Let H be a Hilbert space. In this paper we show among others that,
if the functions f, g : I ⊂ R →[0,∞) are continuous and A, B are selfadjoint
operators with spectra Sp (A) , Sp (B) ⊂ I, then
f2 (A) ⊗ g2 (B) + g2 (A) ⊗ f2 (B)
≥
h
f2(1−λ) (A) g2λ (A)
i
⊗
h
f2λ (B) g2(1−λ) (B)
i
+
h
f2λ (A) g2(1−λ) (A)
i
⊗
h
f2(1−λ) (B) g2λ (B)
i
≥ 2 [f (A) g (A)] ⊗ [f (B) g (B)]
for all λ ∈ [0, 1] . We also have the following inequalities for the Hadamard
product
f2 (A) ◦ g2 (B) + g2 (A) ◦ f2 (B)
≥
h
f2(1−λ) (A) g2λ (A)
i
◦
h
f2λ (B) g2(1−λ) (B)
i
+
h
f2λ (A) g2(1−λ) (A)
i
◦
h
f2(1−λ) (B) g2λ (B)
i
≥ 2 [f (A) g (A)] ◦ [f (B) g (B)]
for all λ ∈ [0, 1] .departmental bulletin pape

Silvestru Sever Dragomir

English

Tokyo University of Science Repository for Academic Resources / 東京理科大学学術リポジトリ

SUT Journal of MathematicsVol. 60, No. 1 (2024), 1–16DOI: 10.55937/sut/1717241617Tensorial and Hadamard product inequalities ofSchwarz type for selfadjoint operators in HilbertspacesSilvestru Sever Dragomir1,2(Received June 14, 2023)Abstract. Let H be a Hilbert space. In this paper we show among others that,if the functions f, g : I ⊂ R → [0,∞) are continuous and A, B are selfadjointoperators with spectra Sp (A) , Sp (B) ⊂ I, thenf2 (A)⊗ g2 (B) + g2 (A)⊗ f2 (B)≥[f2(1−λ) (A) g2λ (A)]⊗[f2λ (B) g2(1−λ) (B)]+[f2λ (A) g2(1−λ) (A)]⊗[f2(1−λ) (B) g2λ (B)]≥ 2 [f (A) g (A)]⊗ [f (B) g (B)]for all λ ∈ [0, 1] . We also have the following inequalities for the Hadamardproductf2 (A) ◦ g2 (B) + g2 (A) ◦ f2 (B)≥[f2(1−λ) (A) g2λ (A)]◦[f2λ (B) g2(1−λ) (B)]+[f2λ (A) g2(1−λ) (A)]◦[f2(1−λ) (B) g2λ (B)]≥ 2 [f (A) g (A)] ◦ [f (B) g (B)]for all λ ∈ [0, 1] .AMS 2020 Mathematics Subject Classification. 47A63, 47A99.Key words and phrases. Tensorial product, Hadamard product, Selfadjoint op-erators, Convex functions, Schwarz inequality.§1. IntroductionLet I1, ..., Ik be intervals from R and let f : I1 × · · · × Ik → R be an es-sentially bounded real function defined on the product of the intervals. Let12 S. S. DRAGOMIRA = (A1, ..., Ak) be a k-tuple of bounded selfadjoint operators on Hilbertspaces H1, ...,Hk such that the spectrum of Ai is contained in Ii for i = 1, ..., k.We say that such a k-tuple is in the domain of f . IfAi =∫IiλidEi (λi)is the spectral resolution of Ai for i = 1, ..., k; by following [1], we define(1.1) f (A1, ..., Ak) :=∫I1· · ·∫Ikf (λ1, · · · , λk) dE1 (λ1)⊗ · · · ⊗ dEk (λk)as a bounded selfadjoint operator on the tensorial product H1 ⊗ · · · ⊗Hk.If the Hilbert spaces are of finite dimension, then the above integrals be-come finite sums, and we may consider the functional calculus for arbitraryreal functions. This construction [1] extends the definition of Korányi [5] forfunctions of two variables and have the property thatf (A1, ..., Ak) = f1(A1)⊗ · · · ⊗ fk(Ak),whenever f can be separated as a product f(t1, ..., tk) = f1(t1) · · · fk(tk) of kfunctions each depending on only one variable.Recall the geometric operator mean for the positive operators A, B > 0A#tB := A1/2(A−1/2BA−1/2)tA1/2,where t ∈ [0, 1] andA#B := A1/2(A−1/2BA−1/2)1/2A1/2.By the definitions of # and ⊗ we haveA#B = B#A and (A#B)⊗ (B#A) = (A⊗B)# (B ⊗A) .In 2007, S. Wada [6] obtained the following Callebaut type inequalities fortensorial product(A#B)⊗ (A#B) ≤ 12[(A#αB)⊗ (A#1−αB) + (A#1−αB)⊗ (A#αB)]≤ 12(A⊗B +B ⊗A)for A, B > 0 and α ∈ [0, 1] .Recall that the Hadamard product of A and B in B(H) is defined to be theoperator A ◦B ∈ B(H) satisfying⟨(A ◦B) ej , ej⟩ = ⟨Aej , ej⟩ ⟨Bej , ej⟩INEQUALITIES OF SCHWARZ TYPE 3for all j ∈ N, where {ej}j∈N is an orthonormal basis for the separable Hilbertspace H.It is known that, see [3], we have the representation(1.2) A ◦B = U∗ (A⊗B)U ,where U : H → H ⊗H is the isometry defined by Uej = ej ⊗ ej for all j ∈ N.We recall also Fiedler inequality(1.3) A ◦A−1 ≥ 1 for A > 0.In this paper we obtain among others some Callebaut type inequalitiesfor tensorial and Hadamard products of operators in Hilbert spaces. Theseprovide refinements of Schwarz inequality as in the scalar case. Functionalgeneralization of Young’s inequality and Schwarz inequality for power seriesare also shown. In the process, some refinements of Fiedler inequality in thesimple form (1.3) and its additive version for n operators (3.1) are given aswell.§2. Main resultsThe following result providing a refinement of Cauchy-Schwarz inequalityholds:Theorem 2.1. Assume that the functions f, g : I ⊂ R → [0,∞) are con-tinuous and A, B are selfadjoint operators with spectra Sp (A) , Sp (B) ⊂ I.Thenf2 (A)⊗ g2 (B) + g2 (A)⊗ f2 (B)(2.1)≥[f2(1−λ) (A) g2λ (A)]⊗[f2λ (B) g2(1−λ) (B)]+[f2λ (A) g2(1−λ) (A)]⊗[f2(1−λ) (B) g2λ (B)]≥ 2 [f (A) g (A)]⊗ [f (B) g (B)]for all λ ∈ [0, 1] .Proof. Using the weighted arithmetic mean-geometric mean inequality for pos-itive numbers, we have for a, b, c, d ≥ 0 and λ ∈ [0, 1] that(1− λ) a2b2 + λc2d2 ≥ a2(1−λ)b2(1−λ)c2λd2λandλa2b2 + (1− λ) c2d2 ≥ a2λb2λc2(1−λ)d2(1−λ).4 S. S. DRAGOMIRIf we add these two inequalities, then we geta2b2 + c2d2 ≥ a2(1−λ)b2(1−λ)c2λd2λ + a2λb2λc2(1−λ)d2(1−λ).By arithmetic mean-geometric mean inequality we also havea2(1−λ)b2(1−λ)c2λd2λ + a2λb2λc2(1−λ)d2(1−λ)=(a1−λb1−λcλdλ)2+(aλbλc1−λd1−λ)2≥ 2a1−λb1−λcλdλaλbλc1−λd1−λ = 2abcdfor a, b, c, d ≥ 0 and λ ∈ [0, 1].Therefore we have(2.2) a2b2 + c2d2 ≥ a2(1−λ)b2(1−λ)c2λd2λ + a2λb2λc2(1−λ)d2(1−λ) ≥ 2abcdfor a, b, c, d ≥ 0 and λ ∈ [0, 1].Take a = f (t) , b = g (s) , c = g (t) and d = f (s) for t, s ∈ I in (2.2) to getf2 (t) g2 (s) + g2 (t) f2 (s)≥ f2(1−λ) (t) g2(1−λ) (s) g2λ (t) f2λ (s) + f2λ (t) g2λ (s) g2(1−λ) (t) f2(1−λ) (s)≥ 2f (t) g (s) g (t) f (s)for all t, s ∈ I and λ ∈ [0, 1].Now, if we take the double integral∫I∫I over dE (t)⊗ dF (s) , then we get∫I∫I[f2 (t) g2 (s) + g2 (t) f2 (s)]dE (t)⊗ dF (s)≥∫I∫If2(1−λ) (t) g2(1−λ) (s) g2λ (t) f2λ (s) dE (t)⊗ dF (s)+∫I∫If2λ (t) g2λ (s) g2(1−λ) (t) f2(1−λ) (s) dE (t)⊗ dF (s)≥ 2∫I∫If (t) g (s) g (t) f (s) dE (t)⊗ dF (s) ,namely, by (1.1),∫If2 (t) dE (t)⊗∫Ig2 (s) dF (s) +∫Ig2 (t) dE (t)⊗∫If2 (s) dF (s)≥∫If2(1−λ) (t) g2λ (t) dE (t)⊗∫Ig2(1−λ) (s) f2λ (s) dF (s)+∫If2λ (t) g2(1−λ) (t) dE (t)⊗∫Ig2λ (s) f2(1−λ) (s) dF (s)≥ 2∫If (t) g (t) dE (t)⊗∫Ig (s) f (s) dF (s)and the inequality (2.1) is thus proved.INEQUALITIES OF SCHWARZ TYPE 5Corollary 2.2. With the assumptions of Theorem 2.1,f2 (A) ◦ g2 (B) + g2 (A) ◦ f2 (B)(2.3)≥[f2(1−λ) (A) g2λ (A)]◦[f2λ (B) g2(1−λ) (B)]+[f2λ (A) g2(1−λ) (A)]◦[f2(1−λ) (B) g2λ (B)]≥ 2 [f (A) g (A)] ◦ [f (B) g (B)]for all λ ∈ [0, 1] .In particular, for B = Af2 (A) ◦ g2 (A) ≥[f2(1−λ) (A) g2λ (A)]◦[f2λ (A) g2(1−λ) (A)](2.4)≥ [f (A) g (A)] ◦ [f (A) g (A)] .Proof. By the inequality (2.1), we deriveU∗[f2 (A)⊗ g2 (B) + g2 (A)⊗ f2 (B)]U≥ U∗{[f2(1−λ) (A) g2λ (A)]⊗[f2λ (B) g2(1−λ) (B)]+[f2λ (A) g2(1−λ) (A)]⊗[f2(1−λ) (B) g2λ (B)]}U≥ 2U∗ {[f (A) g (A)]⊗ [f (B) g (B)]}U ,which is equivalent toU∗[f2 (A)⊗ g2 (B)]U + U∗[g2 (A)⊗ f2 (B)]U≥ U∗{[f2(1−λ) (A) g2λ (A)]⊗[f2λ (B) g2(1−λ) (B)]}U+ U∗{[f2λ (A) g2(1−λ) (A)]⊗[f2(1−λ) (B) g2λ (B)]}U≥ 2U∗ {[f (A) g (A)]⊗ [f (B) g (B)]}U .By utilizing the representation (1.2) we deduce the desired result (2.3).Remark 2.3. We observe that if we take f (t) = t1/2, g (t) = t−1/2, t > 0, in(2.4), then for A > 0 we get(2.5) A ◦A−1 ≥ A1−2λ ◦A2λ−1 ≥ 1for all λ ∈ [0, 1], which is a refinement of Fiedler inequality (1.3).Corollary 2.4. Assume that the functions f, g : I ⊂ R → [0,∞) are contin-uous and Ai, Bi, i ∈ {1, ..., n} are selfadjoint operators with spectra Sp (Ai) ,Sp (Bi) ⊂ I, and pi, qi ≥ 0 for i ∈ {1, ..., n}. Then(n∑i=1pif2 (Ai))⊗ n∑j=1qjg2 (Bj)(2.6)6 S. S. DRAGOMIR+(n∑i=1pig2 (Ai))⊗ n∑j=1qjf2 (Bj)≥[n∑i=1pif2(1−λ) (Ai) g2λ (Ai)]⊗ n∑j=1qjf2λ (Bj) g2(1−λ) (Bj)+[n∑i=1pif2λ (Ai) g2(1−λ) (Ai)]⊗ n∑j=1qjf2(1−λ) (Bj) g2λ (Bj)≥ 2[n∑i=1pif (Ai) g (Ai)]⊗ n∑j=1qjf (Bj) g (Bj)and (n∑i=1pif2 (Ai))◦ n∑j=1qjg2 (Bj)+(n∑i=1pig2 (Ai))◦ n∑j=1qjf2 (Bj)≥[n∑i=1pif2(1−λ) (Ai) g2λ (Ai)]◦ n∑j=1qjf2λ (Bj) g2(1−λ) (Bj)+[n∑i=1pif2λ (Ai) g2(1−λ) (Ai)]◦ n∑j=1qjf2(1−λ) (Bj) g2λ (Bj)≥ 2[n∑i=1pif (Ai) g (Ai)]◦ n∑j=1qjf (Bj) g (Bj)for all λ ∈ [0, 1] .Proof. From (2.1) we getn∑i,j=1piqjf2 (Ai)⊗ g2 (Bj) +n∑i,j=1piqjg2 (Ai)⊗ f2 (Bj)≥n∑i,j=1piqj[f2(1−λ) (Ai) g2λ (Ai)]⊗[f2λ (Bj) g2(1−λ) (Bj)]+n∑i,j=1piqj[f2λ (Ai) g2(1−λ) (Ai)]⊗[f2(1−λ) (Bj) g2λ (Bj)]INEQUALITIES OF SCHWARZ TYPE 7≥ 2n∑i,j=1piqj [f (Ai) g (Ai)]⊗ [f (Bj) g (Bj)] ,which gives (2.6).Remark 2.5. If we take in Corollary 2.4 qi = pi and Bi = Ai, i ∈ {1, ..., n} ,then (n∑i=1pif2 (Ai))⊗(n∑i=1pig2 (Ai))(2.7)+(n∑i=1pig2 (Ai))⊗(n∑i=1pif2 (Ai))≥[n∑i=1pif2(1−λ) (Ai) g2λ (Ai)]⊗[n∑i=1pif2λ (Ai) g2(1−λ) (Ai)]+[n∑i=1pif2λ (Ai) g2(1−λ) (Ai)]⊗[n∑i=1pif2(1−λ) (Ai) g2λ (Ai)]≥ 2[n∑i=1pif (Ai) g (Ai)]⊗[n∑i=1pif (Ai) g (Ai)]and (n∑i=1pif2 (Ai))◦(n∑i=1pig2 (Ai))(2.8)≥[n∑i=1pif2(1−λ) (Ai) g2λ (Ai)]◦[n∑i=1pif2λ (Ai) g2(1−λ) (Ai)]≥[n∑i=1pif (Ai) g (Ai)]◦[n∑i=1pif (Ai) g (Ai)]for all λ ∈ [0, 1] .The inequalities (2.7) and (2.8) are relevant since they provide general-izations for tensorial and Hadamard products of the celebrated Callebaut in-equality [2]n∑i=1pia2in∑i=1pib2i ≥n∑i=1pia2(1−λ)i b2λin∑i=1pia2λi b2(1−λ)i≥(n∑i=1piaibi)2,8 S. S. DRAGOMIRwhere pi, ai, bi ≥ 0, i ∈ {1, ..., n} and λ ∈ [0, 1] ,which is in its turn a refinementof the Cauchy-Schwarz discrete inequality.In the next theorem we provide a generalization of Young’s inequality formonotonic and convex (concave) functions:Theorem 2.6. Assume that f is monotonic nondecreasing and convex on[0,∞) and A, B ≥ 0. If p, q > 1 with 1p +1q = 1, then(2.9) f (A⊗B) ≤ 1pf (Ap)⊗ 1 + 1q1⊗ f (Bq) .In particular,(2.10) f (A⊗B) ≤ 12[f(A2)⊗ 1 + 1⊗ f(B2)].If f is monotonic nonincreasing and concave on [0,∞) , then the reverseinequality holds in (2.9) and (2.10).Proof. Using Young’s inequality and the monotonicity and convexity of f wehavef (ts) ≤ f(1ptp +1qsq)≤ 1pf (tp) +1qf (sq)for all t, s ≥ 0.Now, if we take the double integral∫[0,∞)∫[0,∞) over dE (t)⊗ dF (s) , thenwe get ∫[0,∞)∫[0,∞)f (st) dE (t)⊗ dF (s)(2.11)≤∫[0,∞)∫[0,∞)[1pf (tp) +1qf (sq)]dE (t)⊗ dF (s) .Since∫[0,∞)∫[0,∞)[1pf (tp) +1qf (sq)]dE (t)⊗ dF (s)=1p∫[0,∞)∫[0,∞)f (tp) dE (t)⊗ dF (s) + 1q∫[0,∞)∫[0,∞)f (sq) dE (t)⊗ dF (s)=1pf (Ap)⊗ 1 + 1q1⊗ f (Bq) ,hence by (2.11), we derive (2.9).INEQUALITIES OF SCHWARZ TYPE 9Corollary 2.7. Assume that f is monotonic nondecreasing and operator con-vex on [0,∞) and A, B ≥ 0. If p, q > 1 with 1p +1q = 1, then(2.12) f (A ◦B) ≤[1pf (Ap) +1qf (Bq)]◦ 1.In particular,f (A ◦B) ≤ 12[f(A2)+ f(B2)]◦ 1.Proof. By (1.2) and Davis-Choi-Jensen’s inequality for operator convex func-tions, we have(2.13) f (A ◦B) = f (U∗ (A⊗B)U) ≤ U∗f (A⊗B)U .By (2.9) we getU∗f (A⊗B)U ≤ U∗[1pf (Ap)⊗ 1 + 1q1⊗ f (Bq)]U(2.14)=1pU∗ (f (Ap)⊗ 1)U + 1qU∗ (1⊗ f (Bq))U=1pf (Ap) ◦ 1 + 1q1 ◦ f (Bq) .By making use of (2.13) and (2.14) we derive (2.12).Theorem 2.8. Assume that f is convex on R and A, B > 0. If p, q > 1 with1p +1q = 1, then(2.15) f (ln (A⊗B)) ≤ 1pf (p lnA)⊗ 1 + 1q1⊗ f (q lnB) .In particular,f (ln (A⊗B)) ≤ 12[f (2 lnA)⊗ 1 + 1⊗ f (2 lnB)] .Proof. We observe that for t, s > 0 thatln (ts) = ln t+ ln s =1pln (tp) +1qln (sq) .Now, if we take the function f and use its convexity, then we getf (ln (ts)) = f(1pln (tp) +1qln (sq))≤ 1pf (ln (tp)) +1qf (ln (sq))for t, s > 0.10 S. S. DRAGOMIRNow, if we take the double integral∫[0,∞)∫[0,∞) over dE (t)⊗ dF (s) , thenwe get ∫[0,∞)∫[0,∞)f (ln (ts)) dE (t)⊗ dF (s)(2.16)≤∫[0,∞)∫[0,∞)[1pf (ln (tp)) +1qf (ln (sq))]dE (t)⊗ dF (s) .Observe that∫[0,∞)∫[0,∞)f (ln (ts)) dE (t)⊗ dF (s) = f (ln (A⊗B))and∫[0,∞)∫[0,∞)[1pf (ln (tp)) +1qf (ln (sq))]dE (t)⊗ dF (s)=1p∫[0,∞)∫[0,∞)f (ln (tp)) dE (t)⊗ dF (s)+1q∫[0,∞)∫[0,∞)f (ln (sq)) dE (t)⊗ dF (s)=1pf (lnAp)⊗ 1 + 1q1⊗ f (lnBq) = 1pf (p lnA)⊗ 1 + 1q1⊗ f (q lnB) ,then by (2.16) we derive (2.15).Corollary 2.9. Assume that f is convex on R with f ◦ ln is operator convexon (0,∞) . If A, B > 0 and p, q > 1 with 1p +1q = 1, thenf (ln (A ◦B)) ≤[1pf (p lnA) +1qf (q lnB)]◦ 1.In particular,f (ln (A ◦B)) ≤ 12[f (2 lnA) + f (2 lnB)] ◦ 1.The proof is similar to the one from Corollary 2.7 for the operator convexfunction f ◦ ln .We can also extend some of the inequalities above for functions defined bypower series. The following Schwarz type result holds:Theorem 2.10. Let f (z) =∑∞n=0 anzn be a power series with nonnegativecoefficients and convergent on the open disk D (0, R) ⊂ C, R > 0. Assume that0 ≤ A, B < R1/2. Then(2.17) f2 (A⊗B) ≤ f(A2)⊗ f(B2).If R = ∞, then the inequality (2.17) holds for A, B ≥ 0.INEQUALITIES OF SCHWARZ TYPE 11Proof. For 0 ≤ s, t <√R we get that 0 ≤ st, s2, t2 < R. By Cauchy-Schwarzinequality, we have(2.18)(n∑k=0ak (ts)k)2=(n∑k=0aktksk)2≤(n∑k=0akt2k)(n∑k=0aks2k).Since the series∞∑k=0ak (ts)k ,∞∑k=0akt2k and∞∑k=0aks2kare convergent for 0 ≤ s, t <√R, then by taking the limit over n → ∞ in(2.18) we deduce thatf2 (ts) ≤ f(t2)f(s2)for all s, t ∈ [0,√R).Now, if we take the double integral∫[0,√R)∫[0,√R) over dE (t)⊗ dF (s) , thenwe get ∫[0,√R)∫[0,√R)f2 (ts) dE (t)⊗ dF (s)≤∫[0,√R)∫[0,√R)f(t2)f(s2)dE (t)⊗ dF (s)=(∫[0,√R)f(t2)dE (t))⊗(∫[0,√R)f(s2)dF (s)),which gives (2.17).§3. Some examplesAssume that A, B > 0. If we take f (t) = tp and g (t) = tq, p, q ̸= 0, in (2.1),then we getA2p ⊗B2q +A2q ⊗B2p ≥(A2p(1−λ)+2qλ)⊗(B2pλ+2q(1−λ))+(A2pλ+2q(1−λ))⊗(B2p(1−λ)+2qλ)≥ 2Ap+q ⊗Bp+qand, in particularA2p ⊗A2q +A2q ⊗A2p ≥(A2p(1−λ)+2qλ)⊗(A2pλ+2q(1−λ))+(A2pλ+2q(1−λ))⊗(A2p(1−λ)+2qλ)12 S. S. DRAGOMIR≥ 2Ap+q ⊗Ap+q,where λ ∈ [0, 1] .For q = 1/2 and p = −1/2 we getA−1 ⊗B +A⊗B−1 ≥ A2λ−1 ⊗B1−2λ +A1−2λ ⊗B2λ−1 ≥ 2and, in particularA−1 ⊗A+A⊗A−1 ≥ A2λ−1 ⊗A1−2λ +A1−2λ ⊗A2λ−1 ≥ 2,where λ ∈ [0, 1] .We also have the inequalities for the Hadamard productA2p ◦B2q +A2q ◦B2p ≥(A2p(1−λ)+2qλ)◦(B2pλ+2q(1−λ))+(A2pλ+2q(1−λ))◦(B2p(1−λ)+2qλ)≥ 2Ap+q ◦Bp+qand, in particularA2p ◦A2q ≥(A2p(1−λ)+2qλ)◦(A2pλ+2q(1−λ))≥ Ap+q ◦Ap+q,where λ ∈ [0, 1] .For p = a/2, q = b/2, we getAa ◦Ab ≥ A(1−λ)a+λb ◦Aλa+(1−λ)b ≥ Aa+b2 ◦Aa+b2 ,which, as pointed out by the referee, can be also proved directly by usingoperator geometric averaging.For q = 1/2 and p = −1/2 we getA−1 ◦B +A ◦B−1 ≥ A2λ−1 ◦B1−2λ +A1−2λ ◦B2λ−1 ≥ 2and, in particular the refinement of Fiedler inequality obtained before in (2.5).If Ai, Bi > 0 and pi ≥ 0 for i ∈ {1, ..., n} , then by (2.6) we get(n∑i=1piA2pi)⊗(n∑i=1piB2qi)+(n∑i=1piA2qi)⊗(n∑i=1piB2pi)≥(n∑i=1piA2p(1−λ)+2qλi)⊗(n∑i=1piB2pλ+2q(1−λ)i)+(n∑i=1piA2pλ+2q(1−λ)i)⊗(n∑i=1piB2p(1−λ)+2qλi)INEQUALITIES OF SCHWARZ TYPE 13≥ 2(n∑i=1piAp+qi)⊗(n∑i=1piBp+qi)and (n∑i=1piA2pi)◦(n∑i=1piB2qi)+(n∑i=1piA2qi)◦(n∑i=1piB2pi)≥(n∑i=1piA2p(1−λ)+2qλi)◦(n∑i=1piB2pλ+2q(1−λ)i)+(n∑i=1piA2pλ+2q(1−λ)i)◦(n∑i=1piB2p(1−λ)+2qλi)≥ 2(n∑i=1piAp+qi)◦(n∑i=1piBp+qi).If we take p = 1/2 and q = −1/2 and also assume that∑ni=1 pi = 1, then weget(n∑i=1piAi)◦(n∑i=1piB−1i)+(n∑i=1piA−1i)◦(n∑i=1piBi)≥(n∑i=1piA1−2λi)◦(n∑i=1piB2λ−1i)+(n∑i=1piA2λ−1i)◦(n∑i=1piB1−2λi)≥ 2.In particular, we have(3.1)(n∑i=1piAi)◦(n∑i=1piA−1i)≥(n∑i=1piA1−2λi)◦(n∑i=1piA2λ−1i)≥ 1,where λ ∈ [0, 1] .This inequality can be interpreted as an additive refined version of Fiedlerinequality.Consider the function f (t) = tr, r ≥ 2. This function is monotonic nonde-creasing and convex on [0,∞) . If A, B ≥ 0 and p, q > 1 with 1p +1q = 1, thenby (2.9)(A⊗B)r ≤ 1pArp ⊗ 1 + 1q1⊗Brq.In particular,(A⊗B)r ≤ 12(A2r ⊗ 1 + 1⊗B2r).14 S. S. DRAGOMIRMoreover, if r ∈ [1, 2] , then f is operator convex and by (2.12) we get(A ◦B)r ≤(1pArp +1qBrq)◦ 1.In particular,(A ◦B)r ≤ 12(A2r +B2r)◦ 1.Now, we consider the function f (t) = |t|u that is convex on R for u ≥ 2.By (2.15) we derive|ln (A⊗B)|u ≤ pu−1 |lnA|u ⊗ 1 + qu−11⊗ |lnB|u ,where A, B > 0 and p, q > 1 with 1p +1q = 1.In particular,|ln (A⊗B)|u ≤ 2u−1 (|lnA|u ⊗ 1 + 1⊗ |lnB|u)for A, B > 0.Let h (z) =∑∞n=0 anzn be a power series with complex coefficients andconvergent on the open disk D (0, R) ⊂ C, R > 0. We have the followingexamplesh (z) =∞∑n=11nzn = ln11− z, z ∈ D (0, 1) ;h (z) =∞∑n=01(2n)!z2n = cosh z, z ∈ C;h (z) =∞∑n=01(2n+ 1)!z2n+1 = sinh z, z ∈ C;h (z) =∞∑n=0zn =11− z, z ∈ D (0, 1) .Other important examples of functions as power series representations withnonnegative coefficients are:h (z) =∞∑n=01n!zn = exp (z) , z ∈ C;h (z) =∞∑n=112n− 1z2n−1 =12ln(1 + z1− z), z ∈ D (0, 1) ;INEQUALITIES OF SCHWARZ TYPE 15h (z) =∞∑n=0Γ(n+ 12)√π (2n+ 1)n!z2n+1 = sin−1 (z) , z ∈ D (0, 1) ;andh (z) =∞∑n=112n− 1z2n−1 = tanh−1 (z) , z ∈ D (0, 1)h (z) = F1 (α, β, γ, z) :=∞∑n=0Γ (n+ α) Γ (n+ β) Γ (γ)n!Γ (α) Γ (β) Γ (n+ γ)zn, α, β, γ > 0,z ∈ D (0, 1) ;where Γ is Gamma function.Assume that 0 ≤ A, B < 1, then by (2.17)(1−A⊗B)−2 ≤(1−A2)−1 ⊗ (1−B2)−1 ,[ln (1−A⊗B)]2 ≤ ln(1−A2)⊗ ln(1−B2)and [sin−1 (A⊗B)]2 ≤ sin−1 (A2)⊗ sin−1 (B2) .If A, B ≥ 0, then by (2.17) we getexp (2A⊗B) ≤ exp(A2)⊗ exp(B2),[sinh (A⊗B)]2 ≤ sinh(A2)⊗ sinh(B2)and[cosh (A⊗B)]2 ≤ cosh(A2)⊗ cosh(B2).AcknowledgmentsThe author would like to thank the anonymous referee for valuable suggestionsthat have been implemented in the final version of the manuscript.References[1] H. Araki and F. Hansen, Jensen’s operator inequality for functions of severalvariables, Proc. Amer. Math. Soc. 128 (2000), No. 7, 2075-2084.[2] D. K. Callebaut, Generalization of the Cauchy-Schwarz inequality, J. Math. Anal.Appl. 12 (1965) 491-494.16 S. S. DRAGOMIR[3] J. I. Fujii, The Marcus-Khan theorem for Hilbert space operators, Math. Jpn.41 (1995), 531-535.[4] T. Furuta, J. Mićić Hot, J. Pečarić and Y. Seo, Mond-Pečarić method in operatorinequalities. Inequalities for bounded selfadjoint operators on a Hilbert space,Element, Zagreb, 2005.[5] A. Korányi. On some classes of analytic functions of several variables, Trans.Amer. Math. Soc. 101 (1961), 520-554.[6] S. Wada, On some refinement of the Cauchy-Schwarz inequality, Lin. Alg. &Appl. 420 (2007), 433-440.Silvestru Sever Dragomir1Applied Mathematics Research Group, ISILC,Victoria University, PO Box 14428Melbourne City, MC 8001, Australia.2School of Computer Science& Applied Mathematics, University of the Witwatersrand,Private Bag 3, Johannesburg 2050, South AfricaE-mail : sever.dragomir@ajmaa.org

Tensorial and Hadamard product inequalities of Schwarz type for selfadjoint operators in Hilbert spaces

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Tensorial and Hadamard product inequalities of Schwarz type for selfadjoint operators in Hilbert spaces

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