P, we prove Goralatide Epigenetic Reader Domain Theorem two in two instances. (i) Let us suppose
P, we prove Theorem two in two cases. (i) Let us suppose that D (n, k, c) P c; then, from Lemma three, Q P,n,k,c (sup ||) 0. Hence, by (40), we receive sup ||2 = 0 and Mn is totally umbilical.n In specific, if L1 1 is an Einstein manifold, then (30) indicates that H is also a continuous; Benidipine Technical Information therefore, (39) becomes0=(nH )1 ||2 Q P,n,k,c (||) 0. n-(41)As a result, the inequalities in (41) hold for equalities, which is to say, all of the inequalities that we have obtained are, in actual fact, equalities, also because the curvature condition (2). As a n outcome, (1) and (two) indicate that the Ricci curvature of L1 1 is Ric(e j ) =R ( e j , ei ) R ( e j , e n 1 ) = ( n – 1 ) c two ic1 , nRic(en1 ) =R ( e n 1 , ei ) = c 1 .in 1 Consequently, we have c2 = cn , because of L1 1 being an Einstein manifold, and, by the n hypothesis of geodesic completeness and connectivity, the ambient space L1 1 must be theMathematics 2021, 9,11 ofn de Sitter space S1 1 (c). As a result, by (15) and (30), we know this totally umbilical hypersurface must be a sphere Sn ( R).(ii) When 0 P D (n, k, c), it follows, from Lemma 3 and (40), that either sup ||two = 0 and Mn is entirely umbilical, or Q P,n,k,c (sup ||) 0 with ( P, n, k, c) sup ||2 ( P, n, k, c). If the equality sup ||2 = ( P, n, k, c) holds, then ||two ( P, n, k, c). Making use of Lemma 3, we’ve Q P,n,k,c (||) 0. Inserting this into (39) yields (nH ) 0 on Mn .In addition, because P D (n, k, c) c, then (17) gives n2 H 2 S, which indicates H = 0, by choosing an suitable orientation such that H 0 on Mn , so we’ve nH – i 0; hence, the operator is elliptic. By means of (30), the assumption that sup ||two attains at some points on Mn assures that sup H two also attains at some points on Mn . Therefore, depending on the sturdy maximum principle, H is actually a continual. Furthermore, (41) becomes trivially an equality, which suggests all the inequalities we’ve obtained turn out to be equalities; hence, (22) must be also an equality or, equivalently, | B|2 = n2 | H |2 = 0, that is definitely, Mn is an isoparametric hypersurface. Moreover, (41) assures that the equality in (28) holds, which implies, by (19) and J. Mel dez ([26], Lemma two.two), that the hypersurface has specifically two distinct constant principal curvatures, with multiplicities k and n – k. If the equality ||two = ( P, n, k, c) holds, then Q P,n,k,c (||) = 0 and (41) becomes trivially an equality, by a equivalent way as above; Mn is definitely an isoparametric hypersurface of two distinct constant principal curvatures with multiplicities k and n – k.Inside the following, we classify the isoparametric hypersurface pointed out above which n satisfies sup ||two = ( P, n, k, c) or ||2 = ( P, n, k, c) below the assumption of L1 1 getting a geodesically complete simply-connected Einstein manifold. Considering the fact that we’ve proved that (41) n n becomes trivially an equality in this setting, comparable to (i), we know L1 1 = S1 1 (c). By a n should be isometric to a normal classical congruence theorem (in [29]), we conclude that M n product Hk ( a) Sn-k (b) S1 1 (c), where 2 k n or n k n – two, a 0, b 0 and 2 2 1 1 1 a b = c . Let us denote its principal curvatures by1 = = k = c – a and k1 = = n = c – b. Let = c – a and = c – b; then, with each other with c = by (15),H= and P = c-(42)(n – k) kc k(n – k) c , | |2 = -nn (43)2k(n – k ) k ( k – 1) c2 (n – k)(n – k – 1) 2 – c- . n ( n – 1) n ( n – 1) n ( n – 1) (44)Because 0 c as a result of (42); hence, solving the Equation (44), we receive 0 P D (n, k, c), when two k n ;- P c,whenn k n – two,where D (n, k, c) is provided by (32). With each other using the selection of P in Theorem two,.