0 of (D 0 , K 0 ) is the restriction of h to Ω 0 and the equilibrium potential of (D i , K i ) is the function In the proof of the following lemma we apply Lemma 3.4 on the subcondensers of (D, K). = K.
Proof.Suppose that (3.6) holds.Let (D 0 , K 0 ) and (D i , K i ) be the subcondensers of (D, K).We will show that all the subcondensers of (D, K) are Steiner symmetric with respect to planes parallel to H. Consider the sets and note that (G, L) is a normal condenser such that (see Figure 6) Also, by (3.1), Lemma 3.5 and (3.7), for every i.By Lemma 3.4 we obtain that there exist planes Y 0 and {Y i } parallel to H such that n.e.
= K 0 and S Y i (D i ) n.e.
= D i and S Y i (K i ) n.e.
= K i .We have shown that all the subcondensers of (D, K) are Steiner symmetric.We will now show that in fact (D, K) must be connected.
Suppose that D l \K l = Ω l = ∅, for some l ≥ 1.Without loss of generality we may assume that Y l is a horizontal plane.Let . We apply polarization with respect to the upward oriented horizontal plane H l = H α l to get a contradiction.By (2.3), (3.1) and (3.2),

Therefore
Cap(D, K) = Cap(P H l (D), P H l (K)) and by Theorem 2.8 we obtain that there exists a connected component Ω l of P H l (D) \ P H l (K) such that Ω l n.e.
= R H l (Ω l ).But from the definition of α l we obtain that ) can be nearly everywhere equal to a connected component of P H l (D) \ P H l (K) and we derived a contradiction.Therefore D i \ K i = ∅, i ≥ 1, which means that (D, K) is connected and the assertion follows by Lemma 3.4.
The opposite direction follows by Lemma 3.4.
We need the following lemma to show that, if equality (3.4) holds for a condenser (D, K) and a plane H, then every line which is vertical to H intersects at most one connected component of D.
Let μ and μ be the Green equilibrium measures of K and K with respect to D and D , respectively.Let μ 1 be the restriction of μ on T (K 1 ) and let μ 2 be the restriction of μ on K 2 .Then μ = μ 1 + μ 2 .Consider the measure μ1 = μ 1 • T −1 and the measure μ = μ1 + μ 2 .Then μ is a unit Borel measure on K. From the definition of the equilibrium measure and the relations (3.8), (3.9), (3.10) for the Green functions, and therefore Cap(D , K ) < Cap(D, K).
Proof of Theorem 1.First we show that Then there exists a translation T (x) = x+z, where the vector z is vertical to and consider the condenser (D , K ).By Lemma 3.5 and Lemma 3.7, By (3.1) and Lemma 3.5, Therefore Cap(S H (D i ), S H (K i )) = Cap(D i , K i ) for every i and the assertion of the theorem follows by Lemma 3.6.
The opposite direction is obvious.

4.
Equality statement for Schwarz symmetrization.Let (D, K) be a condenser.Inequality (1.1) for Schwarz symmetrization is (see [21] and Figure 7) The proof of the equality statement for Schwarz symmetrization with respect to a point z 0 ∈ R n is based on Theorem 1.We note that for any open or compact subset E of R n and for any plane H, We need the following characterization of a ball.= A. By Lemma 3.3, Ǎ is convex in any direction and for every plane H there exists a plane H parallel to H such that S H ( Ǎ) = Ǎ.Therefore Ǎ is convex and by simple geometric arguments, similar to those in the proof of the isoperimetric inequality (see e. = Ǎ, the one direction of the lemma follows.The opposite direction is obvious. We proceed to prove the equality statement for Schwarz symmetrization.= B(x 2 , r 2 ).Since for every plane H the plane H in (4.5) is the same for D and K, x 1 = x 2 .Let F be the translation such that F (x 1 ) = z 0 .Since a Borel set of zero capacity in R n has zero n-dimensional Lebesgue measure, D * n.e.

Figure 1 .
Figure 1.An open set D, a compact set K and their Steiner symmetrizations S H (D) and S H (K). The condenser (S H (D), S H (K)) is the Steiner symmetrization of the condenser (D, K).

Theorem 2 . 1 .
and we denote by h(G, φ) the solution of the generalized Dirichlet problem on G with boundary function φ.Note that h(G, φ) ∈ D(G, φ).We shall need the following extended version of the classical Dirichlet principle.[6, p. 414] (Extended Dirichlet Principle).Let G be a Greenian open subset of R n , φ be a continuous real valued function on the boundary of G and h = h(G, φ).Then G |∇h| 2 ≤ G |∇g| 2 , for every g ∈ D(G, φ) and equality occurs if and only if g = h.
We may assume that B ⊂ D \ K since otherwise we may replace B by a ball B 1 ⊂ B \ I(D \ K).If Ω is the connected component of D \ K that intersects the ball B, from Lemma 2.7 we obtain that Ω n.e.= Ω .Also, by Lemma 2.6, Ω ∩ Ω is connected.So, by the identity principle, h = P h on Ω ∩ Ω and from the boundary behavior of the equilibrium potentials h, P h, we obtain that ∂Ω ∩∂P H (K) n.e.= ∂Ω∩∂K and ∂Ω ∩ ∂P H (D) n.e.= ∂Ω ∩ ∂D.In a similar way we can show that, if P h = Rh on an open ball B ⊂ Ω , then Ω n.e.= R H (Ω), ∂Ω ∩∂P H (K) n.e.= R H (∂Ω∩∂K) and ∂Ω ∩ ∂P H (D) n.e.= R H (∂Ω ∩ ∂D).Let G be a connected component of D \ K.In a similar way we can show that: (i) if P h = h on an open ball B ⊂ G, then there is a connected component G of P H (D) \ P H (K) such that G n.e.= R H (G), ∂G ∩ ∂P H (K) n.e.= R H (∂G ∩ ∂K) and ∂G ∩ ∂P H (D) n.e.= R H (∂G ∩ ∂D) and (ii) if P h = Rh on an open ball B ⊂ G, then there is a connected component G of P H (D) \ P H (K) such that G n.e.= R H (G), ∂G ∩ ∂P H (K) n.e.= R H (∂G∩ ∂K) and ∂G ∩ ∂P H (D) n.e.

Lemma 3 . 1 .
Let D be a Greenian domain in R n and let D be a subdomain of D such that C 2 (D \ D ) > 0. Then G D (x, y) < G D (x, y), whenever x, y ∈ D and x = y.

Lemma 3 . 2 .
Let D be a Greenian domain in R n , let D be a subdomain of D such that C 2 (D \ D ) > 0 and let K be a compact subset of D such that C 2 (K) > 0. Then Cap(D, K) < Cap(D , K). Proof.Since D ⊂ D and C 2 (D \ D ) > 0, by Lemma 3.1

Lemma 3 . 4 .
Let (D, K) be a connected condenser with positive capacity and let H be a plane in R n .Then (3.4) Cap(S H (D), S H (K)) = Cap(D, K) if and only if there exists a plane Y parallel to H such that S Y (D) n.e.= D and S Y (K)n.e.

Lemma 3 . 6 .
Let (D, K) be a normal condenser such that the open set D is connected and let H be a plane in R n .Then (3.6) Cap(S H (D), S H (K)) = Cap(D, K) if and only if (D, K) is connected and there exists a plane Y parallel to H such that S Y (D)

Figure 6 .
Figure 6.The domain D is the interior of the polygon minus the small rectangle inside it.D and G have the same Steiner symmetrization.
It states that equality in (1.1) for Steiner symmetrization with respect to H is attained if and only if the orthogonal projections of the connected components of D on H are disjoint and for every component D i of D there is a plane parallel to H relative to which D i and K i = D i ∩ K are Steiner symmetric.Let (D, K) be a normal condenser, let {D i } be the connected components of D, K i = K ∩ D i and let H be a plane in R n .Then That happens because the connected components of P H (D) \ P H (K) have different behavior; that is, some of them are connected components of D \K while others are connected components of R H = R H (D) and P H (K)n.e.= R H (K).Remark 2.10.Equality (2.4) does not imply equations (2.5) or (2.6) if the condenser is not connected.In the example below (see Figure4), equality (2.4) holds for the condenser (D, K) and the plane H but P H (D) is nearly everywhere equal neither to D nor to R H (D).
There exists a horizontal plane H such that S H (A) n.e.= A if and only if for every oriented horizontal plane Y , (3.3) P Y (A) n.e.= A or P Y (A) n.e.= R Y (A).If (3.3) holds for every oriented horizontal plane Y , then Ǎ is vertically convex and there exists a horizontal plane H such that S H ( Ǎ) = Ǎ.Proof.For the case where A is an open subset of R n such that A ∈ G 1 ∪ G 2 ∪ G 3 , the proof was given in [3, Lemma 2, p. 423].
Lemma 4.1.LetA be a bounded open (or closed) subset of R n .There exists an open (or closed) ball B such that A B if and only if for every plane H there exists a plane H parallel to H such that S H (A)Proof.Suppose that for every plane H there exists a plane H parallel to H such that S H (A) =n.e.