Disk to radial segments, respectively, to circles centered in the origin. Lemma 2 ([23] Theorem two.two). For x, y D, sD ( x, y) 0, x and y are collinear.| x -y| . 2-| x y|Equality holds if and only ifLemma 3. The following addition formula holds: if -1 r s t 1, then arctanhsD (r, t) = arctanhsD (r, s) arctanhsD (s, t). The restriction of arctanhsD to each and every diameter of the unit disk is usually a metric. Proof. Recall that arctanh(u) = v 1, then1log 1u for all u (-1, 1). In unique, if -1 u 1- u 1 1 log 2 1-(v-u) 2-(vu) (v-u) 2-(vu)arctanhsD (u, v) ==1 1-u log 2 1-vSymmetry 2021, 13,7 ofLet -1 r s t 1. Then arctanhsD (r, s) arctanhsD (s, t)= =1-r 1 1-s 1 log log two 1-s 2 1-t 1 1-r log = arctanhsD (r, t). two 1-tSince sD is invariant to 20(S)-Hydroxycholesterol Biological Activity rotations about the origin (a lot more typically, sD is invariant to similarities), it SB 271046 Data Sheet suffices to prove that the restriction of arctanhsD for the intersection of your unit disk with all the true axis is a metric. This follows from the above addition formula, observing that 0 max(arctanhsD (r, s), arctanhsD (s, t)) arctanhsD (r, t). We prove that the restriction of the triangular ratio metric on the unit disk sD to every single radial segment with the unit disk requires all values involving 0 and 1. Lemma 4. For each and every (0, 1) and r 0, 1 , there exists s r, 1 such that sD (r, s) = . 21 Proof. Let (0, 1) and r 0, two .Assume that s (r, 1). Then sD (r, s) = if and only if Note that s =(1- )r 1implies s – r =(1-2r ) 1r s -r = , i.e., s = (1-) . 1 1-(sr ) (1-)(2r -1) 1 and s – 2 = 2(1) 0.Proposition 4. Let f : [0, 1) [0, ). (1) In the event the restriction of f sD to some radial segment with the unit disk D is usually a metric, then f tanh is subadditive on [0, ). (two) If f is amenable and f tanh is subadditive and nondecreasing on [0, ), then the restriction of f sD to every diameter in the unit disk D is really a metric. Proof. (1) Let F := f tanh : [0, ) [0, ). As sD is invariant to rotations around the origin, we may possibly assume that the restriction of f sD = F arctanhsD to x iy : x [0, 1), y = 0 is a metric. Considering that F (0) = 0, it suffices to prove that F is subadditive on (0, ). Let a, b (0, ). We prove that F ( a b) F ( a) F (b). Denote tanh a = and tanh b = Repair r 0, 1 . By Lemma 4, there exists s r, 1 two two once again Lemma 4, we get t s,1such that sD (r, s) = . Applyingsuch that sD (s, t) = The addition formula arctanhsD (r, t) = arctanhsD (r, s) arctanhsD (s, t) shows that arctanhsD (r, t) = a b. Since the restriction of F arctanhsD to x iy : x [0, 1), y = 0 is a metric, F (arctanhsD (r, t)) F (arctanhsD (r, s)) F (arctanhsD (s, t)), i.e., F ( a b) F ( a) F (b), q.e.d. (2) The function F := f tanh : [0, ) [0, ) is amenable, subadditive and nondecreasing self-mapping of [0, ); thus, F is metric-preserving. By Lemma 3, the restriction of arctanhsD to just about every diameter from the unit disk D can be a metric. Then the restriction of f sD = F arctanhsD to every diameter on the unit disk D is usually a metric. Remark four. Let f : [0, 1) [0, ). If f is amenable, subadditive and nondecreasing on [0, 1), then f sD is a metric around the entire unit disk D and of course f tanh is amenable, subadditive and nondecreasing on [0, ). We prove that each restriction of arctanhsD to a circle centered at origin and contained in the unit disk is a metric.Symmetry 2021, 13,8 ofIf x, y D with | x | = |y| = 1 it can be identified from ([24] Remark 3.14) that, denoting 0 := 2 arccos and 2 := ( x, 0, y), we’ve got sD ( x, y) = If two = 0 , thensin 12 -2 cossin1 -.