Berezin inequalities for sums of operators and classical inequalities concerning the Berezin radius Текст научной статьи по специальности «Математика»

doi https://doi.org/10.5937/vojtehg72-51843

FIELD: mathematics

ARTICLE TYPE: original scientific paper

Abstract:

Methods: The methods used are standard for operator theory.

Introduction

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Berezin inequalities for sums of operators and classical inequalities concerning the Berezin radius

Mehmet Gurdala, Vuk N. Stojiljkovicb

aSuleyman Demirel University, Department of Mathematics, tS

Isparta, Republic of Turkiye, e-mail: [email protected], n

ORCID iD: ©https://orcid.org/0000-0003-0866-1869 bUniversity of Novi Sad, Faculty of Science, Novi Sad, Republic of Serbia Mathematical Grammar School, Belgrade, Republic of Serbia, e-mail: [email protected], corresponding author, ORCID iD: ©https://orcid.org/0000-0002-4244-4342

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Introduction/purpose: In this article, the author's goal is to seek to obtain .9. new inequalities of the Berezin type. $

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Results: Various inequalities of the type given by Huban et al. and Erkan

have been obtained. ors

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Conclusions: In addition to obtaining various inequalities of the form |

given by Huban et al. and Erkan in particular, the authors sharpened op

the inequalities related to the Berezin norm. of

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Key words: Berezin norm, Berezin number, inequalities. |

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The Berezin norm and the Berezin number of an operator have been a-

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researched for their many applications in numerical analysis, quantum physics, engineering, and other domains in the literature related to operator theory. In order to explain the Berezin number and norm, the authors first go over several ideas and traits of bounded linear operators on a Hilbert space.

Let L (H) denote the C*-algebra of all bounded linear operators defined on a complex Hilbert space H with the an inner product (.,.) and a corresponding norm ||.||. Recall that the functional Hilbert space H = H(2) is §

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a Hilbert space of complex-valued functions on a (nonempty) set 2 such that the evaluation functionals pe(f) = f (g), g e 2, are continuous on H and for every g e 2 there exists a function fe e H such that fe(g) = 0 or, equivalently, there is no g0 e 2 such that f (g0) = 0 for all f e H-The Riesz representation theorem ensures that for each g e 2 there is a g unique element xe e H such that f(g) = (f,xe) for all f e H- The col-

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lection {xe: q e E} is called the reproducing kernel of H. For q e E, let kg := Trii be the normalized reproducing kernel of H. The absolute value

0 of the positive operator is denoted by |3| = (3*3)1/2 ■ ° For a bounded linear operator 3 on H, the function 3 defined on S by ^ 3(e) := {3xe(z), xe(z)) is the Berezin symbol of 3, which firstly have been

1 introduced by Berezin (Berezin, 1972). In other words, the Berezin symbol w 3 is the function on S defined by restriction of the quadratic form {3x1,x1 ) > with x1 e H to the subset of all normalized reproducing kernels of the unit ft sphere in H- It is clear from the Cauchy-Schwarz inequality that 3 is the

bounded function on S whose values lie in the numerical range of the operator 3- The Berezin set (or range) and the Berezin number (or radius) of the operator 3 are defined by

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Ber(Z) := {3(e) : Q € 2} and ber(3) := sup 3(q)

respectively (see (Karaev, 2006)). ¡5 A relevant and important concept is a numerical radius which is the supremum of the absolute values of all numbers in W (3), that is

>

w(3) = sup \{3x,x)\.

x€H,\\x\\ = l

It is obvious that ber(3) < w(3) < \\3\\ and Ber(3) c W(3), where w(3) denotes the numerical radius and W(3) is the numerical range of the operator 3. It is well known that

f < w (3) <M3||

and

ber (3) < w (3) < M3M, (1)

for any 3 e L (H).

In (Huban et al, 2022a), Huban et al. substantially improved the upper bound in (1) by showing that if Z e L (H), then

ber(3) < 2 H3I + |Z* |||ber < 2 ( 11Z H ber + ||3 , | ber

2

(2)

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Another improvement for inequality (1) was provided by Huban et al. (Huban et al, 2021) as

1

ber2 (Z) <- |Z|2 + m

2

>*|2

ber

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which was further improved in (Gurdal & Ba§aran, 2023) by Ba§aran and Gurdal as

ber2 (Z) < 1 |Z|2 + |Z*|2 6

+ ^■ber (Z) |||Z| + |Z*||ber • (4)

ber 3

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The following inequalities for ber2(.) have been obtained in (Huban et al, 2021)

1 1

(5)

2

|Z|2 + |Z*|

b < ber2 (Z) < 2 |Z|2 + |Z*|2

ber 2

ber

Furthermore, Huban et al. (Huban et al, 2022a) established some refinements of (2) and (5), respectively, that can be presented as

and

ber7'(Z) < 2 |||Z|' + |Z*|2J(1"Ç)

ber2-7' (Z) < £ |Z|2' + (1 - £) ^f7

ber

ber

(6) (7)

where Z e L (H), 0 < £ < 1, and j > 1

Another important fact about the Berezin number upper bounds that are of our interest are due to Huban et al. in (Huban et al, 2021): Let Zi, Z2 e

L (H) and r > 1, then

berr(Z2Zi) < 2

|Zi|2r + |Z2|

2r

ber

(8)

For an in-depth exploration of the intricacies surrounding the Berezin symbol, interested readers are strongly encouraged to refer to (Bakherad & Garayev, 2019; Baçaran et al, 2022; Chalendar et al, 2012; Garayev & Alomari, 2021; Garayev et al, 2020; Gunturk & Gurdal, 2024; Garayev

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et al, 2021; Stojiljkovic & Gurdal, 2024a,b; Gurdal et al, 2023; Gurdal & Tapdigoglu, 2023; Tapdigoglu et al, 2021; Yamanciet al, 2020; Gurdal & Stojiljkovic, 2024a; Huban et al, 2022b; Gurdal & Stojiljkovic, 2024b) and the comprehensive references provided therein.

In this paper, motivated by previously reported results (Stojiljkovic & Dragomir, 2024), this work aims to develop new Berezin number upper bounds for reproducing kernel Hilbert space operators by introducing new improvements to the well-known Cauchy-Schwarz inequality.

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We require a few well-known lemmas in order to demonstrate our extended Berezin number inequalities.

According to the traditional Schwarz inequality for positive operator, for

any xi,x2 e H

\(3xi,X2)\2 < (3xi,Xi) (3X2,X2) (9)

if 3 e L (H) is a positive operators. Kato's inequality, sometimes referred to as the combined Cauchy-Schwarz inequality, was initially put out by Kato (Kato, 1952) in 1952 as a companion to the Schwarz inequality (9). It says that

\(3xi,x2)\2 < (\3\2Yxi,xi)(\3*\2(i-Y)x2,x2), Y e [0,1] (10)

for any operator 3 e L (H) and any xi,x2 e H- In order to generalize this result, in 1994 Furuta (Furuta, 1994) obtained the following result:

\{3\3\Y+V-ixi,x2)\2 < {\3\2Yxi,xi){\3*\2Vx2,x2), (11)

for any xi,x2 eH and Y,n e [0,1] with y + n > 1.

Lemma 1. (McCarthy, 1967). Let 3 e L (H), 3 > 0 and let x e H be any unit vector. Then there is

{3x,x)r <{3rx,x) for r > 1, (12)

{3rx,x) <{3x,x)r for 0 < r < 1. (13)

The well-known Buzano's inequality is the following outcome. Lemma 2. Let x,y,e e H with \\e\\ = 1. Then there is

\{x,e){e,y)\< 1(\\x\\ \\y\\ + \{x,y)\). (14)

The following result is found in (Singh Aujla & Silva, 2003) and is related to non-negative convex functions.

Lemma 3. Let f be a non-negative convex function on [0, and 3i, Z2 e L (H) be positive operators. Then

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f

3i + 3; 2

<

f (3i) + f (32)

2

Lemma 4. Let x,y e H and A e L (H) •

\{Ax,y)\2 <{f2(\A\)x,x){g2(\A*\)y,y).

(15)

(16)

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is the case where f and g are two nonnegative continuous functions on [0, that fulfill f (t)g(t) = t,t > 0.

With regard to the mapping p, Stojiljkovic and Gürdal (Stojiljkovic & Gürdal, 2024c) recently acquired the following modification.

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Lemma 5. Let uk,vk, e e H,l,q > 1,p > 1, } + 1 = 1- Let J be a set such that (0,1) c J c R. Let p be a of mapping such that p : J ^ R+, such that the following holds ^n p(a,) = 1- Then the following inequality holds

^p(afc ){uk ,e){e,vk)

k=i

<En=i Pl/2(ak)\{uk, e)\pl +

l

+

ELi Pq/2(ak)\{e,vk)\pq

(17)

Main results

We give our first result, a consequence of Lemma 5 which is instrumental in the development of the later results.

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Corollary 1. Let l,q > 1 such that } + i = 1 and p > 1, 3, e L (H) and

p : J ^

with^ p(®i) = then the following inequality holds:

^p(ak){3kKg,

< Efc=i{\3k\2palKg, Ke)pl/2(ak) +

< 2l +

k=i

En=i{\3k\2pl(i"a)Kg, *Cg)pl/2(ak) , Efc=i Pq/2(ak)\{3kKg, 2g)F

+

2l

+

q

• (18)

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Proof. If we take vk = 3*kKg, uk = 3kKg and e = Kg in Lemma 5 above, one obtains

^^(ak){3kKg, Siey

k=i

<

ELi Vl/2(ak)\{3kKeg, Kig)\p ELi Vq/2(ak)\{3kKg, Kg)\pq

+

<

ELi{\3k\2palKg, Kg)vl/2(ak) , ELi{\3k\2pl(i-a)Kg, Kg)Vl/2(ak)

2l

+

2l

+

+

ELi fq/2(ak)\{3kKg, %e)\pq

□

Remark 1. Taking supremum over all q e 2, one obtains

sup

ges

){3kKg, Kg)2

k=i

< 2l sup E (f(ak ))l/2( (\3k\2pal + \3k\2pl(i-a)) Kg, Kg) +

+ sup ges

g^ k k

ELi yq/2(ak )\{3k Kg, Kg)\pq q '

Setting 3k,pk = 0 for k e {2,3..,n},q = l = 2, we obtain the following inequality which sharpens the one given by Huban et al. (Huban et al, 2022a) (7)

ber2p(3) <

|\3\4«p + \3*\4p(i-a)

L + ber2p(3)

2

< -2

\ 3 \ 4ap + \3*\4p(i-a)

ber

In particular, we obtain a refinement by setting a = 2 of inequality (7) for

s = 2.

Proof.Using (8) on the third term, we obtain the desired inequality.

□

Theorem 1. Letp > 1,l,q> 1, p+± = 1,a e [0,1],p > f, 3i e L (H) andH

be such that H(l) = f 2(\3i\apl)+g2(\3i\pla)+f 2(\3*\pl(i-a))+g2(\3*\pl(i-a)),

p

l

q

q

p

4

then the following inequality holds:

berp V(at)Zt)

< qnJ E pl/2(ai)\3S^Se)\p + Ei Pg/2(ai)|(3i% S*)\1

< »up < +

ees [ l q

(19)

<

E U<pl/2 (ai) + ^/2 (ai)

8 V l

n /

E ^ (ai)

i=i V

+

ber p ( \ 3* \2(1-a) \ Zi \2a)

ber f ( \ 3t \ 2(1-a) \ 3i \ 2a)

Proof. Start from the left-hand side

n p

i S*) I

i=1

n

< E ^(ai) \ <3^^ \ P

i=1

n

< E ^(ai) \ (3iS*, S*) \ T E Pq/2(ai) \ <3i% S*) \ \i=1 n

n

1/q

i=1 n

^n=1 Pl/2(ai)\ <3iS*, S*) \ t + En=1 Pq/2(ai) \ <3i% %) \?

l

q

We focus now on the first part, second one is done analogous. First use Kato's inequality

pi

\ <3iSe, Ke) \T

<< \ 3i \2a p < Se, \ 31 \2(1-a) See)* < 1 (ll\3i\2aSe\\1 ||\3!\2(1-a)5**11f + \<\3i\2aS*, \3H2(1-a)Se\p

< 4 (< \3i\4aSe, Se) ? + < \3! \4(1-a)Se, Se) +1 \ < \3! \2(1-a) \ 3i\2aSe, Se) \ 1

< -< 4

\ plaSe, S*) + < \3!\pl(1-a)S*, Se)) +2 \< \3!\2(1-a)\3i\2a S*, S(

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< 8 ({f 2(\3i\pal )Kg, Kg) + {g2(\3i \pla )Kg, Si,)) + + 1 {{f 2(\3*\pl(i-a))Kg, Kg) + {g2(\3;\pl(i-a))Kg, Kg)) +

+ 2\{\3I\2(i-a)\3i\2aKg, Kg)\* .

The other part is obtained in an analogous way, adding them and taking a supremum over q e 2 we obtain the desired inequality. □

In particulular, by setting 3k = 0, f (t) = g(t) = Vt, = 0 for k e {2,3..., n},q = l = 2, we obtain the refinement of (6), namely

berp (3) < 1 \3\2pa + \3*\2p(i-a) + ^berf (\3*\2(i-a)\3\2a) 4 ber 2

<

\3\2pa + \3*\2p(i-a)

2

ber

In the following we give a variation of the inequality given by Erkan and Gurdal (Erkan & Gurdal, 2024, Theorem 4).

Theorem 2. Let r > 1,a e [0,1],3i,32,3s,34 e L(H) and H(3i) f2( \3i\ 4ra) + g2(\3i\ 4ra) + f2( \3i\ 4r(i-a)) + g2(\3i\ 4r(i-a)) then

berr(3i32 + 3334) < 2r \\H (32) + H (3i) + H (34) + H (3s) \\ber . Proof. We proceed

\ {(3i32 + 3^34)Kg, Kg) \r

(20)

2r

<

<

\ {3i32S^g, Scg) \ + \ {3334Kg, Kg) \

2

\ {3i32 Kg,Kg) \r + \ {3334 Kg,Kg) \7

< 1 { \32 \ 4raKg, Kg) + { \32 \ 4r(i-a)Kg, Kg + { \3i \ ^Kg, Kg) + o

+ { \ 3i \ 4r(i-a) Kg, Kg) + { \ 34 \ 4ra Kg, Kg ) + + { \ 34 \ 4r(i-a) Kg, Kg + { \ 3s \ 4ar Kg , Kg) + { \ 3s \ ^^ Kg, Kg).

Proceeding to use (16) and AG inequality, we obtain the desired inequality.

□

r

2

Corollary 2. Setting f (t) = g(t) = y/t and a = 1 we obtain a variation of the inequality given by Erkan and Gurdal (Erkan & Gurdal, 2024)

berr(3i32 + 3334) < 2r-2 \ 32 \ 2r + \ 33 \ 2r + \ 34 \ 2r + \ 3i \

2r

12r

i2r

12r

ber

Theorem 3. Let r > 1,a e [0,1], 3i, 32 e L (H) and f,g nonnegative such that f (t)g(t) = t also let

H (3i) = f 2( \ 3i \ 2ra) + g2( \ 3i \ 2ra) + f 2( \ 3i \ 2r(i-a))+ g2( \ 3i \ 2r(i-a)) then

berr(3i + 32) < 2r-3 \\H (3i)+ H (32)^ber •

(21)

Proof-

\ {(3i + 32) Kg,Kg) \

2r

<

<

\ {3iKg, Kg) \ + \ {32Kg,Kg) \

\ {3iKg, Kg) \r + \ {32Kg, Kg) \r

< { \ 3i\ 2"Kg, Kg)2 { \3*\ 2(i-a)Kg, Kg)2 + { \32 \ 2aKg, Kg)2 { \32 \ 2(i-a)Kg, Kg)

{ \ 3i \ 2ra Kg, Kg) + { \ 3i \ 2r(i-a) Kg, Kg) +

+

{ \ 32 \ 2ra Kg, Kg) + { \ 32 \ 2r(i-a) Kg, Kg)

The proof is finished with using (16) and AG inequality.

□

Corollary 3. Setting f (t) = g(t) = y/t and replacing the second slot with y and proceeding with the proof for the case ((31 + Z2) Kie, Ke) the proof still holds and then using triangle inequality we obtain the inequality given by Huban et al. (Huban et al, 2022a, Theorem 3.7)

\\3i + 32\\ber < 2

r2

1-> 12ar , 11 12ar \ 3i \ + \32 \

ber

+

\ 3*i \ 2r(i-a) + \ 3*2 \ 2r(i-a)

ber

We now present a variation of the inequality given by Huban et al. (Huban et al, 2022a).

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IJ Theorem 4. Let us define H(3i) = f2(|3l|2ar) + g2(\Zi\2ar), G(3i) = f2(|3*i|2r(i—a)) + g2(|3*i|2r(i—a)) where 3i = 32 + ¿3s r > 2,a e CM I [0, 1],3 i, 32,33 e L (H) where 32,33 are the Cartesian decomposition operators of 3 i where 32,33 are selfadjoint, then one obtains

§ berr(3i) - 22-3 \\H (32) + G (32) + H (3s) + G (3s)\\ber •

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Proof.

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(32Ke, xe)2 + (33% Ke)2

X V 2

W (32 Kg, Kg)r + (33Kg, Kg)1

* < 2 , ^ ^ < K,, Kg) + (|3*2 |2r(i —a) Kg, Kg) +

- 4 +

+ (|33|2ar Kg, Kg) + (|3*з|2r(l — а) Kg, Kg)

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o following a similar procedure to the one given in (21), we obtain the desired g result. □

Theorem 5. Let 3i, 32 e L (H) ,r > 2 and a e [0,1], then

berr(3i + 32) < 2r-3 |\3i\2ar + \31\2r(1"«) + \32\2ar + \32 \2r(1-a)

+ 2r-2(ber( \ 3*i \ r(1-a) \ 3i \ar ) + ber(\3*2\r(1-a)\32\ar ))• (22)

+

ber

Proof.Following a similar principle to the one given in (21) with an exception of using Buzano's inequality, we obtain the desired inequality; one must only realise that in order to use the Mc-Carthy inequality there must

be r > 2. □

Corollary 4. Previously obtained inequality (22) refines the inequality given by Huban et al. in (Huban et al, 2022a), namely Theorem 3.6 (eq. 3.6) for r > 2; it can be seen by using (6) on both Berezin radius terms, from which we obtain

berr(3i + 32) < 2r-3 \ 3i\ 2ar + \ 31 \ 2r(1-a) + \ 32 \ 2ar + \ 32 \ 2r(1-a)

+

ber

+ 2r-2(ber( 13*i | r(1-a) | Zi |ar) + ber( 13*2 I r(1-a) 132 Iar))

< 2

r- 3

I 3i I 2ar + I 31 I 2r(1-a) + I 32 I 2ar + I 32 I 2r(1-a)

ber

+

+ 2

r-3

2

r-2

I31I2«r + I31 I2r(1-a)

I31I2QT + I31I2r(1-a)

ber

+

ber

+

I32I2«r +132 I2r(1-«) I32I2ar + 13 212r(1 a)

ber

ber

If we were to use the triangle inequality on eq. (3.6) (Huban et al, 2022a) we would obtain the right hand side of the above chain of inequalities, which shows that our inequality is sharper than it.

Further, setting 31 = 32 we obtain (6) on the right-hand side which shows that our inequality (22) is a refinement of (6) for r > 2.

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References

Bakherad, M. &Garayev, M.T. 2019. Berezin number inequalities for operators. Concrete Operators, 6(1), pp.33-43. Available at: https://doi.org/10.1515/conop-2019-0003.

Ba§aran, H., Huban, M.B. & Gurdal, M. 2022. Inequalities related to Berezin norm and Berezin number of operators. Bulletin of Mathematical Analysis and Applications, 14(2), pp.1-11 [online]. Available at:

https://www.emis.de/journals/BMAA/repository/docs/BMAA14-2-1.pdf [Accessed: 25 June 2024].

Berezin, F.A. 1972. Covariant and contavariant symbols of operators. Mathematics of the USSR-Izvestiya, 6(5), pp.1117-1151. Available at: https://doi.org/10.1070/IM1972v006n05ABEH001913.

Chalendar, I., Fricain, E., Gurdal, M. & Karaev, M. 2012. Compactness and Berezin symbols. Acta Scientiarum Mathematicarum, 78(1-2), pp.315-329. Available at: https://doi.org/10.1007/BF03651352.

Erkan, G.G. & Gurdal, M. 2024. Operatorlerin Berezin Yarigap E§itsizliklerinin ilerletilmesi. In: 3rd International Conference on Engineering, Natural and Social Sciences ICENSOS, Konya, Turkey, pp.463-470, May 16-17 [online]. Available at: https://www.icensos.com/3rdicensos2024 (in Turkish) [Accessed: 25 June 2024].

Furuta, T. 1994. An extension of the Heinz-Kato theorem. Proceedings of the American Mathematical Society, 120(3), pp.785-787. Available at: https://doi.org/10.2307/2160471.

Garayev, M.T. & Alomari, M.W. 2021. Inequalities for the Berezin number of operators and related questions. Complex Analysis and Operator Theory, 15, art.number:30. Available at: https://doi.org/10.1007/s11785-021-01078-7.

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N cu

cu CO

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Garayev, M., Bouzeffour, F., Gurdal, M. & Yangoz, C.M. 2020. Refinements of Kantorovich type, Schwarz and Berezin number inequalitie. Extracta Mathemati-cae, 35(1), pp.1-20. Available at: https://doi.org/10.17398/2605-5686.35.1.1. Garayev, M.T., Guediri, H., Gurdal, M. & Alsahli, G.M. 2021. On some prob> lems for operators on the reproducing kernel Hilbert space. Linear and Multilinear ^ Algebra, 69(11), pp.2059-2077. Available at:

https://doi.org/10.1080/03081087.2019.1659220.

^ Gunturk, B. & Gurdal, M. 2024. On some refining inequalities via Berezin sym-

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Yamanci, U., Tung, R. & Gurdal, M. 2020. Berezin Number, Gruss-type inequal- o ities and their applications. Bulletin of the Malaysian Mathematical Sciences Soci-

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sigualdades clásicas relativas al radio de Berezin o

Mehmet Gürdala, Vuk N. Stojiljkovicb, autor de correspondencia

a Universidad Suleyman Demirel, Departamento de Matemáticas, Isparta, República de Turquía cu

b Universidad de Novi Sad, Facultad de Ciencias, Novi Sad, República de Serbia; <u

Escuela Secundaria de Matemáticas, Belgrado, República de Serbia

CAMPO: matemáticas i m

TIPO DE ARTÍCULO: artículo científico original "S

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Introducción/objetivo: En este artículo, el objetivo del autor es buscar obtener nuevas desigualdades del tipo Berezin. Métodos: Los métodos utilizados son estándar para la teoría de operadores.

Resultados: Varias desigualdades del tipo dado por Huban et al. y Erkan han sido obtenidas.

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ш Conclusión: Además de obtener en particular diversas desigual-

dades de la forma dada por Huban et al. y Erkan, los autores § agudizaron las desigualdades relacionadas con la norma Bere-

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Palabras claves: norma de Berezin, número de Berezin, des-g igualdades.

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ft Неравенства Березина для сумм операторов и

классические неравенства относительно радиуса Березина

MexMem Гурдала, Вук Н. Соилькович6, корреспондент

— а Университет имени Сулеймана Демиреля,

математический факультет, г Испарта, Турецкая Республика

О 6 Нови-Садский университет, факультет науки,

^ г Нови-Сад, Республика Сербия;

Математическая школа, г. Белград, Республика Сербия

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Ш РУБРИКА ГРНТИ: 27.25.15 Дескриптивная теория функций,

О 27.39.19 Линейные операторы и

операторные уравнения, > 27.39.21 Спектральная теория линейных

операторов ВИД СТАТЬИ: оригинальная научная статья

Резюме:

Введение/цель: Целью данной статьи является поиск новых неравенств типа Березина.

Методы: В исследовании использовались стандартные методы для теории операторов.

Результаты: Были получены различные неравенства типа неравенств, приведенных Хубаном и др. и Эрканом. Выводы: В дополнение к полученным неравенствам типа неравенств, приведенных Хубаном и др. и Эрканом, авторы уточнили неравенства, связанные с нормой Березина.

Ключевые слова: норма Березина, число Березина, нера- ю венства.

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изнели Хубан и др. и Еркан.

Закъучак: Поред доби]ак>а разних не}еднакости облика ко}е

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Березинове не]еднакости за збир оператора и класичне не]еднакости ще се односе на Березинову норму

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Мехмет Гурдала, Вук Н. Сто]и^кови^6, ауторза преписку 50

а Универзитет „Суле]ман Демирел", Катедра за математику, N

Испарта, Република Турска

б Универзитет у Новом Саду, Природно-математички факултет, Катедра за математику, Нови Сад, Република Срби]а; Математичка гимнази]а, Београд, Република Срби]а

ОБЛАСТ: математика

КАТЕГОРША (ТИП) ЧЛАНКА: оригинални научни рад Сажетак:

Увод/цил: Циъ овог рада jесте да се изведу нове не]'една-кости Березиновог типа.

Методе: Примешене су стандардне методе за теорбу го оператора. $

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Резултати: Доб^ене су разне неjеднакости типа ще су о

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су изнели Хубан и др. и Еркан, аутори су пооштрили не}ед- го накости везане за Березинов ради}ус.

Къучне речи: Березинова норма, Березинов броj, не1една- °

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Paper received on: 27.06.2024.

Manuscript corrections submitted on: 16.11.2024.

Paper accepted for publishing on: 18.11.2024.

© 2024 The Authors. Published by Vojnotehnicki glasnik/Military Technical Courier (http://vtg.mod.gov.rs, http://BTr.M0.ynp.cp6). This article is an open access article distributed under n

the terms and conditions of the Creative Commons Attribution license (http://creativecommons.org/licenses/by/3.0/rs/).

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