We finally obtain two k n and Mn is often a solution Hk
We lastly receive 2 k n and Mn is usually a product Hk ( a) Sn-k (b) with two k n . 2 two Now, we give the values in the constants a, b. For any integer two k n , solving the 2 Equation (44), we’ve got n(n – 1)(c – P) – 2k(n – k)c = two(n – k)(n – k – 1),(45)Mathematics 2021, 9,12 ofor, n(n – 1)(c – P) – 2k(n – k)c – = two(n – k)(n – k – 1),(46)exactly where is provided by (34). Substituting (45) and (46) into the second equation of (43) and comparing with (35), we acquire, respectively,||two = ( P, n, k, c), or ||two = ( P, n, k, c).As a result, sup ||two = ( P, n, k, c) or ||2 = ( P, n, k, c) holds if and only if (45) or (46) holds. Solving (45) and with each other with c = and = c – b, we receive(n – 1) (n – 2k)cnP – b= two(n – k)(n – k – 1), (47) .(n – 1) nP – (n – 2k)c a= 2k(k – 1)Similarly, it follows, from (46), that(n – 1) (n – 2k)cnP b= two(n – k)(n – k – 1), (48) .(n – 1) nP – (n – 2k)c – a= 2k(k – 1)Thus, we obtain that Mn is isometric to Hk ( a) Sn-k (b), 2 k n , when the 2 equality sup ||two = ( P, n, k, c) holds having a, b defined by (47), or the equality ||two = ( P, n, k, c) holds having a, b defined by (48). We full the proof of Theorem 2. 6. Proofs of Theorems 1 and three Proof of Theorem 1. By the classical algebraic inequality as a consequence of M. Okumura in ([27], Lemma 2.1), (19) holds automatically for k = 1. Then, Lemmas 1 and 2 imply that(nH )where1 ||two Q P,n,1,c (||), n-(49)Q P,n,1,c ( x ) = (n – two) x2 – (n – two) xx2 n(n – 1)(c – P) n(n – 1) P.It really is quick to determine that Q P,n,1,c ( x ) is decreasing for any x 0, and Q P,n,1,c (0) = n(n – 1) P 0. (i) Let us suppose that n P c; then, we claim that Q P,n,1,c ( x ) 0 for every x 0. Indeed, if there exists a point x0 such that Q P,n,1,c ( x0 ) = 0, a simple computation provides ( n – 1) P2 two x0 = , (50) ( n – two ) n -2 c – P n which indicates that P Q P,n,1,c ( x ), the claim is accurate.( n -2) c n , ( n -2) ca contradiction. By the continuity with the functionUsing Lemma four, there exists a sequence of points q Mn ; evaluating (49) in the sequence q , we acquire 0 lim sup (nH )(q )1 sup ||two Q P,n,1,c (sup ||). n-(51)Mathematics 2021, 9,13 ofSo, we immediately Tyrosine-protein Kinase Lyn Proteins Formulation conclude that sup ||2 = 0 and Mn is entirely umbilical. Moreover, we additional acquire that H is continuous as a result of (30); additional, (49) has to be 0=(nH )1 ||2 Q P,n,1,c (||) 0. n-(52)n That is definitely to say that (52) becomes trivially an equality. So, when L1 1 is often a geodesically complete simply-connected Einstein manifold, together with the same discussion as the proof of n n Theorem 2, L1 1 should be the de Sitter space S1 1 (c). By (15) and (30), we know such entirely umbilical hypersurface has to be the sphere Sn ( R).(ii) Let us suppose that 0 P n ; then, Q P,n,1,c ( x ) = 0 has one positive root given by (50). From (51), we receive that either sup ||2 = 0 and Mn is entirely umbilical, or sup ||2 ( P, n, c) :=( n -2) c( n – 1) P2 . ( n – two ) n -2 c – P nLet us consider the case in which the equality sup ||2 = ( P, n, c) holds; then, ( P, n, c) and Q P,n,1,c (||) 0 on Mn . Considering that sup ||2 attains at some points on n , so does sup H as a result of (30). Apart from, P c guarantees that M is elliptic and, by the sturdy maximum principle, H is usually a continual. Hence, (52) becomes trivially an equality and Mn is definitely an isoparametric hypersurface. Moreover, the inequality in (28) holds for equality. By Lemma 2.1 in [27], we conclude that Mn is definitely an isoparametric hypersurface with two distinct continuous Leukocyte Ig-Like Receptor B4 Proteins custom synthesis principal curvatures, one particular of which can be very simple. n In distinct, if L1 1 is an Einstein manifold with geodesic completeness and.