Involving the Agard Distortion Function

and Applied Analysis 3 in K0,∞ . In particular, the inequality λ K ≥ K holds for all K ∈ 1,∞ with the best possible constant c a. 2 g K logηK t − log t / K − 1 is convex in 1,∞ for fixed t ∈ 0,∞ . 3 If t ≥ 1 and r √ t/ 1 t , then h K logηK t − log t / K − 1/K is strictly increasing from 1,∞ onto 2K r K′ r /π,πK r /K′ r . Theorem 1.2. Let t ∈ 0,∞ , r √ t/ 1 t , a 4/π K 1/√2 , b a/2,A r π2/ 2μ r , B r 8K r K′ r 2 E r − r ′K r /π2, and Fc K K logηK t − log t / K − 1 − c . Then, the following statements are true. 1 Fc K is strictly decreasing from 1,∞ onto −∞, 4K r K′ r /π − c for c > A r . If c A r , then Fc K is strictly decreasing from 1,∞ onto A r − 4 log 2 − log t, 4K r K′ r /π −A r . Moreover, te K−1 A r A r −4 log 2−log t /K < ηK t < te K−1 A r 4K r K ′ r /π−A r /K 1.10 for all t ∈ 0,∞ and K ∈ 1,∞ . In particular, if t 1, then 1.10 becomes e K−1 π π−4 log 2 /K < λ K < e K−1 π a−π /K . 1.11 2 If c ≤ B r , then Fc K is strictly increasing from 1,∞ onto 4K r K′ r /π − c,∞ . Moreover, ηK t > te K−1 B r 4K r K ′ r /π−B r /K 1.12 for all t ∈ 0,∞ and K ∈ 1,∞ . In particular, if t 1, then 1.12 becomes λ K > e K−1 b b/K e K− 1/K . 1.13 3 If B r < c < A r , then there existsK1 ∈ 1,∞ such that Fc K is strictly decreasing on 1, K1 and strictly increasing on K1,∞ . 4 Fc K is convex in 1,∞ . 2. Lemmas In order to prove our main results, we need several formulas and lemmas, which we present in this section. 4 Abstract and Applied Analysis The following formulas were presented in 14, Appendix E, pp. 474-475 . Let t ∈ 0,∞ , K ∈ 0,∞ , r √ t/ 1 t ∈ 0, 1 , and s φK r . Then, dK r dr E r − r ′K r rr ′2 , dE r dr E r −K r r , K r E′ r K′ r E r −K r K′ r π 2 , dμ r dr − π 2 4rr ′2K r 2 , ∂s ∂r ss′K s K′ s rr ′2K r K′ r , ∂s ∂K 2 πK ss′K s K′ s , φK r 2 φ1/K ( r ′ )2 1, ηK t ( s s′ )2 , ∂ηK t ∂K 4 πK ηK t K s K′ s 2 μ r K′ s ηK t . 2.1 Lemma 2.1 see 14, Theorem 1.25 . For −∞ < a < b < ∞, let f, g : a, b → R be continuous on a, b and differentiable on a, b , and let g ′ x / 0 on a, b . If f ′ x /g ′ x is increasing (decreasing) on a, b , then so are f x − f a g x − g a , f x − f b g x − g b . 2.2 If f ′ x /g ′ x is strictly monotone, then the monotonicity in the conclusion is also strict. The following lemma can be found in 14, Theorem 3.21 1 and 7 , Lemma 3.32 1 and Theorem 5.13 2 . Lemma 2.2. 1 E r − r ′K r /r2 is strictly increasing from 0, 1 onto π/4, 1 ; 2 r ′K r is strictly decreasing from 0, 1 onto 0, π/2 if and only if c ≥ 1/2; 3 K r K′ r is strictly decreasing in 0,√2/2 and strictly increasing in √2/2, 1 ; 4 μ r log r is strictly decreasing from 0, 1 onto 0, log 4 . Lemma 2.3. Let r ∈ 1/√2, 1 , K ∈ 1,∞ , and s φK r . Then, G K ≡ {π/ 2K s } μ r /K′ s 2 is strictly decreasing from 1,∞ onto K′ r /K r , π2/ 2K r 2 . Proof. Clearly G 1 π2/ 2K r 2 , G ∞ K′ r /K r . Differentiating G K , one has G′ K 4 πK μ r K s K′ s −2 [ E′ s − s2K′ s ] − π K K s −2K′ s [ E s − s′K s ] 4 πK K s −2K′ s G1 K , 2.3 where G1 K E′ s − s2K′ s K s μ r 2 − π2 E s − s′K s K′ s /4. Abstract and Applied Analysis 5 From Lemma 2.2 1 and 2 , we clearly see that G1 K is strictly decreasing in 1,∞ . Moreover,and Applied Analysis 5 From Lemma 2.2 1 and 2 , we clearly see that G1 K is strictly decreasing in 1,∞ . Moreover, lim K→ 1 G1 K [ E′ r − r2K′ r ] K r μ r 2 − π 2 4 [ E r − r ′K r ] K′ r 3 π2 4 K′ r G2 r , 2.4 whereG2 r K r E′ r −r2K′ r −K′ r E r −r ′K r is also strictly decreasing in 0, 1 . Thus, G2 r ≤ G2 √ 2/2 0 for r ∈ 1/√2, 1 , and G1 K < G1 1 ≤ 0 for K ∈ 1,∞ . Therefore, the monotonicity ofG K follows from 2.3 and 2.4 together with the fact that G1 K < 0 for K ∈ 1,∞ . 3. Proofs of Theorems 1.1 and 1.2 Proof of Theorem 1.1. For part 1 , clearly f 1 1. Let r μ−1 π/ 2K for K ∈ 1,∞ , then λ K r/r ′ , r ∈ 1/√2, 1 , dr dK 2 π rr ′K′ r , dλ K dK 4 π λ K K′ r , 3.1 lim K→ ∞ f K lim r→ 1 r2K′ r r ′2K r ∞. 3.2 Making use of 3.1 , we have K 1f ′ K λ K f1 K ≡ 4 π K′ r K r − c. 3.3 It follows from Lemma 2.2 3 that f1 K is strictly increasing from 1,∞ onto a − c,∞ . Then, from 3.2 and 3.3 , we know that f is strictly increasing from 1,∞ onto 1,∞ for c ≤ a. If c > a, then there existsK0 ∈ 1,∞ such that f is strictly decreasing in 1, K0 and strictly increasing in K0,∞ . Moreover, the inequality λ K ≥ K holds for all K ∈ 1,∞ with the best possible constant c a. For part 2 , denote r √ t/ 1 t . Differentiating g K , we get g ′ K 2K′ s 2 K − 1 /μ r − (logηK t − log t) K − 1 2 . 3.4 Let g1 K 2K′ s 2 K − 1 /μ r − logηK t − log t and g2 K K − 1 , then g1 1 g2 1 0, g ′ K g1 K /g2 K and g1 ′ K g2′ K g3 K ≡ − 2 μ r 2 [ E′ s − s2K′ s ] K′ s . 3.5 6 Abstract and Applied Analysis Clearly, g3 K is strictly increasing in 1,∞ . Then, 3.5 and Lemma 2.1 lead to the conclusion that g ′ K is strictly increasing in 1,∞ . Therefore, g K is convex in 1,∞ . For part 3 , if t ≥ 1, then r ≥ √2/2. Let h1 K logηK t − log t and h2 K K − 1/K, then h1 1 h2 1 0, h K h1 K /h2 K , and h1 ′ K h2 ′ K 2K′ s /μ r 1 K−2 2μ r G K , 3.6 where G K is defined as in Lemma 2.2. Therefore, h K is strictly increasing in 1,∞ for t ≥ 1 follows from Lemmas 2.1 and 2.2 together with 3.6 . Moreover, making use of l’Hôpital’s rule, we have h 1 2K r K′ r /π , h ∞ πK r /K′ r . Proof of Theorem 1.2. Differentiating Fc K gives F ′ c K logηK t − log t K − 1 − c K ⎡ ⎢⎣ ( 2K′ s 2 K − 1 ) /μ r − (logηK t − log t) K − 1 2 ⎤ ⎥⎦ 2K′ s K K − 1 /μ r − (logηK t − log t) K − 1 2 − c. 3.7