converses of convex inequalities in hilbert space

Rend. Circ. Mat. PalermoDOI 10.1007/s12215-013-0137-3

Converses of convex inequalities in Hilbert space

Rozarija Jakšic · Josip Pecaric

Received: 28 June 2012 / Accepted: 1 October 2013© Springer-Verlag Italia 2013

Abstract In this paper we give converses of the operator version for the Jensen inequalityand the scalar Lah–Ribaric inequality. We also apply those results to quasi-arithmetic meansand power means for selfadjoint operators.

Keywords Selfadjoint operators · Jensen’s inequality · Lah–Ribaric’s inequality ·Convex functions · Quasi-arithmetic means · Arithmetic means

Mathematics Subject Classification (1991) 47A63 · 47A64

1 Introduction

Let H be a Hilbert space and let B(H) be the C∗-algebra of all bounded (i.e., continuous)linear operators on H . A bounded linear operator A on a Hilbert space H is said to beselfadjoint if A = A∗.

An operator A ∈ B(H) is selfadjoint if and only if 〈Ax, x〉 ∈ R for every x ∈ H . Wedenote by Bh(H) a semi-space of all selfadjoint operators in B(H).

If A is a selfadjoint operator and f is a real valued continuous function on Sp(A), thenf (t) ≥ 0 for every t ∈ Sp(A) implies that f (A) ≥ 0, i.e., f (A) is a positive operator on H .

Equivalently, if both f and g are real valued continuous function on Sp(A), then thefollowing property holds:

f (t) ≥ g(t) for any t ∈ Sp(A) implies that f (A) ≥ g(A) (1.1)

in the operator order of B(H).

R. Jakšic (B) · J. PecaricFaculty of Textile Technology, University of Zagreb,Zagreb, Croatiae-mail: [email protected]

J. Pecarice-mail: [email protected]

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R. Jakšic and J. Pecaric

The following result that provides an operator version for Jensen’s inequality is due toMond and Pecaric [5] (see also [2] and [3, p.5]):

Theorem 1.1 [5] (Mond-Pecaric) Let A ∈ Bh(H) be a selfadjoint operator with Sp(A) ⊆[m, M] for some scalars m < M. If f is a convex function on [m, M], then

f (〈Ax, x〉) ≤ 〈 f (A)x, x〉 (1.2)

holds for each unit vector x in H.

We also need the following converse for the Mond-Pecaric inequality that generalizesthe scalar Lah–Ribaric inequality for convex functions found in [4, p.9, Tm.7] (see also [3,p.57]):

Theorem 1.2 [4, p.9, Tm.7] Let A be a selfadjoint operator on the Hilbert space H andassume that Sp(A) ⊆ [m, M] for some scalars m and M with m < M. If f is a convexfunction on [m, M], then

〈 f (A)x, x〉 ≤ M − 〈Ax, x〉M − m

f (m) + 〈Ax, x〉 − m

M − mf (M) (1.3)

holds for each unit vector x in H.

In this paper we shall give converses of the Jensen inequality and the scalar Lah–Ribaricinequality for selfadjoint operators. Also, we shall give applications of these results to quasi-arithmetic means and power means of selfadjoint operators.

2 Results

The results in this section are converses of the Jensen and the scalar Lah–Ribaric inequalityfor selfadjoint operators.

Theorem 2.1 Let A ∈ Bh(H) be a selfadjoint operator with Sp(A) ⊆ [m, M] for somescalars m < M. If f is a continuous convex function on an interval of real numbers I such

that [m, M] ⊂◦I , where

◦I is the interior of I , then

0 ≤ 〈 f (A)x, x〉 − f (〈Ax, x〉)≤ (M − 〈Ax, x〉)(〈Ax, x〉 − m)

f ′−(M) − f ′+(m)

M − m

≤ 1

4(M − m)( f ′−(M) − f ′+(m)) (2.1)

holds for each unit vector x in H. If f is concave, the inequalities in (2.1) are reversed.

Proof First we assume that f is convex.The first inequality follows directly from (1.2). If we subtract f (〈Ax, x〉) from both sides

of the inequality (1.3) from Theorem 1.2, we get

〈 f (A)x, x〉 − f (〈Ax, x〉) ≤ M − 〈Ax, x〉M − m

f (m)

+ 〈Ax, x〉 − m

M − mf (M) − f (〈Ax, x〉) =: B.

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Then, by the convexity of f we have the gradient inequality:

f (t) − f (M) ≥ f ′−(M)(t − M)

for any t ∈ [m, M]. If we multiply this inequality by t − m ≥ 0, we deduce

(t − m) f (t) − (t − m) f (M) ≥ f ′−(M)(t − M)(t − m), t ∈ [m, M] (2.2)

Similarly, we get

(M − t) f (t) − (M − t) f (m) ≥ f ′+(m)(t − m)(M − t), t ∈ [m, M] (2.3)

Adding (2.2) to (2.3) and dividing by m − M , we deduce that for any t ∈ [m, M](t − m) f (M) + (M − t) f (m)

M − m− f (t) ≤ (M − t)(t − m)

M − m( f ′−(M) − f ′+(m)). (2.4)

Because Sp(A) ⊆ [m, M], we have 〈Ax, x〉 ∈ [m, M] for each unit vector x in H , so wecan substitute t with 〈Ax, x〉 in (2.4) and obtain

B ≤ (M − 〈Ax, x〉)(〈Ax, x〉 − m)

M − m( f ′−(M) − f ′+(m))

which is the second inequality in (2.1).To prove the third inequality in (2.1), we notice that for every t ∈ [m, M],

(M − t)(t − m)

M − m≤ 1

4(M − m) is valid, and the proof is completed.

If f is concave, then − f is convex, so we can apply (2.1) to function − f and obtainreversed inequalities for f. �

Theorem 2.2 Let the assumptions of Theorem 2.1 hold. If f is a continuous convex function

on an interval of real numbers I such that [m, M] ⊂◦I , where

◦I is the interior of I , then

0 ≤ M − 〈Ax, x〉M − m

f (m) + 〈Ax, x〉 − m

M − mf (M) − 〈 f (A)x, x〉

≤ f ′−(M) − f ′+(m)

M − m〈(M1H − A)(A − m1H )x, x〉

≤ f ′−(M) − f ′+(m)

M − m(M − 〈Ax, x〉)(〈Ax, x〉 − m)

≤ 1

4(M − m)( f ′−(M) − f ′+(m)) (2.5)

holds for each unit vector x in H. If f is concave, the inequalities in (2.5) are reversed.

Proof Let us assume that f is convex.The first inequality in (2.5) follows directly from (1.3). Due to property (1.1), we can

replace t with operator A in (2.4) and obtain

1

M − m((M1H − A) f (m) + (A − m1H ) f (M)) − f (A)

≤ f ′−(M) − f ′+(m)

M − m(M1H − A)(A − m1H )

(2.6)

in the operator order of B(H).

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Because scalar product is linear in the first argument and ‖x‖ = 1, by applying it to (2.6)we have

M − 〈Ax, x〉M − m

f (m) + 〈Ax, x〉 − m

M − mf (M) − 〈 f (A)x, x〉

≤ f ′−(M) − f ′+(m)

M − m〈(M1H − A)(A − m1H )x, x〉,

and the second inequality in (2.5) is proved. Since h(t) = (M − t)(t − m) is concave, fromthe Jensen inequality we have

〈(M1H − A)(A − m1H )x, x〉 ≤ (M − 〈Ax, x〉)(〈Ax, x〉 − m),

which gives us the third inequality in (2.5). To prove the last inequality in (2.5), we noticethat for every t ∈ [m, M], (M−t)(t−m)

M−m ≤ 14 (M − m) is valid, and the proof is completed.

If f is concave, then we apply the inequalities (2.5) to − f which is convex, and obtainthe reversed inequalities. � Remark 2.1 Let the assumptions from the previous two theorems hold and let l be a linearfunction through (m, f (m)) and (M, f (M)). Since f is convex on [m, M] we have that

f (〈Ax, x〉) ≤ 〈 f (A)x, x〉 ≤ l(〈Ax, x〉)holds for every unit vector x in H .

From Theorem 2.1 and Theorem 2.2 we can see that both of the differences

〈 f (A)x, x〉 − f (〈Ax, x〉) and l(〈Ax, x〉) − 〈 f (A)x, x〉have the same estimation, so one can see that, in a weak sense, 〈 f (A)x, x〉 is almost the midpoint between f (〈Ax, x〉) and l(〈Ax, x〉).Remark 2.2 For more results in inequalities for functions of selfadjoint operators related toMond-Pecaric method and applications, see [1] and the references therein.

3 Applications

3.1 Quasi-arithmetic means

Let A be a positive invertible operator on a Hilbert space such that Sp(A) ⊆ [m, M] for somescalars m < M and x a unit vector in H . Let f be a strictly monotone continuous functionon [m, M]. Quasi-arithmetic mean of the operator A with respect to f is defined by

M f (A, x) = f −1〈 f (A)x, x〉. (3.1)

Theorem 3.1 Let A ∈ Bh(H) be a selfadjoint operator with Sp(A) ⊆ [m, M] for somescalars m < M, and let f, g be strictly monotone continuous functions on an interval of real

numbers I such that [m, M] ⊂◦I , where

◦I is the interior of I . If the function f ◦g−1 is convex,

then

0 ≤ f (M f (A, x)) − f (Mg(A, x))

≤ (Mg − 〈g(A)x, x〉)(〈g(A)x, x〉 − mg)[ f ◦ g−1]′−(Mg) − [ f ◦ g−1]′+(mg)

Mg − mg

≤ 1

4(Mg − mg)([ f ◦ g−1]′−(Mg) − [ f ◦ g−1]′+(mg)), (3.2)

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holds for each unit vector x in H, where [mg, Mg] = g([m, M]). If f ◦g−1 is concave, thenthe inequalities in (3.2) are reversed.

Proof Function f ◦ g−1 is obviously continuous. Let us assume that f ◦ g−1 is convex.If g is strictly increasing, then mg = g(m), Mg = g(M); if g is strictly decreasing, then

mg = g(M), Mg = g(m). The conditions of Theorem 2.2 are satisfied. We obtain (3.3) bysubstituting f ↔ f ◦ g−1 and A ↔ g(A) in (2.1).

Now let us assume that f ◦ g−1 is concave. Then the function − f ◦ g−1 is convex, so wecan obtain reversed inequalities by replacing f ◦ g−1 with − f ◦ g−1 in (3.2). � Theorem 3.2 Under the same assumptions, if the function f ◦ g−1 is convex, then the fol-lowing inequalities are valid:

0 ≤ Mg − 〈g(A)x, x〉Mg − mg

f (m) + 〈g(A)x, x〉 − mg

Mg − mgf (M) − f (M f (A, x))

≤ [ f ◦ g−1]′−(Mg) − [ f ◦ g−1]′+(mg))

Mg − mg〈(Mg1H − g(A))(g(A) − mg1H )x, x〉

≤ [ f ◦ g−1]′−(Mg) − [ f ◦ g−1]′+(mg))

Mg − mg(Mg − 〈g(A)x, x〉)(〈g(A)x, x〉 − mg)

≤ 1

4(Mg − mg)([ f ◦ g−1]′−(Mg) − [ f ◦ g−1]′+(mg)), (3.3)

for each unit vector x in H, where [mg, Mg] = g([m, M]). If f ◦ g−1 is concave, theinequalities in (3.3) are reversed.

Proof Similarly to the proof of the previous theorem, let us assume that f ◦ g−1 is convex(obviously, f ◦ g−1 is continuous).

If g is strictly increasing, then mg = g(m), Mg = g(M); if g is strictly decreasing, thenmg = g(M), Mg = g(m). The conditions of Theorem 2.2 are satisfied. We obtain (3.3) bysubstituting f ↔ f ◦ g−1 and A ↔ g(A) in (2.5).

Now let us assume that f ◦ g−1 is concave. Then the function − f ◦ g−1 is convex, so wecan obtain reversed inequalities by replacing f ◦ g−1 with − f ◦ g−1 in (3.3). � Remark 3.1 Let f, g and A be as in Theorem 3.2. Then directly from the second inequalityin Theorem 3.2 we have

(g(M)−g(m)) f (M f (A, x))−( f (M)− f (m))g(Mg(A, x)) ≤ g(M) f (m) − g(m) f (M)

(3.4)

if f ◦ g−1 is convex. If f ◦ g−1 is concave, the inequality in (3.4) is reversed.

3.2 Power means

Let A be a positive invertible operator on a Hilbert space and x a unit vector in H . For r ∈ R,the power mean Mr (A, x) is defined by

Mr (A, x) = (〈Ar x, x〉)1/r .

In [3, p.69] has been shown that if r → 0, then (〈Ar x, x〉)1/r converges monotone toexp〈log Ax, x〉, so we can extend the definition of the power mean to the case r = 0.

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Theorem 3.3 Let A ∈ Bh(H) be a selfadjoint operator with Sp(A) ⊆ [m, M] for somescalars m < M. Let x be a unit vector in H and r, s ∈ R such that r < s.

If r < 0 < s or 0 < r < s, then

0 ≤ (Ms(A, x))s − (Mr (A, x))s

≤ s

r(Mr − 〈Ar x, x〉)(〈Ar x, x〉 − mr )

Ms−r − ms−r

Mr − mr

≤ s

4r(Mr − mr )(Ms−r − ms−r ). (3.5)

If r < s < 0, then

0 ≥ (Ms(A, x))s − (Mr (A, x))s

≥ s

r(Mr − 〈Ar x, x〉)(〈Ar x, x〉 − mr )

Ms−r − ms−r

Mr − mr

≥ s

4r(Mr − mr )(Ms−r − ms−r ). (3.6)

If r = 0 and s > 0, then

0 ≤ (Ms(A, x))s − (M0(A, x))s

≤ (log M − 〈log Ax, x〉)(〈log Ax, x〉 − log m)s(Ms − ms)

log M − log m

≤ s

4(Ms − ms) log

M

m. (3.7)

If r < 0 or s = 0, then

0 ≤ log(M0(A, x)) − log(Mr (A, x))

≤ −1

r(Mr − 〈Ar x, x〉)(〈Ar x, x〉 − mr )

1

Mr mr

≤ 1

4r(Mr − mr )

(1

Mr− 1

mr

). (3.8)

Proof The function f (t) = t s/r is continuous, and convex for 0 < r < s and r < 0 < s. Ifr > 0, then Sp(Ar ) ⊆ [mr , Mr ] and the conditions of Theorem 2.1 are satisfied, so we canobtain (3.5) by replacing m ↔ mr , M ↔ Mr , f (t) = t s/r and A ↔ Ar in (2.1). If r < 0,then Sp(Ar ) ⊆ [Mr , mr ] and the conditions of Theorem 2.1 are satisfied, so we can obtain(3.5) by replacing m ↔ Mr , M ↔ mr , f (t) = t s/r and A ↔ Ar in (2.1).

In case r < s < 0, the function f (t) = t s/r is concave, so we obtain (3.6) from Theorem2.1 by making substitutions M ↔ mr , m ↔ Mr , f (t) ↔ −t s/r (which is convex) andA ↔ Ar in (2.1).

In case r < 0 and s = 0, the function f (t) = 1

rlog t is continuous and convex, and

Sp(Ar ) ⊆ [Mr , mr ], so the conditions of Theorem 2.1 are satisfied and we can obtain (3.8)

by making substitutions M ↔ mr , m ↔ Mr , f (t) ↔ 1

rlog t and A ↔ Ar in (2.1).

In case r = 0, s > 0, the function f (t) = est is continuous and convex. The inequalities(3.7) are now obtained by replacing m ↔ log m, M ↔ log M , f (t) ↔ est and A ↔ log Ain (2.1). �

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Theorem 3.4 Under the same hypothesis as in the previous theorem,if 0 < r < s or r < 0 < s, then:

0 ≤ Mr − 〈Ar x, x〉Mr − mr

ms + 〈Ar x, x〉 − mr

Mr − mrMs − (Ms(A, x))s

≤ s

r

Ms−r − ms−r

Mr − mr〈(Mr 1H − Ar )(Ar − mr 1H )x, x〉

≤ s

r

Ms−r − ms−r

Mr − mr(Mr − 〈Ar x, x〉)(〈Ar x, x〉 − mr )

≤ s

4r(Mr − mr )(Ms−r − ms−r ). (3.9)

If r < s < 0 then:

0 ≥ Mr − 〈Ar x, x〉Mr − mr

ms + 〈Ar x, x〉 − mr

Mr − mrMs − (Ms(A, x))s

≥ s

r

Ms−r − ms−r

Mr − mr〈(Mr 1H − Ar )(Ar − mr 1H )x, x〉

≥ s

r

Ms−r − ms−r

Mr − mr(Mr − 〈Ar x, x〉)(〈Ar x, x〉 − mr )

≥ s

4r(Mr − mr )(Ms−r − ms−r ). (3.10)

If s = 0 and r < 0, then:

0 ≤ Mr − 〈Ar x, x〉Mr − mr

log m + 〈Ar x, x〉 − mr

Mr − mrlog M − log(M0(A, x))

≤ −1

r

〈(Mr 1H − Ar )(Ar − mr 1H )x, x〉Mr mr

≤ −1

r

(Mr − 〈Ar x, x〉)(〈Ar x, x〉 − mr )

Mr mr

≤ − 1

4r(Mr − mr )

(1

Mr− 1

mr

). (3.11)

If r = 0 and s > 0, then:

0 ≤ log M − 〈log Ax, x〉log M − log m

ms + 〈log Ax, x〉 − log m

log M − log mMs − (Ms(A, x))s

≤ ses M − esm

log M − log m〈(log M1H − log A)(log A − log m1H )x, x〉

≤ ses M − esm

log M − log m(log M − 〈log Ax, x〉)(〈log Ax, x〉 − log m)

≤ s

4(es M − esm) log

M

m. (3.12)

Proof All the inequalities can be obtained directly from (2.5) by making the same substitu-tions as in the proof of the previous theorem. � Remark 3.2 The following inequalities are obtained directly from Theorem 3.4:

(Mr − mr )(Ms(A, x))s − (Ms − ms)(Mr (A, x))r ≤ Mr ms − mr Ms (3.13)

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if 0 < r < s or r < 0 < s. In case r < s < 0, the inequality in (3.13) is reversed. Similarly,for r = 0 we have:

(Ms(A, x))s logM

m− (Ms − ms) log(M0(A, x)) ≤ ms log M − Ms log m (3.14)

Corollary 3.1 Let A ∈ Bh(H) be a selfadjoint operator with Sp(A) ⊆ [m, M] for somescalars m < M. Let x be a unit vector in H and r, s ∈ R such that r < s.

If r < 0 < s or r < s < 0 then:

0 ≤ (Mr (A, x))r − (Ms(A, x))r

≤ r

s(Ms − 〈As x, x〉)(〈As x, x〉 − ms)

Mr−s − mr−s

Ms − ms

≤ r

4s(Ms − ms)(Mr−s − mr−s). (3.15)

If 0 < r < s then:

0 ≥ (Mr (A, x))r − (Ms(A, x))r

≥ r

s(Ms − 〈As x, x〉)(〈As x, x〉 − ms)

Mr−s − mr−s

Ms − ms

≥ r

4s(Ms − ms)(Mr−s − mr−s). (3.16)


0 ≤ (Mr (A, x))r − (M0(A, x))r

≤ (log M − 〈log Ax, x〉)(〈log Ax, x〉 − log m)r(Mr − mr )

log M − log m

≤ r

4(Mr − mr ) log

M

m. (3.17)


0 ≥ log(M0(A, x)) − log(Ms(A, x))

≥ −1

s(Ms − 〈As x, x〉)(〈As x, x〉 − ms)

1

Msms

≥ 1

4s(Ms − ms)

(1

ms− 1

Ms

). (3.18)

Proof It is easy to see that Mr (A−1, x) = (M−r (A, x))−1 holds for every unit vector x ∈ H .Using that result, we can obtain Corolarry 3.1 from Theorem 3.3 by replacing A ↔ A−1,−r ↔ s and −s ↔ r . �

Corollary 3.2 Under the same hypothesis as in the previous theorem, if r < s < 0 orr < 0 < s, then:

0 ≤ Ms − 〈As x, x〉Ms − ms

mr + 〈As x, x〉 − ms

Ms − msMr − (Mr (A, x))r

≤ r

s

Mr−s − mr−s

Ms − ms〈(Ms1H − As)(As − ms1H )x, x〉

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≤ r

s

Mr−s − mr−s

Ms − ms(Ms − 〈As x, x〉)(〈As x, x〉 − ms)

≤ r

4s(Ms − ms)(Mr−s − mr−s). (3.19)

If 0 < r < s then:

0 ≥ Ms − 〈As x, x〉Ms − ms

mr + 〈As x, x〉 − ms

Ms − msMr − (Mr (A, x))r

≥ r

s

Mr−s − mr−s

Ms − ms〈(Ms1H − As)(As − ms1H )x, x〉

≥ r

s

Mr−s − mr−s

Ms − ms(Ms − 〈As x, x〉)(〈As x, x〉 − ms)

≥ r

4s(Ms − ms)(Mr−s − mr−s). (3.20)


0 ≤ log M − 〈log Ax, x〉log M − log m

mr + 〈log Ax, x〉 − log m

log M − log mMr − (Mr (A, x))r

≤ r(Mr − mr )

log M − log m〈(log M1H − log A)(log A − log m1H )x, x〉

≤ r(Mr − mr )

log M − log m(log M − 〈log Ax, x〉)(〈log Ax, x〉 − log m)

≤ r

4(Mr − mr ) log

M

m. (3.21)


0 ≥ Ms − 〈As x, x〉Ms − ms

log m + 〈As x, x〉 − ms

Ms − mslog M − log(M0(A, x))

≥ −1

s

1

Msms〈(Ms1H − As)(As − ms1H )x, x〉

≥ −1

s

1

Msms(Ms − 〈As x, x〉)(〈As x, x〉 − ms)

≥ 1

s(Ms − ms)

1

Msms. (3.22)

Proof Corollary 3.2 follows directly from Theorem 3.4 by making the same substitutions asin the proof of the previous theorem. �

References

1. Dragomir, S.S: Hermite-Hadamard’s type inequalities for convex functions of selfadjoint operators inHilbert spaces. LAA. 436, 1503–1515 (2012)

2. Dragomir, S.S.: Some reverses of the Jensen inequality for the functions of selfadjoint operators in Hilbertspaces. J. Inequal. Appl. 2010(2008), Article ID 496821

3. Furuta, T., Micic Hot, J., Pecaric, J.E., Seo, Y.: Mond-Pecaric method in operator inequalities. Inequalitiesfor bounded, selfadjoint operators on a Hilbert space element, Zagreb (2005)

4. Mitrinovic, D.S., Pecaric, J.E, Fink, A.M.: Classical and new inequalities in analysis. Kluwer Academic,Dordrecht; 1993

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converses of convex inequalities in hilbert space

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