0 Introduction

The classical Bohr inequality^[1] for scalars asserts that for complex numbers a, b and real numbers p, q>1 with ${1 \over p}$+${1 \over q}$=1, the inequality

|a-b|2≤p|a|2+q|b|2

(1)

holds. Over the years, various generalizations of Bohr inequality for scalars, vectors, matrices and operators have been obtained^{[2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]}. In 1989, Pecari and Dragomir generalized Bohr inequality to the context of normed vector spaces in reference^[7]. It is showed that if (X ;‖·‖)is a normed vector space and p, q>1 are real numbers such that $\frac{1}{p}$+$\frac{1}{q}$=1, then for all v, w∈X,

‖v+w‖2≤p‖v‖2+q‖w‖2.

In 2003, Hirzallah^[2] further generalized the inequality to the context of operator algebras.

Let B(H) be the unital C^*-algebra of all bounded linear operators on a complex separable Hilbert space H with H≠0. As usual, I denotes the identity operator on H. For all A∈B(H), denote |A|=(AA^*)^{$\frac{1}{2}$}, where A^* is the adjoint operator of A. Note that |A|=0 if and only if A=0.We write A≥0 if A∈B(H) such that 〈Ax, x〉≥0 for all x∈H, and A≥B if A and B are self-adjoint elements in B(H) such that A-B≥0.

An operator version of the Bohr inequality was obtained by Hirzallah^[2].

Theorem 1^[2] Let A, B∈B(H), p, q>1 such that $\frac{1}{p}$+$\frac{1}{q}$=1 and p≤q, then

|A-B|2+|(1-p)A-B|2≤p|A|2+q|B|2.

(2)

It is worthwhile noting that in reference ^[2], only the situation where q≥p>1 is considered. Equivalently, the conjugate exponents p, q are only restricted to q≤2 and 1<p≤2, while other combinations are left alone. In reference^[3], Cheunga and Pecaricb continued working in the setting as that in reference^[2], but without restriction to the conjugate exponents p, q. Meanwhile, they also investigated the situation of equality in detail and made connection with the parallelogram law for the Banach algebra B(H).

Recently, reference^[4] proved some related operator inequalities by means of 2×2 (block) operator matrices, and nally gived a generalization of the operator Bohr inequality for multiple operators. Some very interesting operator identities were also established.

Theorem 2^[4]  Let A, B∈B(H), p, q>1 such that $\frac{1}{p}$+$\frac{1}{q}$=1.Then

|A-B|2+$\sqrt{\frac{p}{q}}$A+$\sqrt{\frac{q}{p}}$B2=p|A|2+q|B|2.

(3)

By using the identity (3), the inequality (2) deduced the condition that p≤q in reference^[4].

In 2010, reference ^[10] generalized the classical Bohr inequality from Hilbert space operators to the context of C^*-algebras and some extensions and related inequalities were obtained. For each inequality, the necessary and suffcient condition for the equality was also obtained.

In this note, by using an isometric *-representation between C^*-algebra and B(H), where H is a Hilbert space, we give some necessary and sufficient conditions for four generalized Bohr inequalities in C^*-algebras.

1 Main results

Let S be a unital C^*-algebra with a unit I. For all A∈S, put |A|=(A^*A)^1/2.

Theorem 3  Let p, q∈R⁺, and $\frac{1}{p}$+$\frac{1}{q}$=1. Then the following statements are equivalent.

(a) For all A, B∈S,

|A-B|2+|(1-p)A-B|2≤p|A|2+q|B|2.

(4)

(b) p≤2.

Proof  Let (a) hold, and A=I, B=0 in (4), then we have (p-2)(p-1)=p²-3p+2≤0.Since p>1, we see that p≤2.

Let (b) hold. Then p≤q.To prove (a), i.e. for all A, B∈S, (4) holds. Let Φ:S→B(H) be an isometric*-representation of S. By Theorem 1, we have

|Φ(A)-Φ(B)|2+|(1-p)Φ(A)-Φ(B)|2≤p|Φ(A)|2+q|Φ(B)|2, A, B∈S.

Since Φ is an isometric isomorphism of S, we have |Φ(X)|=Φ(|X|) for all X∈S. Hence,

|Φ(A)-Φ(B)|2+|(1-p)Φ(A)-Φ(B)|2= |Φ(A-B)|2+|Φ((1-p)A-B)|2=Φ(|A-B|2+|(1-p)A-B|2)

and

p|Φ(A)|2+q|Φ(B)|2=Φ(p|A|2+q|B|2).

This shows that

Φ(|A-B|2+|(1-p)A-B|2)≤Φ(p|A|2+q|B|2).

From the fact that Φ(X)≤Φ(Y)⇔X⇔Y, we see

|A-B|2+|(1-p)A-B|2≤p|A|2+q|B|2.

for all A, B∈S. This theorem is proved.

Lemma 1 Let l, m∈R. Then lx²+my²≥2xy holds for all x, y∈R，if and only if l>0, m>0 and lm≥1.

Proof  Put f(x, y)=lx²+my²-2xy. Suppose that f(x, y)≥0 for all x, y∈R.Then l=f(1, 0)≥0, m=f(0, 1)≥0.Clearly, lm≠0. Thus, m>0 and l>0. Since

$2\sqrt{lm}-2=f\left( {{\left( \frac{m}{l} \right)}^{\frac{1}{4}}}, {{\left( \frac{m}{l} \right)}^{\frac{1}{4}}} \right)\ge 0, $

we see that lm≥1.

Conversely, we assume that l>0, m>0 and lm≥1. Then for all x, y∈R, we have

f(x, y)≥2$\sqrt{l{{x}^{2}}\cdot m{{y}^{2}}}$-2xy≥2(|xy|-xy)≥0.

This lemma is proved.

Theorem 4 Let α, β, u，v∈R be real numbers and p, q∈R⁺ be positive real numbers. Then the following statements are equivalent.

(a) For all A, B∈S,

|αA+βB|2+|uA+vB|2≤p|A|2+q|B|2.

(5)

(b) p≥α²+u², q≥β²+v² and

(p-(α2+u2))(q-(β2+v2))≥(αβ+uv)2.

(6)

Proof  Let (a) hold. Then for all real numbers x, y, taking A=xI, B=yI in (5) yields that

(αx+βy)2+(ux+vy)2≤px2+qy2,

that is

[p-(α2+u2)]x2+[q-(β2+v2)]y2≥2(αβ+uv)xy.

(7)

Taking x=1, y=0 and x=0, y=1 in (7) respectively, we get p≥α²+u² and q≥β²+v². Thus, in the case where αβ+uv=0, the inequality (6) is clearly valid.

Hence, we may assume that αβ+uv>0. Therefore, (7) implies that

[p-(α2+u2)](αβ+uv)-1x2+[q-(β2+v2)](αβ+uv)-1y2≥2xy,

for all x, y∈R. Lemma 1 shows that

[p-(α2+u2)](αβ+uv)-1·[q-(β2+v2)](αβ+uv)-1≥1,

which is as same as (6).

Let (b) hold and Φ:S→B(H) be an isometric*-representation of S. By Theorem 6  of reference^[3], we have

|αΦ<(A)/i>+βΦ(B)|2+|uΦ(A)+vΦ(B)|2=p|Φ(A)|2+q|Φ(B)|2.

Since Φ is an isometric isomorphism of S, on the C^*-algebra Φ(S), we have |Φ(X)|=Φ(|X|) for all X∈S. Hence,

|αΦ(A)+βΦ(B)|2+|uΦ(A)+vΦ(B)|2= |Φ(αA+βB)|2+|Φ(uA+vB)|2= Φ(|αA+βB|2+|uA+vB|2)

and

p|Φ(A)|2+q|Φ(B)|2=Φ(p|A|2+q|B|2).

This shows that

Φ(|αA+βB|2+|uA+vB|2)≤Φ(p|A|2+q|B|2).

From the fact that Φ(X)≤Φ(Y)⇔X⇔Y, we see

|αA+βB|2+|uA+vB|2≤p|A|2+q|B|2,

for all A, B∈S. This theorem is proved.

Theorem 5  Let a, b∈R⁺ be positive real numbers and c∈C. Then the following statements are equivalent.

(b) For all A, B∈S,

a|A|2+b|B|2+cA*B+B*A≥0.

(8)

(b) ab≥|c|².

Proof  Let (a) hold and c≠0.Write c=|c|exp(iθ) and for all x, y∈R, using (8) for A=exp(iθ) xI and B=-yI yieds that

ax2+by2≥2|c|xy.

Hence, Lemma 1 yields that ab≥|c|².

Let (b) hold and Φ:S→B(H) be an isometric *-representation of S. Write c=|c|exp(iθ), then by using Lemma 1 of reference ^[3] for exp(-iθ)Φ(A) and Φ(B), we have

a|exp (-iθ)Φ(A)|2+b|Φ(B)|2+|c|(exp(iθ)Φ(A)*Φ(B)+exp(-iθ)Φ(B)*Φ(A))≥0.

Thus,

a|exp(-iθ)Φ(A)|2+b|Φ(B)|2+|c|(exp(iθ)Φ(A)*Φ(B)+exp(-iθ)Φ(B)*Φ(A))= Φ(a|exp(-iθ)A|2+b|B|2+|c|(exp(iθ)A*B+exp(-iθ)B*A))≥0.

From the fact that Φ(X)≥0⇔X≥0, we see that (8) holds. This theorem is proved.

Theorem 6  Let α, β∈R and x, y∈R⁺ be positive numbers. Then the following statements are equivalent.

(a) For all A, B∈S,

|αA+βB|2≤x|A|2+y|B|2.

(9)

(b) x≥α², y≥β² and (x-α²)(y-β²)≥α²β².

Proof  Let (a) hold. Using (8) for A=I, B=0 and A=0, B=I respectively, we have x≥α², y≥β². For all real numbers s, t, letting A=sI and B=tI in (9) yields that

(x-α2)s2+(y-β2)t2≥2αβst.

Clearly, we may assume that αβ>0. Thus, Lemma 1 implies that (x-α²)(y-β²)≥α²β².

Conversely, let Φ:S→B(H) be an isometric *-representation of S. Then from reference ^[4] we can see that for all A, B∈S,

|αΦ(A)+βΦ(B)|2≤x|Φ(A)|2+y|Φ(B)|2.

This shows that

Φ(|αA+βB|2)≤Φ(x|A|2+y|B|2).

From the fact that Φ(X)≤Φ(Y)⇔X⇔Y, we see

|αA+βB|2≤x|A|2+y|B|2, ∀A, B∈S.

This theorem is proved.