AN OPERATOR INEQUALITY

Derming Wang

California State University–Long Beach

1. Introduction. In this paper inequalities for positive operators acting on a Hilbert space are considered. For positive operators A and B, the following conjecture [1] was posed:

(1)  If 0 ≤ B ≤ A, then (AB^{2}A)^{1/2} ≤ A^{2}.

This conjecture was answered affirmatively by Furuta [3]. Indeed, Furuta proved a more general inequality which contains inequality (1) as a special case:

If 0 ≤ B ≤ A, then (A^{r}B^{p}A^{r})^{1/q} ≤ A^{(p+2r)/q}
for p, r ≥ 0, q ≥ 1 with p + 2r ≤ (1 + 2r)q.

Setting p = q = 2 and r = 1, Furuta’s inequality becomes (1). More interestingly, if one sets p = 2r and q = 2, Furuta’s inequality becomes a generalization of (1):

(2)  If 0 ≤ B ≤ A, then (A^{r}B^{2r}A^{r})^{1/2} ≤ A^{2r} for r ≥ 0.

If A and B are positive invertible operators with B ≤ A, then it is known that log B ≤ log A. In [2], a result of Ando was rephrased: For positive invertible operators A and B, log B ≤ log A if and only if (A^{r}B^{2r}A^{r})^{1/2} ≤ A^{2r} holds for all r ≥ 0. Thus, Ando’s result also establishes (2) as a corollary.

In this paper we establish (2) directly by elementary means. Our approach is inspired by the work of Pedersen and Takesaki [5].

2. Preliminary. We are interested in (bounded, linear) operators acting on a Hilbert space ℍ with inner product < ., . >. An operator A is said to be self–adjoint if A = A^{∗}, where A^{∗} is the adjoint of A. A self-adjoint operator A is said to be positive, in notation A ≥ 0 (or 0 ≤ A), if < Ax, x >≥ 0 for every vector x ∈ ℍ. For positive operators A and B we write A ≥ B (or B ≤ A) if A − B ≥ 0. The relation “≥” defines a partial order on the set of positive operators. The following properties of positive operators are well known:

(a) If A ≥ 0, then A^{r} ≥ 0 for every real number r ≥ 0.

(b) If A ≥ 0 and A is invertible, then A^{−1} ≥ 0.

(c) If A, B ≥ 0, then ABA ≥ 0. Thus, if 0 ≤ B ≤ A, then CBC ≤ CAC for every C ≥ 0.

(d) If 0 ≤ B ≤ A, then B^{r} ≤ A^{r} for 0 ≤ r ≤ 1. In particular, if 0 ≤ B ≤ A, then B^{1/2} ≤ A^{1/2}.

(e) In general, 0 ≤ B ≤ A does not necessarily imply B^{2} ≤ A^{2}.

(f) If 0 ≤ T ≤ I, the identity operator, then T^{2} ≤ I. Properties (c), (d) and (e) led the authors of [1] to pose the conjecture (1). Now we state a weakened version of a result of Pedersen and Takesaki [5] that is suitable for our purpose. For the sake of completeness, a proof is given.

\underline{Theorem A}. Suppose A and B are positive operators. Also, assume A is invertible. Then, if (A^{1/2}BA^{1/2})^{1/2} ≤ A, then there is a unique positive operator T satisfying 0 ≤ T ≤ I and T AT = B.

\underline{Proof}. Let T = A^{−1/2}(A^{1/2}BA^{1/2})A^{−1/2}. Since (A^{1/2}BA^{1/2})^{1/2} ≤ A, multiplying both sides of this inequality on the left and on the right by A^{−1/2} yields 0 ≤ T ≤ I. A simple calculation shows that T AT = B. This establishes the existence of T. To prove the uniqueness, assume S is a positive operator with the property SAS = B. Since SAS = T AT, multiplying both sides of this equality on the left and on the right by A^{1/2} we obtain (A^{1/2}SA^{1/2})^{2} = (A^{1/2}T A^{1/2})^{2}. Now, taking square roots of both sides and then multiplying both sides on the left and on the right by A^{−1/2} produces the desired S = T.

3. The Main Result. We are ready to present our proof of (2).

\underline{Lemma}. If 0 ≤ B ≤ A, then (A^{2^{n−1}}B^{2^{n}}A^{2^{n−1}})^{1/2} ≤ A^{2^{n}} for n = 0, 1, 2, . . ..

\underline{Proof}. First note that if the inequality can be proven with the extra assumption that the operator A is invertible, then the result follows. For if A is not invertible, then for any ϵ > 0, the operator A_{ϵ} = A + ϵI is a positive invertible operator. If the inequality in the lemma is derived with A_{ϵ} in place of A, taking limits as ϵ tends to 0 will produce the desired inequality. Therefore, we may, without loss of generality, assume that A is invertible.

Since B ≤ A, (A^{1/2}BA^{1/2})^{1/2} ≤ A. This proves the lemma for n = 0. Now, Theorem A implies there is a positive operator T_{1} ≤ I such that T_{1}AT_{1} = B. Thus,

AB^{2}A = A(T_{1}AT_{1})^{2}A = (AT_{1}A)T_{1}^{2}(AT_{1}A) ≤ (AT_{1}A)^{2}.

Taking square roots, we have (AB^{2}A)^{1/2} ≤ AT_{1}A ≤ A^{2}. This establishes the lemma for n = 1. Again, Theorem A implies there is a positive operator T_{2} ≤ I such that T_{2}A^{2}T_{2} = B^{2}. Similar arguments give (A^{2}B^{4}A^{2})^{1/2} ≤ A^{4}. This establishes the lemma for n = 2. It is now apparent that the lemma follows by induction.

Notice that for the case n = 1, the inequality of the Lemma is (1). \underline{Theorem}. If 0 ≤ B ≤ A, then (A^{r}B^{2r}A^{r})^{1/2} ≤ A^{2r} for r ≥ 0. \underline{Proof}. For each r ≥ 0, there is a smallest nonnegative integer k such that r/2^{k} ≤ 1. Let

A_{1} = A^{r/2^{k}} and B_{1} = B^{r/2^{k}}. We have 0 ≤ B_{1} ≤ A_{1}, A_{1}^{2^{k}} = A^{r} and B_{1}^{2^{k}} = B^{r}. The result follows by applying the Lemma with n = k + 1 to the operators A_{1} and B_{1}. This completes the proof.

We now use the Theorem to amplify Theorem A when 0 ≤ B ≤ A. \underline{Corollary}. Suppose 0 ≤ B ≤ A with A invertible. Then, for each r ≥ 0 there is a

unique operator T_{r} satisfying 0 ≤ T_{r} ≤ I and T_{r}A^{r}T_{r} = B^{r}.

\underline{Proof}. Let T_{r} = A^{−r}(A^{r}B^{2r}A^{r})^{1/2}A^{−r}. Clearly we have T_{r}A^{r}T_{r} = B^{r}. Since B ≤ A, the Theorem implies 0 ≤ T_{r} ≤ I. Arguments similar to those of Theorem A show that the existence of T_{r} is unique. The proof is complete.

In conclusion, we note that a result of Hansen [4] may be applied to show that the family {T_{r}}_{r≥0} in the Corollary is a decreasing family.

\underline{References}

1. N. N. Chan and M. K. Kwong, “Hermitian Matrix Inequalities and a Conjecture,” American Mathematical Monthly, 92 (1985), 533–541.

2. M. Fujii and E. Kamei, “Furuta’s Inequality for Chaotic Order,” Mathematica Japonica, 36 (1991), 603–606.

3. T. Furuta, “A ≥ B ≥ 0 Assures (B^{r}A^{p}B^{r})^{1/q} ≥ B^{(p+2r)/q} for r ≥ 0, p ≥ 0, q ≥ 1 with (1 + 2r)q ≥ p + 2r,” Proceedings of the American Mathematical Society, 101 (1987), 85–88.

4. F. Hansen, “An Operator Inequality,” Mathematische Annalen, 246 (1980), 249–250.

5. G. K. Pedersen and M. Takesaki, “The Operator Equation T HT = K,” Proceedings of the American Mathematical Society, 36 (1972), 311–312. æ