# Isoperimetric inequalities for Lp-norms of the distance function to the boundary.

код для вставкиСкачатьУЧЕНЫЕ ЗАПИСКИ КАЗАНСКОО ОСУДАСТВЕННОО УНИВЕСИТЕТА Физико-математические науки 2006 Том 148, кн. 2 UDK 517.5 ISOPERIMETRIC INEQUALITIES FOR Lp -NORMS OF THE DISTANCE FUNCTION TO THE BOUNDARY R.G. Salahudinov Abstrat The main goal of the paper is to prove that Lp -norms of dist(x, ?G) and dist?1 (x, ?G) are dereasing funtions of p , where G is a domain in Rn (n ? 2) . We also obtain a sharp estimation of the rate of dereasing for these norms using Lp norms of the distane funtion for a onsistent ball. We prove a new isoperimetri inequality for Lp norms of dist(x, ?G) , this inequality is analogous to the inequality of Lp norms of the onformal radii (see Avkhadiev F.G., Salahudinov R.G. // J. of Inequal.& Appl. 2002. V. 7, No 4. P. 593601). Note that L2 -norm of dist(x, ?G) plays an important role to investigate the torsional rigidity in Mathematial Physis (see, for instane, Avkhadiev F.G. // Sbornik: Math. 1998. V. 189, No 12. P. 17391748; Ba nuelos R., van den Berg M., Carroll T. // J. London Math. So. 2002. V. 66, No 2. P. 499512). As a onsequene we get new inequalities in the torsional rigidity problem. Also we generalize the n -dimensional isoperimetri inequality. Introdution Let G be a domain in n R (n ? 2) . tional [1, 2? I(p, G) = Let us onsider the following geometrial fun- Z dist p (x, ?G) dA. (1) G Here dist(x, ?G) denotes the distane funtion from x ? G to the boundary ?G , dA dx1 , . . . , dxn , and p ? ?1 . In [1? it was noted, that it might be I(p, G) the p -order eulidean moment of G with respet to its boundary. is the volume element justied to all First we remark some appliations of (1), and further we note onnetion of (1) with a new oneption of isoperimetrial monotoniity. 0 -order moment is the volume of G . Also, d(G) = max dist(x, ?G) . We also found that for a It is lear that we an get as a limit ase of (1), wide lass of domains x?G (?1) -order moment is, up to a onstant fator, the surfae area of G . Further, we remark a new geometrial funtional of G that reently appears in the elasti torsion problem. This is the seond order eulidean moment or, otherwise, the eulidean moment of inertia with respet to the boundary. Consider the boundary value problem ?u = ?1 where ?? in G, u=0 on ?G, is the Dirihlet Laplaian, and let P(G) := 4 Z u(x) dA, G P(G) is exatly the torsional rigidity of a simply onneted domain all P(G) the torsional rigidity of G, even if n>2 and/or G G . Just as in [3? we is not simply onneted. R.G. SALAHUDINOV 152 We begin from the ase n = 2. In 1995 F.G. Avkhadiev [1? proved the two-sided inequality I(2, G) ? P(G) ? 64 I(2, G). (2) Later in [4? the left-side hand of the inequality was improved to Let us remark that an n -dimensional 3I(2, G) < 2 P(G) . generalization of was reently proved in [5? 2 I(2, G) ? P(G) ? CG I(2, G), n under the additional restrition that onstant (see [6?), where CG G satises a strong Hardy inequality with some is a funtional on G. In partiular, the two-sided inequali- When is the torsional rigidity of a simply onneted ties (2), (3) answered the question: Rn , however for domain in (3) bounded? n?3 the question is still open. Another appliation of (1) was disovered by F.G. Avkhadiev in [2?, this is also n -dimentional generalize (2) on an Kp,q (G) = ase. Consider the funtional sup f ?C0? (G) ? ? Kp,q (G) ? ?1/q ZZ q ? |f | (x) dA? G ZZ G ?1/p . |grad f |p (x) dA? appears in the Poinare Sobolev's inequality. In [2? it was proved the two sided estimations for Kp,q (G) using (1). The rst property of isoperimetri monotoniity was onjetured by J. Hersh [7?, and it was proved by M.-Th. Kohler-Jobin [8?. We shall onsider the following boundary value problem [9? ?v + ?v + 1 = 0 in G, v=0 on ?G ( ? ? < ? < ?1 (G)) (4) and the orresponding funtional Q(?) := Z v dA. G In partiular, we have A(G) Q(0) = P(G)/4 , Q(?1 ) = ?j04 (8?21 (G)) , and (??)Q(?) ?? ???? (see, for example, [12?). By Q(?) we denote the radius of a ball in Rn with same Q(?) . Let G be a bounded domain and not a ball in Rn , then Q(?) is a dereasing funtion on ? . If G is a ball, then Q(?) is a onstant funtion. Theorem A [8, 10?. This monotoniity property ontains several well-known plane isoperimetri inequal- ?1 (G) , P(G) , and A(G) . On the other hand, ?1 (G) are expressing in terms of I(?, G) (see [1, ities (see [12?), whih inlude funtionals existene theorems for P(G) , and 11?). Further, other monotoniity properties were disovered by C. Bandle [13?, and by J. Hersh [12?. All of these isoperimetri monotoniity properties were onneted with the solutions of the dierential equations (4). So, the onjeture on the isoperimetri monotoniity property of (1) is a geometrial analog of Theorem A. On the other hand, this work was intended as an attempt to bind ontinuously the well-known geometrial Rn suh as the surfae area, the volume, the radius of the quantities of a domain in largest ball ontained in the domain. ISOPERIMETRIC INEQUALITIES FOR LP -NORMS OF THE DISTANCE . . . 1. Let ? > n ? 1. B1 := [I(? ? n, B1 )] Rn , is a unit ball in Theorem 1. < ?. Z Then Main results and orollaries Further it would be suitable to use a onstant ?,n where 153 and ?1 ?(·) ?(n /2 )?(? + 1) , 2? n/2 ?(? ? n + 1)?(n) = (5) is Euler's Gamma funtion. Let ? ? ? ? n ? 1 , and G is a domain in Rn suh that I(? ? n, G) < ?+??n dist (x, ?G) dA ? G ?,n ?,n ?+?,n Z dist ??n (x, ?G) dAЧ G Ч Z dist ??n (x, ?G) dA (6) G The equality holds i G is a ball in Rn . This kind of inequality was rst proved in [14? for onformal moments of a simply onneted plane domain. Also, in [14? a hain of plane isoperimetri inequalities was obtained in order to get sharp lower bound in the torsional problem. This hain is similar to ` disrete isoperimetri monotoniity.' three geometrial harateristis of Note that the inequality (6) ontains G. Corollary 1. Let n = 2 and ? = ? = 1 in Theorem 1 , then inequality (6) turns to the plane lassial isoperimetri inequality A(G) ? L2 (G) . 4? In the next two assertions we prove isoperimetri monotoniity properties for integral funtionals whih depend on dist(x, ?G) . Theorem 2. i) Let ||dist(x, ?G)||p < ? , then ||dist(x, ?G)||p? ||dist(x, ?D1 )||p? ? , ||dist(x, ?G)||p?? ||dist(x, ?D1 )||p?? (7) where p? ? p ? p?? ? 0 , p? > p?? , and D1 is a ball suh that ||dist(x, ?D1 )||p := ||dist(x, ?G)||p . Equality holds only for a ball in Rn . ?1 ii) Let ||dist (x, ?G)||p < ? , then ||dist?1 (x, ?G)||p? ||dist?1 (x, ?D2 )||p? ? , ?1 ||dist (x, ?G)||p?? ||dist?1 (x, ?D2 )||p?? where 0 ? p? ? p ? p?? < 1 , p? < p?? , and D2 is a ball suh that ||dist?1 (x, ?D2 )||p := ||dist?1 (x, ?G)||p . Equality holds i G a ball in Rn . In the plane ase for p = p?? = 0 the inequality (7) is an analogous of famous p = p?? = 0 , and for p? = 1 the ?? inequality (7) was proved by J. Leavitt and P. Ungar [15?, and in the ase p = p = 0 , ? and p ? 0 in [16?. St Venant's and Polya's inequality for P(G) . In the ase Using a result from [4? we an get the following lower estimations for the torsional rigidity of a plane domain. R.G. SALAHUDINOV 154 Corollary 2. Let G be a simply onneted plane domain, and let P(G) is bounded, then we have for ? ? 4 ? Z 3? ? ?(? ? 1) ?(? ? 1) 2? ??2 dist G ?4/? (x, ?G) dA? Now we shall onsider a normalized version of I(? ? I? (G) = ??,n where ? > n ? 1. Z dist ??n G < P(G). ? n, G) ?1/? (x, ?G) dA? , (8) In?1 (G) does not exist, however we give a ? = n ? 1 . In that ase we suppose that G has a 1 boundary ?G , whih belongs to C (G) , and we put ? ?1/(n?1) Z ?(n /2 ) In?1 (G) := lim I? (G) = ? dS ? , (9) ??(n?1)+0 2? n/2 In the Lebesgue's sense meaningful denition for the ase smooth ?G where dS is the volume element of ?G . On the other hand, we have lim I? (G) = d(G). (10) ??? Therefore, d(G) < ? is the neessary ondition to apply our theorems. Further we prove that the evaluations given in (9) and (10) are well dened. Theorem 3. Let G is a domain in R , and suppose that I?0 (G) is bounded for some ?0 ? n ? 1 . Then n i) If G is not a ball in R , then I? (G) is a stritly dereasing funtion of ? for ? ? ?0 and I? (G) = d(G) . n ii) If G is a ball in R , then I? (G) is the radius of the ball G for all ? ? n ? 1 . n Corollary 3. Let G is a domain in R suh that I? (G) < ? for ? ? [n ? 1, n) , then In (G) ? I? (G), and I? (G) ? In?1 (G). In the both equalities the equalities hold only for a ball in Rn . n The last inequalities are a generalization of the lassial n -dimentional isoperimetri inequality, whih was proved by E. Shmidt [17?, moreover, our assertion is a more sharper result. 2. Proofs of theorems We begin by proving some simple important properties of the funtional I(?, G) for ? > ?1 . Throughout the proof we will use the following notations G? (?) := {x ? G |dist? (x, ?G) > ?} , ?? (?) := {x ? G |dist? (x, ?G) = ?} a(G) := Z G dA, (11) ISOPERIMETRIC INEQUALITIES FOR LP -NORMS OF THE DISTANCE . . . and denote by [·]S the Shwarz symmetrization of a domain in ? in G? := G? (?) . over the domain. We will usually x a parameter brevity, we will drop ?, for example, Rn , 155 and of a funtion our proof, and in those ases, for Let ? > 0 , and G be an unbounded domain in Rn suh that I(?, G) < < ? . Then: 1) a(G? ) < ? for 0 < ? ? d? (G) , in partiular, d(G) < ? ; 2) ? a(G? ) ?? 0 . Lemma 1. ??0 ? > 0 , then we have a hain of inequalities Z ? ? dist (x, ?G) dA ? a(G? ) inf dist (x, ?G) = ?a(G? ), I(?, G) ? Proof. Let x?G? G? so a(G? ) ? ??1 I(?, G) < ? . Further, note that I(?, G) = Z ? dist (x, ?G) dA Z + G? dist ? (x, ?G) dA. G\G? Using the denition of integral by Lebesgues, and integration by parts, we obtain Z ? a(G? ) dist ? (x, ?G) dA = Z d ?(a) da = ? a(G? ) + 0 G? Z(G) a(Gt ) dt. ? The last integral bounded by I(?, G) . Hene a(G? ) is integrable on a(G? ) ?0 ?. for all admissible [0, d? (G)] beause Therefore 0 ? lim ? a(G? ) ? lim ??0 ??0 Z? a(Gt ) dt = 0. 0 In the sequel we will use some well-known properties of the level sets (see, for example, [7?), in partiular, we will frequently make use of the relations Z ? d ? dist Z(G) (x, ?G) dA = a(G? (?))d? ? > 0, 0 G Z for ? dist ? (x, ?G) dA = a(G)d (G) + G da(G? (1)) =? d? Z Z? a(G? (?)) d? for ? < 0, (12) d? (G) dS. ?? (1) Let G be not a ball in Rn , and [·]S is the Shwarz symmetrization of a domain G? . Then: Lemma 2. 1) Z dist ? (x, ?G) dA < G? 2) Z G? Z ? dist (x, ?[G]S ) dA for ? > 0; [G? ]S dist ? (x, ?G) dA > Z [G? ]S ? dist (x, ?[G]S ) dA for ? 1 < ? < 0. R.G. SALAHUDINOV 156 Proof. Let x ? G? , then it is easy to hek that dist(x, ?G) = dist(x, ?G? ) + ?1/? . (13) The assertion follows by means of the Shwarz symmetrization of [dist(·, ?G? )]S (x) < have dist(x, ?[G? ]S ) for x ? [G? ]S , G? . ? and a ([G? ]S ) Indeed, we a (([G]S )? ) (see [7? ). Using (13) we obtain [dist(·, ?G)]S (x) = h dist(·, ?G? ) + ?1/? i S (x) = [dist(·, ?G? )]S (x) + ?1/? < < dist(x, ?[G? ]S ) + ?1/? ? dist(x, ? ([G]S )? ) + ?1/? = dist(x, ?[G]S ). Using the basi properties of the Shwarz symmetrization, for Z ? dist Z (x, ?G) dA = G? ? ([dist(·, ?G)]S (x)) dA < [G? ]S In the ase ?1 < ? < 0 Z ? > 0, dist ? we get (x, ?[G]S ) dA. [G? ]S we get the same inequality, but with the revers sign. D(? Rn ), whih orresponds to the domain G . From (5) we an onlude that I(?, D) , for xed ? , is a stritly inreasing funtion of the radius of D , and runs from 0 to ? with the radius of D . Therefore there is exatly one ball D , up to an Eulidean motion, suh that I(?, D) = I(?, G) . Note that D depends of ? and G , and it plays very important role in our proof. Aording to Lemma 2 we shall distinguish two ases: 1) ? ? 0 , and 2) ?1 < ? < 0 . We now dene a ball Proposition 1. Let ? ? 0 , and D is the ball dened as above, then 1) I(?, G) ? I(?, D) for ? > ?; (14) 2) I(?, G) ? I(?, D) for 0 ? ? < ?. (15) In the both ases the equalities hold if and only if G is a ball. Proof. To prove the assertion we apply the M.-Th. Kohler-Jobin symmetrization with a slight modiation. This method was introdued in [8, 10?, and it was applied to study isoperimetri properties of the solutions of (4), ?1 (G) , and the rst eigenfuntion of (4) (see [3, 8, 10, 12?). ? = 0 is the speial ase of ? > 0 . Further note that the ase ? = 0 with ? = 0 . Indeed, we have First of all we note that the ase ? = 0. Therefore let us x interpretation of Lemma 2 I(?, D) and using (5), we obtain I(0, D) Thus we an suppose = I(?, G) ? I(?, [G]S ), = a(D) ? a([G]S ) = a(G) = I(0, G) . ? > 0. Set i(?) = Z ? dist (x, ?G) dA ? ? a(G? ). G? The orresponding value for D i? (?? ) . we denote by From (12) we obtain ? d i(?) = Z(G) a(Gt ) dt, ? Lemma 2 with also is another ISOPERIMETRIC INEQUALITIES FOR LP -NORMS OF THE DISTANCE . . . and sine i(d? (G)) = 0 , 157 we have di(?) = a(G? ). d? ? An analogous omputation for D?? ? (16) leads to di? (?? ) = a(D?? ). d?? Now let us dene a orrespondene between ? By the denition of D , for ? = ? = = i? (?? ) . = i? (0) . G? and D?? by requiring that i(?) = 0 we have i(0) = I(?, G) = I(?, D) = G? is bounded, by Lemma 1, applying Lemma 2 to G? and D?? , we obtain ? a(G? ) . Let ?(i) be the inverse of i(?) . Hene by the dened orrespondene Beause a(D?? ) we must have the inequality d?? (i) d?(i) ?? . di di ?? (I(?, D)) = 0 = ?(I(?, G)) , we obtain ? Again, using by integration from i0 (> 0) to I(?, G) ?? (i) ? ?(i). In partiular, we have the inequality d(D) ? d(G) , (17) whih, indeed, follows immediately D. form Lemma 2 and the denition of G? (?) = G??/? , where ? > 0 . ?di(?) = a(G? )d? , by (16), we get From the denition (11) easily follows Lemma 1, (11), (12), and the equality Z ? a(G) ? dist (x, ?G)dA = Z 0 G ? d ?(a)da = d Z(G) a(Gt (?)) dt = 0 ? ? 0 0 d Z(G) a(Gt?/? ) dt = 0 dZ(G) I(?,G) Z(G) Z ? ? ? ?/??1 ?/??1 t a(Gt ) dt = ? t di(t) = ??/??1 (i) di. = ? ? ? In the both ases ? ? ?, and 0 < ? < ?, in the seond ase we suppose that I(?, G) Using (18) 0 using (17), we obtain (14), and (15), but < ?. The ases of equality immediately follows from Lemma 2. Proposition 2. way like above, then Let ?1 < ? < 0 , ? ? 0 , and D is the ball dened in the same 1) I(?, G) ? I(?, D) for 0 ? ? > ?; 2) I(?, G) ? I(?, D) for ? > ?. Proof. We will use same ideas as in Proposition 1, but in quite dierent situation. ? , then I(?, D) is an inreasing funtion of the radius D . Moreover, the diret alulation shows that the same assertion remains true for a(D? ) with xed ? and ? . We will use this remark bellow. In partiular, from Lemma 2, and the denition of D , follow First, we note that if we x of a(G) As in Proposition 1 the ase Therefore let ? < 0. ? = 0 = a([G]S ) ? a(D). is the partiular ase of Lemma 2 with (19) ? = 0. R.G. SALAHUDINOV 158 z z=? z=d?(G) y G ?1/? G? x Fig. 1 Set i(?) = Z? ? where b a(G? ) = ? ?a(G) ? a(G? ) b a(Gt ) dt, for 0 ? ? ? d? (G), for ? > d? (G). In partiular, using (12) we get ? i(0) = a(G)d (G) + Z? a(Gt ) dt = I(?, G). d? (G) For the onveniene of the reader we give an illustration for the plane ase (see Fig. 1). ? ? The orresponding value for D we denote by i (? ) . Further, almost everywhere we have ? An analogous omputation for D?? ? di(?) =b a(G? ). d? leads to di? (?? ) =b a(D?? ). d?? G? and D?? by requiring that i(?) = ? = ?? = 0 we have i(0) = I(?, G) = I(?, D) = Let us dene a orrespondene between = i? (?? ) . = i? (0) . By the denition of D, (20) for ISOPERIMETRIC INEQUALITIES FOR LP -NORMS OF THE DISTANCE . . . 159 G? and D?? , the note at the beginning of the proof, and ?b a(G? ) . Let ?(i) be the inverse of i(?) . Hene by the dened Applying Lemma 2 to (19), we obtain b a(D?? ) orrespondene we must have the inequality ? Using d?? (i) d?(i) ?? . di di ?? (I(?, D)) = 0 = ?(I(?, G)) , we obtain by integration from i0 (> 0) to I(?, G) ?? (i) ? ?(i). (21) G? (?) = G??/? , where ? < 0 . ?di(?) = b a(G? )d? , by (20), we get From the denition (11) easily follows Lemma 1, (11), (12), and the equality Z ? dist (x, ?G) dA = a(G)d? (G) + a(Gt (?)) dt = a(G) (d? (G)) ?/? + d? (G) G + Z? ? d Z? Z(G) ? ?/??1 a(G) dt + t?/??1 a(Gt ) dt = t ? ? a(Gt?/? ) dt = ? d? (G) ? = ? d? (G) 0 Z? t ?/??1 0 Let Z? Using ? < ? < 0, then ? a(Gt ) dt = ? b ? ?/? ? 1 < 0 . Z? I(?,G) t ?/??1 0 Z Z (?(i))?/??1 di. (22) 0 From (21) and (22) we easy obtain I(?,G) ? I(?, G) = ? ? di(t) = ? I(?,D) ?/??1 (?(i)) ? di ? ? 0 Z (?? (i))?/??1 di = I(?, D), 0 whih is the rst inequality of our proposition. If we interhange ? and ?, then the seond inequality follows from the rst. This ompletes the proof of Proposition 2. Proof of Theorem I(?, D) Proof of Theorem I(p, D) = I(p, D1 ) . = I(p, D2 ) . We just apply Proposition 1 and Proposition 2 for the ase 2. To prove the rst part we apply Proposition 1 for the ase The seond part follows from Proposition 2 for the ase I(p, D) Proof of Theorem for 1. = I(?, G) . ? ? n ? ? 3. First note that the isoperimetri inequality = I? (G) ? I? (G) follows easily from Lemma 2, and moreover, we have the same lower and the upper bound In (G) for I? (G) and I? (G) respetively. Now we prove that the denition (9) is well-dened for domains with smooth boundaries. For the brevity we will use the denotation lim := ? lim ??(n?1)+0 . Indeed, using (5), (11), and (12), and R.G. SALAHUDINOV 160 applying a integration by parts, we obtain Z lim(I? (G))? = lim ?,n ? ? ??n (x, ?G) dA dist = G a(G) = lim ?,n ? Z d(G) ??n ? (a(G? (1))) da(G? (1)) = lim ? ?,n ??n+1 0 0 = lim ? ? d(G) dS Z Note that, from (5) and (8) we an see G ??n+1 ? 0 ?d(G) (1) the radius of the ball Z dSd???n+1 = ?? (1) ?(? + 1)?(n/2) Ч 2? n/2 ?(? ? n + 2)?(n) Z ? Ч ?d(G)??n+1 Let Z ? ? d? ?? Z ?? (1) I? (D) = R ?? ?(n /2 ) dS ?? = 2? n/2 for all ? ? n ? 1, Z dS. ?G where R is D. is not a ball in Rn , and x ?(? ?0 ) , then I? (G) is bounded, by Proposition 1 D(?) in Rn , and Proposition 2. Like above we an show that there is exatly one ball I? (G) = I? (D(?)) . Further we onlude from (5) ? and ? . Therefore for a small ? > 0 , from Proposition 1 and Proposition 2 follow again that I?+? (G) is bounded, and the ball D(?) gives a larger I?+? (·) than the domain G , that is up to an Eulidean motion, suh that and (6) that I? (D(?)) = I? (D(?)) for all admissible I?+? (G) < I?+? (D(?)) = I? (D(?)) = I? (G), (23) whih is the desired onlusion. To omplete the proof we have to prove (10). From the monotoni property of I? (G) with respet to the domain we get the inequality be enough for us to establish the reverse inequality. Let Then, applying (18) for ? > ?0 , 1/? I? (G) = (?,n I(? ? n)) ? ? = d(G) ? where ? we an write the equality ? Z ? ?(???0 )/(?0 ?n) (i) di? 0 I(?0 ?n,G) (? ? n)?,n (?0 ? n) [d(G)] ?1/? I(?0 ?n,G) ??n ? = ??,n ?0 ? n ??0 Z 0 ? ?0 ?n d (G) (???0 )/(?0 ?n) = ?1/? ? (i) di? , ?0 ?n orresponds to the level sets of dist (x, ?G) . Thus we easily get I? (G) ? d(G) Tending I? (G) ? d(G) . Thus it would I?0 (G) < ? for some ?0 ? n . ? 1/? (? ? n) ?,n I(?0 ? n, G) . (?0 ? n) [d(G)]??0 to the innity, we get I? (G) ? d(G) . This nishes the proof of the theorem. In the onlusion we have to make a remark about reverse inequalities in the theorems. Note that, using Proposition 1 we also an 'go bak' from I? (G) to I??? (G) , but ISOPERIMETRIC INEQUALITIES FOR LP -NORMS OF THE DISTANCE . . . we need to know that G I??? (G) 161 is bounded. We now give a simple example of domain on the plane, suh that it has I? (G) = ? and I? (G) G, ? > ? . In parI? (G) > cI? (G) for ? > ? is bounded for tiular, that means, we annot prove the reverse inequality c > 0 is a universal onstant. ? > 0 , and onsider the domain plaed ?1/? between urves x = 1 , y = 0 , and y = x . There exists X ? R suh that dist(z, ?G) = y/2 , where z = x + iy , and x > X . Denote by GX the subdomain of G between x = 1 and x = X , we obtain without additional restritions on here Without loss of generality we suppose Z dist ? (x, ?G) dA ? G Z ? dist 1 (x, ?G) dA + 2 GX ZX0 dx ?? ?, x X0 ?? X nevertheless Z ?+? dist G where ? > 0. (x, ?G) dA ? Z ?+? dist (x, ?G) dA + GX Z? dx < ?, x1+?/? X The generalization to an n -dimensional ase is obvious. The author is thankful to Professor F.G. Avkhadiev for interesting disussions on the mathematial physis and useful advies. The work was also supported by Russian Foundation of Basi Researh (grant No. 05-01-00523). езюме Изопериметрические неравенства для Lp -норм ункции расстояния до границы области. .. Салахудинов. Доказана изопериметрическая монотонность евклидовых степенных моментов области относительно своей границы. Доказанное свойство, эквивалентно изопериметрическим неравенствам для Lp -норм ункции расстояния до границы области для различных значений p . Literature 1. Avkhadiev F.G. Solution of generalizated St Venant problem // Sborn.: Math. ? 1998. ? V. 189, No 12. ? P. 1739?1748. 2. Avkhadiev F.G. Geometric characteristics of domains equivalent to the norms of some embedding operators // Proc. of the Intern. Conf. and Chebyshev Lectures. ? M.: Moscow State University, 1996. ? V. 1. ? P. 12?14 (in Russian). 3. Kohler-Jobin M.-Th. Symmetrization with equal Dirichlet integrals // SIAM J. Math. Anal. ? 1982. ? V. 13. ? P. 153?161. 4. Salahudinov R.G. Isoperimetric inequality for torsional rigidity in the complex plane // J. of Inequal. & Appl. ? 2001. ? V. 6. ? P. 253?260. 5. Ban?uelos R., van den Berg M., Carroll T. Torsional Rigidity and Expected Lifetime of Brownian motion // J. London Math. Soc. ? 2002. ? V. 66. No 2. ? P. 499?512. 6. Davies E.B. A review of Hardy Inequalities // Operator Theory: Adv. Appl. ? 1989. ? V. 110. ? P. 55-67. 162 R.G. SALAHUDINOV 7. Bandle C. Isoperimetric inequalities and applications. ? Boston: Pitman Advanced Publishing Program, 1980. ? 228 p. 8. Kohler-Jobin M.-Th. Une proprie?te? de monotonie isope?rime?trique qui contient plusieurs the?ore?mes classiques // C. R. Acad. Sci. Paris. ? 1977. ? V. 284, No 3. ? P. 917?920. 9. Bandle C. Bounds of the solutions of boundary value problems // J. of Math. Anal. and Appl. ? 1976. ? V. 54. ? P. 706?716. 10. Kohler-Jobin M.-Th. Isoperimetric monotonicity and isoperimetric inequalities of Payne ? Rayner type for the first eigenfunction of the Helmholtz problem // Z. Angew. Math. Phys. ? 1981. ? V. 32. ? P. 625?646. 11. Hayman W.K. Some bounds for the principal frequency // Appl. Anal. ? 1978. ? V. 7. ? P. 247?254. 12. Hersch J. Isoperimetric monotonicity ? some properties and conjectures (connection between Isoperimetric Inequalities) // SIAM REV. ? 1988. ? V. 30, No 4. ? P. 551?577. 13. Bandle C. Estimates for the Green?s functions of Elliptic Operators // SIAM J. Math. Anal. ? 1978. ? V. 9. ? P. 1126?1136. 14. Avkhadiev F.G., Salahudinov R.G. Isoperimetric inequalities for conformal moments of plane domains // J. of Inequal. & Appl. ? 2002. ? V. 7, No 4. ? P. 593?601. 15. Leavitt J., Ungar P. Circle supports the largest sandpile // Comment. Pure Appl. Math. ? 1962. ? V. 15. ? P. 35?37. 16. Avkhadiev F.G., Salahudinov R.G. Bilateral Isoperimetric inequalities for boundary moments of plane domains // Lobachevskii J. of Mathematics. ? 2001. ? V. 9. ? P. 3?5 (URL: http://ljm.ksu.ru). 17. Schmidt E. U?ber das isoperimetrische Problem im Raum von n Dimensionen // Math. Z. ? 1938. ? V. 44. ? P. 689?788. Поступила в редакцию 21.03.06 Салахудинов устем умерович кандидат изико-математических наук, доцент, старший научный сотрудник НИИ математики и механики им. Н.. Чеботарева Казанского государственного университета. E-mail: rsalahudinovksu.ru

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