Some Improvements of the Cauchy-Schwarz Inequality Using the Tapia Semi-Inner-Product

Article Some Improvements of the Cauchy-Schwarz Inequality Using the Tapia Semi-Inner-Product

Nicu¸sorMinculete 1,* and Hamid Reza Moradi 2

1 Department of Mathematics and Computer Science, Transilvania University of Bra¸sov, 500091 Bra¸sov, Romania 2 Department of Mathematics, Payame Noor University (PNU), Tehran P.O. Box 19395-4697, Iran; [email protected] * Correspondence: [email protected]

Received: 24 October 2020; Accepted: 23 November 2020; Published: 26 November 2020

Abstract: The aim of this article is to establish several estimates of the triangle inequality in a normed space over the ﬁeld of real numbers. We obtain some improvements of the Cauchy–Schwarz inequality, which is improved by using the Tapia semi-inner-product. Finally, we obtain some new inequalities for the numerical radius and norm inequalities for Hilbert space operators.

Keywords: inner product space; triangle inequality; Cauchy–Schwarz inequality

JEL Classiﬁcation: Primary 46C05; secondary 26D15; 26D10

1. Introduction In an inner product space an important inequality is the inequality of Cauchy–Schwarz [1,2], namely:

|hx, yi| ≤ kxkkyk, (1) for all x, y ∈ X, where X is a complex inner product space. Aldaz [3] and Dragomir [4] studied the Cauchy–Schwarz inequality in the complex case. Another inequality that plays a central role in a normed space is the triangle inequality,

kx + yk ≤ kxk + kyk, (2) for all x, y ∈ X, where X is a complex normed space. Peˇcari´cand Raji´cin [5] proved other results about the triangle inequality. In [6], Maligranda showed the following inequality:

A · min{kxk, kyk} ≤ kxk + kyk − kx + yk ≤ A · max{kxk, kyk}, (3) x y where A = 2 − + , and x and y are nonzero vectors in a normed space X = (X, k · k). kxk kyk Using this inequality we obtain an estimate for the norm-angular distance or Clarkson distance x y (see, e.g., [7]) between nonzero vectors x and y, α[x, y] = − , thus [8]: kxk kyk

kx − yk − |kxk − kyk| kx − yk + |kxk − kyk| ≤ α[x, y] ≤ . (4) min{kxk, kyk} max{kxk, kyk}

Mathematics 2020, 8, 2112; doi:10.3390/math8122112 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 2112 2 of 13

In [6], Maligranda generalized the norm-angular distance to the p-angular distance in a normed space given by: p−1 p−1 αp[x, y] := kkxk x − kyk yk, where p ∈ R. In [9], Dragomir studied new bounds for this distance. Other results for bounds for the angular distance, named Dunkl–Williams type theorems (see [10]), are given by Moslehian et al. [11] and Krnićand Minculete [12,13]. Dehghan [14] presented a new refinement of the triangle inequality and defined the skew angular x y distance between nonzero vectors x and y by β[x, y] = − . In [15], we remarked on kyk kxk several estimates of the triangle inequality using integrals and in [16] a characterization is given for a generalized triangle inequality in normed spaces. The aim of this article is to establish several estimates of the triangle inequality in a normed space over the field of real numbers. This study is presented in Section2. We also obtain, in Section3, some improvements of the Cauchy–Schwarz inequality, which is improved by using the Tapia semi-inner-product. In Section4, we obtain some new inequalities for the numerical radius and norm inequalities for Hilbert space operators.

2. Inequalities Related to the Triangle Inequality We present some results regarding the several estimates of the triangle inequality in a normed space over the ﬁeld of real numbers R.

Theorem 1. If X = (X, k · k) is a normed vector space over the ﬁeld R, then the following inequality holds

a b min{ , }(ckxk + dkyk − kcx + dyk) ≤ akxk + bkyk − kax + byk c d a b ≤ max{ , }(ckxk + dkyk − kcx + dyk), (5) c d for all vectors x and y in X and a, b, c, d ∈ R+, c, d 6= 0.

a b Proof. Without loss of generality, we may assume that 0 ≤ ≤ . Then, by calculations and using c d the triangle inequality, we have

a b akxk + bkyk − min{ , }(ckxk + dkyk − kcx + dyk) = c d ad ad akxk + bkyk − akxk − kyk + kax + yk) = c c ad ad ad ad (b − )kyk + kax + yk = k(b − )yk + kax + yk ≥ kax + byk. c c c c Therefore, we obtain the ﬁrst inequality of the statement. a b In the same way, according to inequality ≤ , we have ( bc − a)kxk = k bc x − axk and c d d d we deduce

a b max{ , }(ckxk + dkyk − kcx + dyk) + kax + byk c d bc bc = kxk + bkyk − k x + byk + kax + byk d d bc bc = akxk + bkyk + − a kxk + kax + byk − k x + byk d d Mathematics 2020, 8, 2112 3 of 13

bc bc = akxk + bkyk + k x − axk + kax + byk − k x + byk ≥ akxk + bkyk. d d Because, using the triangle inequality, we have

bc bc k x − axk + kax + byk ≥ k x + byk, d d for all vectors x and y in X and a, b, c, d ∈ R+, c, d 6= 0. b a For 0 ≤ ≤ , we make the similar calculations. Therefore, the inequalities of the statement d c are true.

Remark 1. (a) If c = d in inequality (5), then we have

min{a, b}(kxk + kyk − kx + yk) ≤ akxk + bkyk − kax + byk ≤ max{a, b}(kxk + kyk − kx + yk), (6) for all vectors x and y in X and a, b ∈ R+. Another extension of this result can be found in [13].

→ x → y (b) For x kxk and y kyk in relation (6), we obtain the following inequalities:

x y a b min{a, b} 2 − + ≤ a + b − x + y kxk kyk kxk kyk x y ≤ max{a, b} 2 − + , (7) kxk kyk for all nonzero vectors x and y in X and a, b ∈ R+. For a = kxk and b = kyk in inequality (7), we ﬁnd inequality (3).

(c) For the nonzero vectors x and y in X, in Theorem 1, we make the following substitutions: a = d = 1 = = 1 kxk , b c kyk . We obtain

kxk kyk kxk kyk x y x y min{ , } + − + ≤ 2 − + ≤ kyk kxk kyk kxk kyk kxk kxk kyk kxk kyk kxk kyk x y max{ , } + − + . (8) kyk kxk kyk kxk kyk kxk

In relation (8), if we replace y by −y, then we have an inequality which uses the angular distance and the skew angular distance, thus:

2 kxk kyk kxk − kyk min{ , } − β[x, y] ≤ 2 − α[x, y] ≤ kyk kxk kxkkyk 2 kxk kyk kxk − kyk max{ , } − β[x, y] , (9) kyk kxk kxkkyk for all nonzero vectors x and y in X.

Theorem 2. If X = (X, k · k) is a normed vector space over the ﬁeld of real numbers R, with kxk = kyk = 1, then we have

ax + by − |a − b| ax + by + |a − b| ≤ x + y ≤ (10) min{a, b} max{a, b} Mathematics 2020, 8, 2112 4 of 13

∗ for every vector x and y in X and a, b ∈ R+.

Proof. From relation (6), using the ﬁrst inequality, we ﬁnd,

a + b − |a − b| − min{a, b} x + y ≤ a + b − ax + by , which implies

ax + by − |a − b| ≤ min{a, b} x + y . Using the second inequality of relation (6), we ﬁnd the inequality

a + b − ax + by ≤ a + b + |a − b| − max{a, b} x + y , which becomes

max{a, b} x + y ≤ ax + by + |a − b|, ∗ for all vectors x and y in X and a, b ∈ R+.

Therefore, we deduce the relation of the statement.

Remark 2. If we replace y by −y in Theorem 2, then we obtain an inequality for Clarkson distance, in the case kxk = kyk = 1, given by:

ax − by − |a − b| ax − by + |a − b| ≤ x − y ≤ . (11) min{a, b} max{a, b}

x Inequality (11) represents a generalization of Maligranda’s inequality, because, replacing x by kxk , y by y = k k = k k kyk and taking a x , b y , we deduce the inequalities from relation (3).

3. Some Inequalities Related to the Tapia Semi-Inner-Product Next, we study estimates of the Cauchy–Schwarz inequality using the Tapia semi-inner-product. The Tapia semi-inner-product on the normed space X (see [17]) is the function (·, ·)T : X × X → R, deﬁned by ϕ(x + ty) − ϕ(x) (x, y)T := lim , t→0 t t>0 1 2 where ϕ(x) = 2 kxk , x ∈ X. The above limit exists for any pair of elements x, y ∈ X. The semi-inner-product have been used by many authors in several contexts (see, e.g., [18]). The Tapia semi-inner-product is positive homogeneous in each argument and satisﬁes inequality

|(x, y)T| ≤ kxkkyk, (12) for all x, y ∈ X. In the case when the norm k · k is generated by an inner product h·, ·i, then (x, y)T = hx, yi, for all x, y ∈ X, thus we ﬁnd the Cauchy–Schwarz inequality.

Theorem 3. If X = (X, k · k) is a normed vector space over the ﬁeld of real numbers R, then we have

kxkkyk + (x, y)T ≤ xkyk + ykxk . (13) Mathematics 2020, 8, 2112 5 of 13 for every vector x and y in X.

Proof. If x = 0 or y = 0, then the inequality of the statement is true. We consider x 6= 0 and y 6= 0. If in inequality (7) we replace a and b by kxk and tkyk with t > 0, then we obtain

x y 2 − + min{kxk, tkyk} ≤ kxk + tkyk − kx + tyk kxk kyk

x y ≤ 2 − + max{kxk, tkyk}, kxk kyk which is equivalent to

x y 1 kx + tyk − kxk 2 − + min{kxk, tkyk} ≤ kyk − kxk kyk t t

x y 1 ≤ 2 − + max{kxk, tkyk}. kxk kyk t Thus, by passing to limit for t → 0, t > 0, we deduce

x y 1 kx + tyk − kxk 2 − + lim min{kxk, tkyk} ≤ kyk − lim kxk kyk t→0 t t→0 t t>0 t>0

x y 1 ≤ 2 − + lim max{kxk, tkyk}. kxk kyk t→0 t t>0 Because we have

kx + tyk − kxk kx + tyk2 − kxk2 (x, y) lim = lim = T t→0 t t→0 t(kx + tyk + kxk) kxk t>0 t>0

1 and for t → 0, t > 0, we obtain min{kxk, tkyk} = tkyk, equivalent to t min{kxk, tkyk} = kyk. Then, we deduce the inequality

x y 2 − + kxkkyk ≤ kxkkyk − (x, y) , (14) kxk kyk T which means that x y kxkkyk + (x, y) ≤ + kxkkyk = xkyk + ykxk . T kxk kyk

This inequality is equivalent with the inequality of the statement.

∈ x y Remark 3. For nonzero elements x, y X, if we replace x by kxk and y by kyk in inequality (13), then we ﬁnd the following inequality x y x y , ≤ + − 1. (15) kxk kyk T kxk kyk Let X = (X, h·, ·i) be a vector space with inner product; then, for nonzero elements x, y, inequality (14) becomes x y 2 − + kxkkyk ≤ kxkkyk − hx, yi. (16) kxk kyk

This inequality represents an improvement of Cauchy–Schwarz inequality. Mathematics 2020, 8, 2112 6 of 13

x y For nonzero elements x, y ∈ X denote v(x, y) := + , where X is a normed space. Then, inequality kxk kyk (14) becomes (see [15]): (x, y)T ≤ kxkkyk kv(x, y)k − 1 ≤ kxkkyk, (17) for all nonzero vectors x, y ∈ X.

Theorem 4. If X = (X, k · k) is a normed vector space over the ﬁeld of real numbers R, then we have 0 ≤ kxk kxk + kyk − kx + yk ≤ kxkkyk − (x, y)T (18) for every vector x and y in X.

Proof. Using the triangle inequality, we have 0 ≤ kxkkxk + kyk − kx + yk. From relation (6), for a = 1, b = t, we deduce

min{1, t}kxk + kyk − kx + yk ≤ kxk + tkyk − kx + tyk

≤ max{1, t}kxk + kyk − kx + yk, which divided by t becomes

min{1, t} kxk − kx + tyk max{1, t} kxk + kyk − kx + yk ≤ + kyk ≤ kxk + kyk − kx + yk. t t t Thus, by passing to limit, we ﬁnd the relation

min{1, t} kxk − kx + tyk lim kxk + kyk − kx + yk ≤ lim + kyk. t→0 t t→0 t t>0 t>0

Consequently, we obtain the inequality

(x, y) kxk + kyk − kx + yk ≤ kyk − T , kxk which implies the inequality of the statement.

Remark 4. (a) From inequality (18), we get:

2 kxk + (x, y)T ≤ kxkkx + yk. (19)

(b) If X = (X, h·, ·i) is a vector space with inner product ((x, y)T = hx, yi), then inequality (18) is in fact a characterization between the triangle inequality and Cauchy–Schwarz inequality:

0 ≤ kxkkxk + kyk − kx + yk ≤ kxkkyk − hx, yi, (20) for every vector x and y in X.

Relation (20) suggests the following:

Theorem 5. If X = (X, h·, ·i) is an inner product space over the ﬁeld of complex numbers C and the norm k · k is generated by h·, ·i, then we have

0 ≤ kxkkxk + kyk − kx + yk ≤ kxkkyk − Rehx, yi (21) for every vector x and y in X. Mathematics 2020, 8, 2112 7 of 13

Proof. Inequality (21) becomes

kxkkx + yk ≥ kxk2 + Rehx, yi.

By squaring, we obtain kxk2 kxk2 + hx, yi + hy, xi + kyk2 ≥ kxk4 + 2kxk2Rehx, yi + (Rehx, yi)2 which is equivalent to kxk2kyk2 − (Rehx, yi)2 ≥ 0, which is true, for every vector x and y in X.

Remark 5. Since |hx, yi| + Rehx, yi ≥ 0, from inequality (21), we deduce

0 ≤ kxkkxk + kyk − kx + yk ≤ kxkkyk + |hx, yi|, (22) for every vector x and y in X.

Below, we give a connection between the triangle inequality and the Cauchy–Schwarz inequality:

Theorem 6. If X = (X, h·, ·i) is an inner product space over the ﬁeld of complex numbers C and the norm k · k is generated by h·, ·i, then we have

kxkkyk − |hx, yi| kxk + kyk − kx + yk ≥ (23) kxk + kyk for every vector x and y in X, x 6= 0 or y 6= 0 .

Proof. From the equality 2 (kxkkyk − Rehx, yi) = 2kxkkyk − hx, yi + hx, yi =

kxk2 + 2kxkkyk + kyk2 − kxk2 + hx, yi + hy, xi + kyk2 = (kxk + kyk)2 − kx + yk2, we deduce (kxk + kyk + kx + yk)(kxk + kyk − kx + yk) ≥ 2 (kxkkyk − |hx, yi|) , because |hx, yi| ≥ Rehx, yi, which implies the following inequality

2 (kxk + kyk)(kxk + kyk − kx + yk) ≥ 2 (kxkkyk − |hx, yi|) , for every vector x and y in X. Consequently, we obtain the statement.

4. Estimates for Numerical Radii via Cauchy–Schwarz and Triangle Inequalities In this section, we employ the above results to obtain some new inequalities for the numerical radius and norm inequalities for Hilbert space operators. ∗ Let B (H) denote the C -algebra of all bounded linear operators on a complex Hilbert space H with inner product h·, ·i. For A ∈ B (H), let ω (A) and kAk denote the numerical radius and the operator norm of A, respectively. Recall that ω (A) = sup |hAx, xi|. It is well-known that ω (·) kxk=1 deﬁnes a norm on B (H), which is equivalent to the operator norm k·k. In fact, for every A ∈ B (H),

1 kAk ≤ ω (A) ≤ kAk . (24) 2 Mathematics 2020, 8, 2112 8 of 13

In [19], Kittaneh gave the following estimate of the numerical radius which reﬁnes the ﬁrst inequality in (24): 1 |A|2 + |A∗|2 ≤ ω2 (A) . (25) 4 For several other results of this kind, we refer the reader to papers [5,11,20–22]. We recall the following inequality for ω (A) which is known in the literature as the power inequality ω (An) ≤ ωn (A) (26)

A 2 for every n ≥ 1. We denote by ζλ (a, b) = infkxk=1k(aA + bλI) xk .

Theorem 7. Let A ∈ B (H). Then

(0 ≤) kAk2 − ω2 (A) ≤ kAk kA + A∗k . (27)

Proof. From the inequality (3.11), we obtain

kxk2 ≤ kxkkx + yk + |hx, yi|. (28)

Putting x = Ax and y = A∗x with kxk = 1 in the inequality (28), we get D E 2 ∗ 2 kAxk ≤ kAxk k(A + A ) xk + A x, x . (29)

Taking the supremum in (29) over x ∈ H with kxk = 1, we infer that kAk2 ≤ kAk kA + A∗k + ω A2 . (30)

The inequality (27) follows by combining (26) and (30).

Combining the second inequality in (24) and the inequality (30), we conclude the following result:

Theorem 8. Let A ∈ B (H). Then (0 ≤) ω2 (A) − ω A2 ≤ kAk kA + A∗k . (31)

Corollary 1. Let A ∈ B (H) be an invertible operator and let 0 6= λ ∈ C. Then,

kAk − kA + λIk kAk ≤ ω (A) . (32) |λ|

Proof. Replacing x = Ax and y = λx with kxk = 1 and 0 6= λ ∈ C in the inequality (28), we obtain

kAxk2 ≤ kAxkk (A + λI) xk + |λ||hAx, xi|.

Now, taking supremum over unit vector x ∈ H with kxk = 1, we get

kAk2 ≤ kAkkA + λIk + |λ|ω (A) . (33)

Now, if we divide (33) by |λ| > 0, then we get

kAk − kA + λIk kAk ≤ ω (A) . |λ| Mathematics 2020, 8, 2112 9 of 13

This completes the proof.

Theorem 9. Let A ∈ B (H) and let 0 6= λ ∈ C. Then

(kAk + |λ|)2 − ζ A(1, 1) kAk − ω(A) ≤ λ , (34) |λ|

A 2 where ζλ (1, 1) = infkxk=1k(A + λI) xk .

Proof. It follows from Theorem 5 that

kxk kyk − |hx, yi| ≤ (kxk + kyk)2 − (kxk + kyk) kx + yk .

From the triangle inequality, we infer that

kxk kyk − |hx, yi| ≤ (kxk + kyk)2 − kx + yk2.

Replacing x = Ax and y = λx with kxk = 1 and 0 6= λ ∈ C in the above inequality, we get

kAxk |λ| − |hAx, xi| |λ| ≤ (kAxk + |λ|)2 − k(A + λI) xk2.

The above inequality implies,

(kAxk + |λ|)2 − ζ A(1, 1) kAxk ≤ λ + |hAx, xi| . |λ|

Taking the supremum over all unit vectors x ∈ H gives

(kAk + |λ|)2 − ζ A(1, 1) kAk ≤ λ + ω (A) |λ| as required.

The following lemma contains a norm inequality for sums of positive operators that is sharper than the triangle inequality (see [23]).

Lemma 1. Let A, B ∈ B (H) be two positive operators; then, r ! 1 2 1 1 2 kA + Bk ≤ kAk + kBk + (kAk − kBk) + 4 A 2 B 2 . (35) 2

The following theorem provides a reﬁnement of the triangle inequality for general (i.e., not necessarily positive) operators.

Theorem 10. Let A, B ∈ B (H). Then v u r ! u 1 2 kA + Bk ≤ t kAk2 + kBk2 + kAk2 − kBk2 + 4kAB∗k2 + 2ω (B∗ A). (36) 2 Mathematics 2020, 8, 2112 10 of 13

Proof. We have for any x, y ∈ H,

kx + yk2 = hx + y, x + yi = hx, xi + hx, yi + hy, xi + hy, yi = kxk2 + kyk2 + 2 Re hx, yi ≤ kxk2 + kyk2 + 2 |hx, yi| i.e., kx + yk2 ≤ kxk2 + kyk2 + 2 |hx, yi| . (37)

Replacing x by Ax and y by Bx with kxk = 1 in the inequality (37), we infer that

k(A + B) xk2 ≤ kAxk2 + kBxk2 + 2 |hAx, Bxi| = hAx, Axi + hBx, Bxi + 2 |hB∗ Ax, xi| D E D E = |A|2x, x + |B|2x, x + 2 |hB∗ Ax, xi| D E = |A|2 + |B|2 x, x + 2 |hB∗ Ax, xi| .

Thus, D E k(A + B) xk2 ≤ |A|2 + |B|2 x, x + 2 |hB∗ Ax, xi| .

By taking supremum over x ∈ H with kxk = 1, we get

2 2 2 ∗ kA + Bk ≤ |A| + |B| + 2ω (B A) . (38)

On the other hand, from Lemma 1, replacing A with |A|2 and B with |B|2 , we have r ! 1 2 |A|2 + |B|2 ≤ kAk2 + kBk2 + kAk2 − kBk2 + 4kAB∗k2 , (39) 2 where we use the following two identities

k|A| |B|k = kAB∗k , and

2 2 |A| = kAk . On making use of (38) and (39), we get v u r ! u 1 2 kA + Bk ≤ t kAk2 + kBk2 + kAk2 − kBk2 + 4kAB∗k2 + 2ω (B∗ A), (40) 2 as required. Mathematics 2020, 8, 2112 11 of 13

Remark 6. To show the inequality (40) improves the triangle inequality, we can write v u r ! u 1 2 kA + Bk ≤ t kAk2 + kBk2 + kAk2 − kBk2 + 4kAB∗k2 + 2ω (B∗ A) 2 v u r ! u 1 2 ≤ t kAk2 + kBk2 + kAk2 − kBk2 + 4kAB∗k2 + 2 kB∗ Ak 2

(since ω (T) ≤ kTk, for any T ∈ B (H)) v u r ! u 1 2 ≤ t kAk2 + kBk2 + kAk2 − kBk2 + 4kAk2kBk2 + 2 kAk kBk 2

(by the submultiplicativity property of the operator norm) q = kAk2 + kBk2 + 2 kAk kBk q = (kAk + kBk)2 = kAk + kBk .

If, in the above inequalities, we take B = A∗, then we deduce the following:

Proposition 1. Let A ∈ B (H). Then, q kA + A∗k ≤ kAk2 + kA2k + 2ω (A2). (41)

Remark 7. Proposition 1 easily implies

1 kA + A∗k2 − kAk2 + A2 ≤ ω A2 . (42) 2 Now, on making use of the inequalities (42) and (26), we get

1 kA + A∗k2 − kAk2 + A2 ≤ ω2 (A) . (43) 2 It is worth mentioning here that, if A is a self-adjoint operator, then (43) is a sharper inequality than (25).

Theorem 11. Let A ∈ B (H) and let 0 6= λ ∈ R. Then,

1 1 min{ , }(ckAk + d|λ| − kcA + dλIk) ≤ kAk + |λ| − ζ A(1, 1), (44) c d λ and 1 1 kAk + |λ| − kA + λIk ≤ max{ , }(ckAk + d|λ| − ζ A(c, d)), (45) c d λ A 2 where ζλ (c, d) = infkxk=1k(cA + dλI) xk .

Proof. For a = b in inequality (5), we have

1 1 min{ , }(ckxk + dkyk − kcx + dyk) ≤ kxk + kyk − kx + yk ≤ c d 1 1 ≤ max{ , }(ckxk + dkyk − kcx + dyk), (46) c d Mathematics 2020, 8, 2112 12 of 13

∗ for all vectors x and y in X and c, d ∈ R+.

If we take the substitutions x by Ax and y by λx, λ ∈ R, with kxk = 1, then

1 1 min{ , }(ckAxk + d|λ| − k(cA + dλI)xk) ≤ kAxk + |λ| − k(A + λI)xkk c d 1 1 ≤ max{ , }(ckAxk + d|λ| − k(cA + dλI)xk), (47) c d ∗ for all vectors x and y in X and c, d ∈ R+.

Taking the supremum in (47) over all unit vectors x ∈ H gives

1 1 1 1 min{ , }(ckAk + d|λ|) ≤ min{ , }kcA + dλIk + kAk + |λ| − ζ A(1, 1), (48) c d c d λ ∗ for all vectors x and y in X and c, d ∈ R+.

From inequality (46), we take the substitutions x by Ax and y by λx, λ ∈ R, with kxk = 1, then

1 1 kAxk + |λ| ≤ k(A + λI)xk + max{ , }(ckAxk + d|λ| − k(cA + dλI)xk), (49) c d ∗ for all vectors x and y in X and c, d ∈ R+.

Taking the supremum over all unit vectors x ∈ H, we deduce the statement.

Remark 8. If A is an invertible operator, then, for λ = ω(A) in inequalities (44) and (45), we have

1 1 min{ , }(ckAk + dω(A) − kcA + dω(A)Ik) ≤ kAk + ω(A) − ζ A (1, 1), (50) c d ω(A) and 1 1 kAk + ω(A) − kA + ω(A)Ik ≤ max{ , }(ckAk + dω(A) − ζ A (c, d)). (51) c d ω(A) For c = ω(A) and d = kAk in the above inequalities, we obtain

A kAkζω(A)(1, 1) − kω(A)A + kAkω(A)Ik ≤ kAk(kAk − ω(A)), (52) and 1 (ζ A (ω(A), kAk) − kA + ω(A)Ik ≤ kAk(kAk − ω(A)). (53) ω(A) ω(A)

Author Contributions: The work presented here was carried out in collaboration between N.M. and H.R.M.. N.M. and H.R.M. contributed equally and significantly in writing this article. N.M. and H.R.M. have contributed to the manuscript. All authors have read and agreed to the published version of the manuscript. Funding: This research received no external funding. Acknowledgments: The authors thank the reviewers for their pertinent remarks, which led to an improvement of the paper. Conflicts of Interest: The authors declare no conflict of interest. Mathematics 2020, 8, 2112 13 of 13

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