Two points and $n$ th derivatives norm inequalities for analytic functions in Banach algebras
UDC 517.5 Let $\mathcal{B}$ be a unital Banach algebra, let $a \in \mathcal{B},$ $G$ be a convex domain of $\mathbb{C}$ with $\sigma (a) \subset G,$ let $\alpha, \beta \in G,$ and let $f \colon G \rightarrow \mathbb{C}$ be analytic on $G.$By using the analytic functional calculus, we obtain among ot...
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Institute of Mathematics, NAS of Ukraine
2022
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860512272504324096 |
|---|---|
| author | Dragomir, S. S. Dragomir, Silvestru Sever Dragomir, S. S. |
| author_facet | Dragomir, S. S. Dragomir, Silvestru Sever Dragomir, S. S. |
| author_sort | Dragomir, S. S. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2022-10-24T17:55:16Z |
| description | UDC 517.5
Let $\mathcal{B}$ be a unital Banach algebra, let $a \in \mathcal{B},$ $G$ be a convex domain of $\mathbb{C}$ with $\sigma (a) \subset G,$ let $\alpha, \beta \in G,$ and let $f \colon G \rightarrow \mathbb{C}$ be analytic on $G.$By using the analytic functional calculus, we obtain among others the following result:\begin{gather*}\left\| f(a) - \frac{1}{2}\sum_{k=0}^{n}\frac{1}{k!}\left[f^{(k) }(\alpha) (a-\alpha)^{k}+(-1)^{k}f^{(k) }(\beta) (\beta-a)^{k}\right] \right\|\leq\\\leq \frac{1}{2(n+1) !}\left[\|a-\alpha\|^{n+1}+\|\beta - a\|^{n+1}\right]\times\\\times \max \left\{\sup_{s\in [0,1] }\left\| f^{(n+1) }[(1-s) \alpha+sa] \right\|,\sup_{s\in [0,1] }\left\| f^{(n+1) }[(1-s) a+s\beta] \right\| \right\}.\end{gather*}Some examples for the exponential function on Banach algebras are also given. |
| doi_str_mv | 10.37863/umzh.v74i8.6116 |
| first_indexed | 2026-03-24T03:26:09Z |
| format | Article |
| fulltext |
DOI: 10.37863/umzh.v74i8.6116
UDC 517.5
S. S. Dragomir1 (College Engineering and Sci., Victoria Univ., Melbourne, Australia; School Comput. Sci. and Appl.
Math., Univ. Witwatersrand, Johannesburg, South Africa)
TWO POINTS AND \bfitn TH DERIVATIVES NORM INEQUALITIES
FOR ANALYTIC FUNCTIONS IN BANACH ALGEBRAS
НЕРIВНОСТI ДЛЯ НОРМИ ДВОХ ТОЧОК I \bfitn -Ї ПОХIДНОЇ
ДЛЯ АНАЛIТИЧНИХ ФУНКЦIЙ У БАНАХОВИХ АЛГЕБРАХ
Let \scrB be a unital Banach algebra, let a \in \scrB , G be a convex domain of \BbbC with \sigma (a) \subset G, let \alpha , \beta \in G, and let f :
G \rightarrow \BbbC be analytic on G. By using the analytic functional calculus, we obtain among others the following result:\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| f(a) - 1
2
n\sum
k=0
1
k!
\Bigl[
f (k)(\alpha )(a - \alpha )k + ( - 1)kf (k)(\beta )(\beta - a)k
\Bigr] \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \leq
\leq 1
2(n+ 1)!
\bigl[
\| a - \alpha \| n+1 + \| \beta - a\| n+1\bigr] \times
\times \mathrm{m}\mathrm{a}\mathrm{x}
\Biggl\{
\mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| , \mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| \Biggr\} .
Some examples for the exponential function on Banach algebras are also given.
Нехай \scrB — унiтальна алгебра Банаха, a \in \scrB , G — опукла область в \BbbC з \sigma (a) \subset G, \alpha , \beta \in G, а f : G \rightarrow \BbbC
є аналiтичною на G. Використовуючи аналiтичне функцiональне числення, ми отримуємо серед iнших такий
результат: \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| f(a) - 1
2
n\sum
k=0
1
k!
\Bigl[
f (k)(\alpha )(a - \alpha )k + ( - 1)kf (k)(\beta )(\beta - a)k
\Bigr] \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \leq
\leq 1
2(n+ 1)!
\bigl[
\| a - \alpha \| n+1 + \| \beta - a\| n+1\bigr] \times
\times \mathrm{m}\mathrm{a}\mathrm{x}
\Biggl\{
\mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| , \mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| \Biggr\} .
Наведено також деякi приклади для експоненцiальних функцiй на алгебрах Банаха.
1. Introduction. Let \scrB be an algebra. An algebra norm on \scrB is a map \| \cdot \| : \scrB \rightarrow [0,\infty ) such that
(\scrB , \| \cdot \| ) is a normed space, and, further, \| ab\| \leq \| a\| \| b\| for any a, b \in \scrB . The normed algebra
(\scrB , \| \cdot \| ) is a Banach algebra if \| \cdot \| is a complete norm. We assume that the Banach algebra is
unital, this means that \scrB has an identity 1 and that \| 1\| = 1.
Let \scrB be a unital algebra. An element a \in \scrB is invertible if there exists an element b \in \scrB
with ab = ba = 1. The element b is unique; it is called the inverse of a and written a - 1 or
1
a
.
The set of invertible elements of \scrB is denoted by \mathrm{I}\mathrm{n}\mathrm{v}(\scrB ). If a, b \in \mathrm{I}\mathrm{n}\mathrm{v}(\scrB ) then ab \in \mathrm{I}\mathrm{n}\mathrm{v}(\scrB ) and
(ab) - 1 = b - 1a - 1.
For a unital Banach algebra we also have:
1e-mail: sever.dragomir@vu.edu.au.
c\bigcirc S. S. DRAGOMIR, 2022
1086 ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
TWO POINTS AND nTH DERIVATIVES NORM INEQUALITIES FOR ANALYTIC FUNCTIONS . . . 1087
(i) if a \in \scrB and \mathrm{l}\mathrm{i}\mathrm{m}n\rightarrow \infty \| an\| 1/n < 1, then 1 - a \in \mathrm{I}\mathrm{n}\mathrm{v}(\scrB );
(ii) \{ a \in \scrB : \| 1 - b\| < 1\} \subset \mathrm{I}\mathrm{n}\mathrm{v}(\scrB );
(iii) \mathrm{I}\mathrm{n}\mathrm{v}(\scrB ) is an open subset of \scrB ;
(iv) the map \mathrm{I}\mathrm{n}\mathrm{v}(\scrB ) \ni a \mapsto - \rightarrow a - 1 \in \mathrm{I}\mathrm{n}\mathrm{v}(\scrB ) is continuous.
For simplicity, we denote z1, where z \in \BbbC and 1 is the identity of \scrB , by z. The resolvent set of
a \in \scrB is defined by
\rho (a) :=
\bigl\{
z \in \BbbC : z - a \in \mathrm{I}\mathrm{n}\mathrm{v}(\scrB )
\bigr\}
;
the spectrum of a is \sigma (a), the complement of \rho (a) in \BbbC , and the resolvent function of a is
Ra : \rho (a) \rightarrow \mathrm{I}\mathrm{n}\mathrm{v}(\scrB ), Ra(z) := (z - a) - 1. For each z, w \in \rho (a), we have the identity
Ra(w) - Ra(z) = (z - w)Ra(z)Ra(w).
We also get that
\sigma (a) \subset
\bigl\{
z \in \BbbC : | z| \leq \| a\|
\bigr\}
.
The spectral radius of a is defined as
\nu (a) = \mathrm{s}\mathrm{u}\mathrm{p}
\bigl\{
| z| : z \in \sigma (a)
\bigr\}
.
Let \scrB a unital Banach algebra and a \in \scrB . Then:
(i) the resolvent set \rho (a) is open in \BbbC ;
(ii) for any bounded linear functionals \lambda : \scrB \rightarrow \BbbC , the function \lambda \circ Ra is analytic on \rho (a);
(iii) the spectrum \sigma (a) is compact and nonempty in \BbbC ;
(iv) for each n \in \BbbN and r > \nu (a), we have an =
1
2\pi i
\int
| \xi | =r
\xi n(\xi - a) - 1d\xi ;
(v) we have \nu (a) = \mathrm{l}\mathrm{i}\mathrm{m}n\rightarrow \infty \| an\| 1/n.
Let \scrB be a unital Banach algebra, a \in \scrB and G be a domain of \BbbC with \sigma (a) \subset G. If
f : G \rightarrow \BbbC is analytic on G, we define an element f(a) in \scrB by
f(a) :=
1
2\pi i
\int
\delta
f(\xi )(\xi - a) - 1d\xi , (1.1)
where \delta \subset G is taken to be close rectifiable curve in G and such that \sigma (a) \subset \mathrm{i}\mathrm{n}\mathrm{s}(\delta ), the inside of \delta .
It is well-known (see, for instance, [6, p. 201 – 204]) that f(a) does not depend on the choice of
\delta and the Spectral Mapping Theorem (SMT)
\sigma (f(a)) = f(\sigma (a))
holds.
Let \frakH ol(a) be the set of all the functions that are analytic in a neighborhood of \sigma (a). Note that
\frakH ol(a) is an algebra where if f, g \in \frakH ol(a) and f and g have domains D(f) and D(g), then fg
and f + g have domain D(f) \cap D(g). \frakH ol(a) is not, however a Banach algebra.
The following result is known as the Riesz Functional Calculus Theorem [6, p. 201 – 203].
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1088 S. S. DRAGOMIR
Theorem 1. Let \scrB a unital Banach algebra and a \in \scrB .
(a) The map f \mapsto \rightarrow f(a) of \frakH ol(a) \rightarrow \scrB is an algebra homomorphism.
(b) If f(z) =
\sum \infty
k=0
\alpha kz
k has radius of convergence r > \nu (a), then f \in \frakH ol(a) and f(a) =
=
\sum \infty
k=0
\alpha ka
k.
(c) If f(z) \equiv 1, then f(a) = 1.
(d) If f(z) = z for all z, then f(a) = a.
(e) If f, f1, . . . , fn, . . . are analytic on G, \sigma (a) \subset G and fn(z) \rightarrow f(z) uniformly on compact
subsets of G, then
\bigm\| \bigm\| fn(a) - f(a)
\bigm\| \bigm\| \rightarrow 0 as n \rightarrow \infty .
(f) The Riesz Functional Calculus is unique and if a, b are commuting elements in \scrB and
f \in \frakH ol(a), then f(a)b = bf(a).
For some recent norm inequalities for functions on Banach algebras, see [4, 5, 9 – 15].
By using the analytic functional calculus in Banach algebra \scrB and function f \in \frakH ol(a), we
establish in this paper some norm error estimates in approximation the element f(a) by some simpler
expressions such as
(1 - \lambda )f(\alpha ) + \lambda f(\beta ) +
n\sum
k=1
1
k!
\Bigl[
(1 - \lambda )f (k)(\alpha )(a - \alpha )k + ( - 1)k\lambda f (k)(\beta )(\beta - a)k
\Bigr]
,
1
\beta - \alpha
[(\beta - a)f(\alpha ) + (a - \alpha )f(\beta )] +
(\beta - a)(a - \alpha )
\beta - \alpha
\times
\times
n\sum
k=1
1
k!
\Bigl\{
(a - \alpha )k - 1f (k)(\alpha ) + ( - 1)k(\beta - a)k - 1f (k)(\beta )
\Bigr\}
and
1
\beta - \alpha
[(a - \alpha )f(\alpha ) + (\beta - a)f(\beta )] +
+
1
\beta - \alpha
n\sum
k=1
1
k!
\Bigl\{
(a - \alpha )k+1f (k)(\alpha ) + ( - 1)k(\beta - a)k+1f (k)(\beta )
\Bigr\}
,
where \alpha , \beta \in D and \lambda \in \BbbC .
2. Scalar identities. Let f : D \subseteq \BbbC \rightarrow \BbbC be an analytic function on the convex domain D and
\xi , \alpha \in D. Then we have the following Taylor expansion with integral remainder:
f(\xi ) =
n\sum
k=0
1
k!
f (k)(\alpha )(\xi - \alpha )k +
+
1
n!
(\xi - \alpha )n+1
1\int
0
f (n+1)[(1 - s)\alpha + s\xi ](1 - s)nds (2.1)
for n \geq 0 (see, for instance, [24]).
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
TWO POINTS AND nTH DERIVATIVES NORM INEQUALITIES FOR ANALYTIC FUNCTIONS . . . 1089
Consider the function f(\xi ) = \mathrm{L}\mathrm{o}\mathrm{g}(\xi ), where \mathrm{L}\mathrm{o}\mathrm{g}(\xi ) = \mathrm{l}\mathrm{n}| \xi | + i\mathrm{A}\mathrm{r}\mathrm{g}(\xi ) and \mathrm{A}\mathrm{r}\mathrm{g}(\xi ) is such that
- \pi < \mathrm{A}\mathrm{r}\mathrm{g}(\xi ) \leq \pi . \mathrm{L}\mathrm{o}\mathrm{g} is called the “principal branch” of the complex logarithmic function. The
function f is analytic on all of \BbbC \ell := \BbbC \setminus \{ \alpha + i\beta : \alpha \leq 0, \beta = 0\} and
f (k)(\xi ) =
( - 1)k - 1(k - 1)!
\xi k
, k \geq 1, \xi \in \BbbC \ell .
By using the representation (2.1), we have
\mathrm{L}\mathrm{o}\mathrm{g}(\xi ) = \mathrm{L}\mathrm{o}\mathrm{g}(\alpha ) +
n\sum
k=0
( - 1)k - 1
k
\biggl(
\xi - \alpha
\alpha
\biggr) k
+
+ ( - 1)n(\xi - \alpha )n+1
1\int
0
(1 - s)nds
[(1 - s)\alpha + s\xi ]n+1
for all \xi , \alpha \in \BbbC \ell with (1 - s)\alpha + s\xi \in \BbbC \ell for s \in [0, 1].
Consider the complex exponential function f(\xi ) = \mathrm{e}\mathrm{x}\mathrm{p}(\xi ), then by (2.1) we get
\mathrm{e}\mathrm{x}\mathrm{p}(\xi ) =
n\sum
k=0
1
k!
(\xi - \alpha )k \mathrm{e}\mathrm{x}\mathrm{p}(\alpha ) +
+
1
n!
(\xi - \alpha )n+1
1\int
0
(1 - s)n \mathrm{e}\mathrm{x}\mathrm{p}[(1 - s)\alpha + s\xi ]ds
for all \xi , \alpha \in \BbbC .
For various inequalities related to Taylor’s expansions for real functions, see [1 – 3, 16 – 23].
Lemma 1. Let f : D \subseteq \BbbC \rightarrow \BbbC be an analytic function on the convex domain D and \xi , \alpha ,
\beta \in D. Then, for all \lambda \in \BbbC and n \geq 1, we have
f(\xi ) = (1 - \lambda )f(\alpha ) + \lambda f(\beta ) +
+
n\sum
k=1
1
k!
\Bigl[
(1 - \lambda )f (k)(\alpha )(\xi - \alpha )k + ( - 1)k\lambda f (k)(\beta )(\beta - \xi )k
\Bigr]
+ Sn,\lambda (\xi , \alpha , \beta ), (2.2)
where the remainder Sn,\lambda (\xi , \alpha , \beta ) is given by
Sn,\lambda (\xi , \alpha , \beta ) :=
1
n!
\left[ (1 - \lambda )(\xi - \alpha )n+1
1\int
0
f (n+1)[(1 - s)\alpha + s\xi ](1 - s)nds +
+ ( - 1)n+1\lambda (\beta - \xi )n+1
1\int
0
f (n+1)[(1 - s)\xi + s\beta ]sn ds
\right] . (2.3)
Proof. If we replace in (2.1) \alpha by \beta , then we get
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1090 S. S. DRAGOMIR
f(\xi ) =
n\sum
k=0
1
k!
f (k)(\alpha )(\xi - \beta )k +
+
1
n!
(\xi - \beta )n+1
1\int
0
f (n+1)[(1 - s)\beta + s\xi ](1 - s)nds =
=
n\sum
k=0
( - 1)k
k!
f (k)(\alpha )(\beta - \xi )k +
+
( - 1)n+1
n!
(\beta - \xi )n+1
1\int
0
f (n+1)[(1 - s)\beta + s\xi ](1 - s)nds =
=
n\sum
k=0
( - 1)k
k!
f (k)(\alpha )(\beta - \xi )k +
+
( - 1)n+1
n!
(\beta - \xi )n+1
1\int
0
f (n+1)[(1 - s)\xi + s\beta ]sn ds. (2.4)
Assume that \lambda \not = 1, 0. If we multiply (2.1) by 1 - \lambda and (2.4) by \lambda we get the desired representa-
tion (2.2) with the remainder Sn,\lambda (\xi , \alpha , \beta ) given by (2.3).
If either \lambda = 1 or \lambda = 0, then the theorem also holds by the use of Taylor’s usual expansion.
Lemma 1 is proved.
Remark 1. We observe that for n = 0 the representation from Lemma 1 becomes
f(\xi ) = (1 - \lambda )f(\alpha ) + \lambda f(\beta ) + S\lambda (\xi , \alpha , \beta ), (2.5)
where the remainder S\lambda (\xi , \alpha , \beta ) is given by
S\lambda (\xi , \alpha , \beta ) := (1 - \lambda )(\xi - \alpha )
1\int
0
f \prime ((1 - s)\alpha + s\xi )ds -
- \lambda (\beta - \xi )
1\int
0
f \prime ((1 - s)\xi + s\beta )ds.
Corollary 1. With the assumptions in Lemma 1 we have, for each distinct \xi , \alpha , \beta \in D with
\beta \not = \alpha ,
f(\xi ) =
1
\beta - \alpha
[(\beta - \xi )f(\alpha ) + (\xi - \alpha )f(\beta )] +
(\beta - \xi )(\xi - \alpha )
\beta - \alpha
\times
\times
n\sum
k=1
1
k!
\Bigl\{
(\xi - \alpha )k - 1f (k)(\alpha ) + ( - 1)k(\beta - \xi )k - 1f (k)(\beta )
\Bigr\}
+ Ln(\xi , \alpha , \beta ), (2.6)
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
TWO POINTS AND nTH DERIVATIVES NORM INEQUALITIES FOR ANALYTIC FUNCTIONS . . . 1091
where
Ln(\xi , \alpha , \beta ) :=
(\beta - \xi )(\xi - \alpha )
n!(\beta - \alpha )
\left[ (\xi - \alpha )n
1\int
0
f (n+1)((1 - s)\alpha + s\xi )(1 - s)nds+
+( - 1)n+1(\beta - \xi )n
1\int
0
f (n+1)((1 - s)\xi + s\beta )sn ds
\right]
and
f(\xi ) =
1
\beta - \alpha
\bigl[
(\xi - \alpha )f(\alpha ) + (\beta - \xi )f(\beta )
\bigr]
+
+
1
\beta - \alpha
n\sum
k=1
1
k!
\Bigl\{
(\xi - \alpha )k+1f (k)(\alpha ) + ( - 1)k(\beta - \xi )k+1f (k)(\beta )
\Bigr\}
+ Pn(\xi , \alpha , \beta ), (2.7)
where
Pn(\xi , \alpha , \beta ) :=
1
n!(\beta - \alpha )
\left[ (\xi - \alpha )n+2
1\int
0
f (n+1)((1 - s)\alpha + s\xi )(1 - s)nds +
+ ( - 1)n+1(\beta - \xi )n+2
1\int
0
f (n+1)((1 - s)\xi + s\beta )sn ds
\right] ,
respectively.
The proof is obvious, by choosing \lambda = (\xi - \alpha )/(\beta - \alpha ) and \lambda = (\beta - \xi )/(\beta - \alpha ), respectively,
in Lemma 1. The details are omitted.
The case n = 0 produces the following simple identities for each distinct \xi , \alpha , \beta \in D:
f(\xi ) =
1
\beta - \alpha
\bigl[
(\beta - \xi )f(\alpha ) + (\xi - \alpha )f(\beta )
\bigr]
+ L(\xi , \alpha , \beta ),
where
L(\xi , \alpha , \beta ) :=
(\beta - \xi )(\xi - \alpha )
\beta - \alpha
\left[ 1\int
0
f \prime ((1 - s)\alpha + s\xi )ds -
1\int
0
f \prime ((1 - s)\xi + s\beta )ds
\right]
and
f(\xi ) =
1
\beta - \alpha
[(\xi - \alpha )f(\alpha ) + (\beta - \xi )f(\beta )] + P (\xi , \alpha , \beta ),
where
P (\xi , \alpha , \beta ) :=
1
\beta - \alpha
\left[ (\xi - \alpha )2
1\int
0
f \prime ((1 - s)\alpha + s\xi )ds - (\beta - \xi )2
1\int
0
f \prime ((1 - s)\xi + s\beta )ds
\right] .
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1092 S. S. DRAGOMIR
3. Identities in Banach algebras. We have the following two point representation of an analytic
function on Banach algebras.
Theorem 2. Let \scrB be a unital Banach algebra, a \in \scrB , G be a convex domain of \BbbC with
\sigma (a) \subset G and \alpha , \beta \in G. If f : G \rightarrow \BbbC is analytic on G, then, for all \lambda \in \BbbC and n \geq 1, we have
f(a) = (1 - \lambda )f(\alpha ) + \lambda f(\beta ) +
+
n\sum
k=1
1
k!
\Bigl[
(1 - \lambda )f (k)(\alpha )(a - \alpha )k + ( - 1)k\lambda f (k)(\beta )(\beta - a)k
\Bigr]
+ Sn,\lambda (a, \alpha , \beta ), (3.1)
where the remainder Sn,\lambda (a, \alpha , \beta ) is given by
Sn,\lambda (a, \alpha , \beta ) :=
1
n!
\left[ (1 - \lambda )(a - \alpha )n+1
1\int
0
f (n+1)[(1 - s)\alpha + sa](1 - s)nds +
+ ( - 1)n+1\lambda (\beta - a)n+1
1\int
0
f (n+1)[(1 - s)a+ s\beta ]sn ds
\right] . (3.2)
In particular, for \lambda =
1
2
, we obtain the trapezoid type identity
f(a) =
f(\alpha ) + f(\beta )
2
+
+
1
2
n\sum
k=1
1
k!
\Bigl[
f (k)(\alpha )(a - \alpha )k + ( - 1)kf (k)(\beta )(\beta - a)k
\Bigr]
+ Tn(a, \alpha , \beta ), (3.3)
where the remainder Tn(a, \alpha , \beta ) is given by
Tn(a, \alpha , \beta ) :=
1
2n!
\left[ (a - \alpha )n+1
1\int
0
f (n+1)[(1 - s)\alpha + sa](1 - s)nds +
+ ( - 1)n+1(\beta - a)n+1
1\int
0
f (n+1)[(1 - s)a+ s\beta ]sn ds
\right] .
Proof. Assume that \delta \subset G is taken to be close rectifiable curve in G and such that \sigma (a) \subset
\subset \mathrm{i}\mathrm{n}\mathrm{s}(\delta ). By using the analytic functional calculus (1.1) and Lemma 1, we get
1
2\pi i
\int
\delta
f(\xi )(\xi - a) - 1d\xi = [(1 - \lambda )f(\alpha ) + \lambda f(\beta )]
1
2\pi i
\int
\delta
(\xi - a) - 1d\xi +
+
n\sum
k=1
1
k!
(1 - \lambda )f (k)(\alpha )
1
2\pi i
\int
\delta
(\xi - \alpha )k(\xi - a) - 1d\xi +
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TWO POINTS AND nTH DERIVATIVES NORM INEQUALITIES FOR ANALYTIC FUNCTIONS . . . 1093
+
n\sum
k=1
1
k!
( - 1)k\lambda f (k)(\beta )
1
2\pi i
\int
\delta
(\beta - \xi )k(\xi - a) - 1d\xi +
+
1
n!
(1 - \lambda )
1
2\pi i
\int
\delta
(\xi - \alpha )n+1
\left( 1\int
0
f (n+1)[(1 - s)\alpha + s\xi ](1 - s)nds
\right) (\xi - a) - 1d\xi +
+
1
n!
( - 1)n+1\lambda
1
2\pi i
\int
\delta
(\beta - \xi )n+1
\left( 1\int
0
f (n+1)[(1 - s)\xi + s\beta ]sn ds
\right) (\xi - a) - 1d\xi =
=
1
n!
(1 - \lambda )
1\int
0
\left( 1
2\pi i
\int
\delta
(\xi - \alpha )n+1f (n+1)[(1 - s)\alpha + s\xi ](\xi - a) - 1d\xi
\right) (1 - s)nds +
+
1
n!
( - 1)n+1\lambda
1\int
0
\left( 1
2\pi i
\int
\delta
(\beta - \xi )n+1f (n+1)[(1 - s)\xi + s\beta ](\xi - a) - 1d\xi
\right) sn ds, (3.4)
where for the last equality we used Fubini’s theorem.
By using the functional calculus for the analytic functions
G \ni \xi \mapsto \rightarrow (\xi - \alpha )n+1f (n+1)[(1 - s)\alpha + s\xi ] \in \BbbC
and
G \ni \xi \mapsto \rightarrow (\beta - \xi )n+1f (n+1)[(1 - s)\xi + s\beta ] \in \BbbC ,
we have
1
2\pi i
\int
\delta
(\xi - \alpha )n+1f (n+1)[(1 - s)\alpha + s\xi ](\xi - a) - 1d\xi =
= (a - \alpha )n+1f (n+1)[(1 - s)\alpha + sa]
and
1
2\pi i
\int
\delta
(\beta - \xi )n+1f (n+1)[(1 - s)\xi + s\beta ](\xi - a) - 1d\xi =
= (\beta - a)n+1f (n+1)[(1 - s)a+ s\beta ].
Therefore,
1\int
0
\left( 1
2\pi i
\int
\delta
(\xi - \alpha )n+1f (n+1)[(1 - s)\alpha + s\xi ](\xi - a) - 1d\xi
\right) (1 - s)nds =
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1094 S. S. DRAGOMIR
=
1\int
0
(a - \alpha )n+1f (n+1)[(1 - s)\alpha + sa](1 - s)nds =
= (a - \alpha )n+1
1\int
0
f (n+1)[(1 - s)\alpha + sa](1 - s)nds
and
1\int
0
\left( 1
2\pi i
\int
\delta
(\beta - \xi )n+1f (n+1)[(1 - s)\xi + s\beta ](\xi - a) - 1d\xi
\right) sn ds =
=
1\int
0
(\beta - a)n+1f (n+1)[(1 - s)a+ s\beta ]sn ds =
= (\beta - a)n+1
1\int
0
f (n+1)[(1 - s)a+ s\beta ]sn ds.
Since
1
2\pi i
\int
\delta
(\xi - a) - 1d\xi = 1,
1
2\pi i
\int
\delta
(\xi - \alpha )k(\xi - a) - 1d\xi = (a - \alpha )k
and
1
2\pi i
\int
\delta
(\beta - \xi )k(\xi - a) - 1d\xi = (\beta - a)k
for k = 1, . . . , n, hence by (3.4) we get the representation (3.1) with the remainder (3.2).
Theorem 2 is proved.
Remark 2. Withe the assumptions from Theorem 2 and by using the scalar identity (2.5) we
have, for n = 0, that
f(a) = (1 - \lambda )f(\alpha ) + \lambda f(\beta ) + S\lambda (a, \alpha , \beta ), (3.5)
where the remainder S\lambda (a, \alpha , \beta ) is given by
S\lambda (a, \alpha , \beta ) := (1 - \lambda )(a - \alpha )
1\int
0
f \prime ((1 - s)\alpha + sa)ds -
- \lambda (\beta - a)
1\int
0
f \prime \bigl( (1 - s)a+ s\beta
\bigr)
ds.
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TWO POINTS AND nTH DERIVATIVES NORM INEQUALITIES FOR ANALYTIC FUNCTIONS . . . 1095
In particular, we have
f(a) =
f(\alpha ) + f(\beta )
2
+ T (a, \alpha , \beta ), (3.6)
where the remainder T (a, \alpha , \beta ) is given by
T (a, \alpha , \beta ) :=
1
2
\left[ (a - \alpha )
1\int
0
f \prime \bigl( (1 - s)\alpha + sa
\bigr)
ds - (\beta - a)
1\int
0
f \prime \bigl( (1 - s)a+ s\beta
\bigr)
ds
\right] .
We also have the following theorem.
Theorem 3. Let \scrB be a unital Banach algebra, a \in \scrB , G be a convex domain of \BbbC with
\sigma (a) \subset G and \alpha , \beta \in G with \alpha \not = \beta . If f : G \rightarrow \BbbC is analytic on G, then
f(a) =
1
\beta - \alpha
\bigl[
f(\alpha )(\beta - a) + f(\beta )(a - \alpha )
\bigr]
+
(\beta - a)(a - \alpha )
\beta - \alpha
\times
\times
n\sum
k=1
1
k!
\Bigl\{
f (k)(\alpha )(a - \alpha )k - 1 + ( - 1)kf (k)(\beta )(\beta - a)k - 1
\Bigr\}
+ Ln(a, \alpha , \beta ), (3.7)
where
Ln(a, \alpha , \beta ) :=
(\beta - a)(a - \alpha )
n!(\beta - \alpha )
\left[ (a - \alpha )n
1\int
0
f (n+1)((1 - s)\alpha + sa)(1 - s)nds +
+ ( - 1)n+1(\beta - a)n
1\int
0
f (n+1)((1 - s)a+ s\beta )sn ds
\right]
and
f(a) =
1
\beta - \alpha
\bigl[
f(\alpha )(a - \alpha ) + f(\beta )(\beta - a)
\bigr]
+
+
1
\beta - \alpha
n\sum
k=1
1
k!
\Bigl\{
f (k)(\alpha )(a - \alpha )k+1 + ( - 1)kf (k)(\beta )(\beta - a)k+1
\Bigr\}
+ Pn(a, \alpha , \beta ), (3.8)
where
Pn(a, \alpha , \beta ) :=
1
n!(\beta - \alpha )
\left[ (a - \alpha )n+2
1\int
0
f (n+1)((1 - s)\alpha + sa)(1 - s)nds+
+( - 1)n+1(\beta - a)n+2
1\int
0
f (n+1)((1 - s)a+ s\beta )sn ds
\right] .
The proof follows in a similar way to the one from Theorem 2 by utilising the functional calculus
for analytic functions (1.1) and the scalar identities (2.6) and (2.7).
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1096 S. S. DRAGOMIR
The case n = 0 produces the following simple identities for each distinct \alpha , \beta \in G:
f(a) =
1
\beta - \alpha
\bigl[
f(\alpha )(\beta - a) + f(\beta )(a - \alpha )
\bigr]
+ L(a, \alpha , \beta ),
where
L(a, \alpha , \beta ) :=
(\beta - a)(a - \alpha )
\beta - \alpha
\left[ 1\int
0
f \prime ((1 - s)\alpha + sa)ds -
1\int
0
f \prime \bigl( (1 - s)a+ s\beta
\bigr)
ds
\right] ,
and
f(a) =
1
\beta - \alpha
\bigl[
f(\alpha )(a - \alpha ) + f(\beta )(\beta - a)
\bigr]
+ P (a, \alpha , \beta ),
where
P (a, \alpha , \beta ) :=
1
\beta - \alpha
\left[ (a - \alpha )2
1\int
0
f \prime ((1 - s)\alpha + sa)ds - (\beta - a)2
1\int
0
f \prime \bigl( (1 - s)a+ s\beta
\bigr)
ds
\right] .
4. Norm inequalities. The following result providing norm error estimates, holds.
Theorem 4. Let \scrB be a unital Banach algebra, a \in \scrB , G be a convex domain of \BbbC with
\sigma (a) \subset G and \alpha , \beta \in G. If f : G \rightarrow \BbbC is analytic on G, then, for all \lambda \in \BbbC and n \geq 1, we have
the representation (2.3) and the remainder Sn,\lambda (a, \alpha , \beta ) satisfies the norm inequalities
\| Sn,\lambda (a, \alpha , \beta )\| \leq 1
n!
\left[ | 1 - \lambda | \| a - \alpha \| n+1
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| (1 - s)nds +
+ | \lambda | \| \beta - a\| n+1
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| sn ds
\right] \leq
\leq 1
n!
| 1 - \lambda | \| a - \alpha \| n+1
\left\{
1
n+ 1
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| ,
1
(nq + 1)1/q
\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| ds
+
+
1
n!
| \lambda | \| \beta - a\| n+1
\left\{
1
n+ 1
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| ,
1
(nq + 1)1/q
\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| ds.
(4.1)
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TWO POINTS AND nTH DERIVATIVES NORM INEQUALITIES FOR ANALYTIC FUNCTIONS . . . 1097
In particular, we get the representation (3.3) and the remainder satisfies the norm inequalities
\| Tn(a, \alpha , \beta )\| \leq 1
2n!
\left[ \| a - \alpha \| n+1
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| (1 - s)nds +
+ \| \beta - a\| n+1
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| sn ds
\right] \leq
\leq 1
2n!
\| a - \alpha \| n+1
\left\{
1
n+ 1
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| ,
1
(nq + 1)1/q
\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| ds
+
+
1
2n!
\| \beta - a\| n+1
\left\{
1
n+ 1
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| ,
1
(nq + 1)1/q
\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| ds.
Proof. By using the representation (2.4), we obtain
\| Sn,\lambda (a, \alpha , \beta )\| \leq 1
n!
\left[ | 1 - \lambda |
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| (a - \alpha )n+1
1\int
0
f (n+1)[(1 - s)\alpha + sa](1 - s)nds
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| +
+ | \lambda |
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| (\beta - a)n+1
1\int
0
f (n+1)[(1 - s)a+ s\beta ]sn ds
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\|
\right] \leq
\leq 1
n!
\left[ | 1 - \lambda |
\bigm\| \bigm\| (a - \alpha )n+1
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\|
1\int
0
f (n+1)[(1 - s)\alpha + sa](1 - s)nds
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| +
+ | \lambda |
\bigm\| \bigm\| (\beta - a)n+1
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\|
1\int
0
f (n+1)[(1 - s)a+ s\beta ]sn ds
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\|
\right] \leq
\leq 1
n!
\left[ | 1 - \lambda | \| a - \alpha \| n+1
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| (1 - s)nds +
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1098 S. S. DRAGOMIR
+ | \lambda | \| \beta - a\| n+1
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| sn ds
\right] =: A. (4.2)
This proves the first inequality in (4.1).
By using Hölder’s integral inequality, we have
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| (1 - s)nds \leq
\leq
\left\{
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \int 1
0
(1 - s)nds,\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p\biggl( \int 1
0
(1 - s)qnds
\biggr) 1/q
for p, q > 1, where
1
p
+
1
q
= 1,
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1](1 - s)n
\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| ds
=
=
\left\{
1
n+ 1
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| ,
1
(nq + 1)1/q
\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| ds.
Similarly,
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| sn ds \leq
\leq
\left\{
1
n+ 1
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| ,
1
(nq + 1)1/q
\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| ds.
Therefore,
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TWO POINTS AND nTH DERIVATIVES NORM INEQUALITIES FOR ANALYTIC FUNCTIONS . . . 1099
A \leq 1
n!
| 1 - \lambda | \| a - \alpha \| n+1
\left\{
1
n+ 1
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| ,
1
(nq + 1)1/q
\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| ds
+
+
1
n!
| \lambda | \| \beta - a\| n+1
\left\{
1
n+ 1
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| ,
1
(nq + 1)1/q
\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| ds.
By using (4.2) we get the second part of (4.1).
Theorem 4 is proved.
Remark 3. In the case n = 0 we have the representations (3.5) and (3.6) and the remainders
S\lambda (a, \alpha , \beta ) and T (a, \alpha , \beta ) satisfy the bounds
\| S(a, \alpha , \beta )\| \leq | 1 - \lambda | \| a - \alpha \|
1\int
0
\bigm\| \bigm\| f \prime [(1 - s)\alpha + sa]
\bigm\| \bigm\| ds +
+ | \lambda | \| \beta - a\|
1\int
0
\bigm\| \bigm\| f \prime [(1 - s)a+ s\beta ]
\bigm\| \bigm\| ds \leq
\leq | 1 - \lambda | \| a - \alpha \|
\left\{
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]\| f \prime [(1 - s)\alpha + sa]\| ,\biggl( \int 1
0
\bigm\| \bigm\| f \prime [(1 - s)\alpha + sa]
\bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,
+
+ | \lambda | \| \beta - a\|
\left\{
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]\| f \prime [(1 - s)a+ s\beta ]\| ,\biggl( \int 1
0
\| f [(1 - s)a+ s\beta ]\| pds
\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,
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1100 S. S. DRAGOMIR
and
\| T (a, \alpha , \beta )\| \leq 1
2
\left[ \| a - \alpha \|
1\int
0
\| f [(1 - s)\alpha + sa]\| ds +
+ \| \beta - a\|
1\int
0
\bigm\| \bigm\| f \prime [(1 - s)a+ s\beta ]
\bigm\| \bigm\| ds
\right] \leq
\leq 1
2
\| a - \alpha \|
\left\{
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]\| f \prime [(1 - s)\alpha + sa]\| ,\biggl( \int 1
0
\bigm\| \bigm\| f \prime [(1 - s)\alpha + sa]
\bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,
+
+
1
2
\| \beta - a\|
\left\{
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]\| f \prime [(1 - s)a+ s\beta ]\| ,\biggl( \int 1
0
\bigm\| \bigm\| f \prime [(1 - s)a+ s\beta ]
\bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1.
Corollary 2. With the assumptions of Theorem 4 we have
\| Sn,\lambda (a, \alpha , \beta )\| \leq 1
(n+ 1)!
\bigl[
| 1 - \lambda | \| a - \alpha \| n+1 + | \lambda | \| \beta - a\| n+1
\bigr]
\times
\times \mathrm{m}\mathrm{a}\mathrm{x}
\Biggl\{
\mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| , \mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| \Biggr\}
and, in particular,
\| Tn(a, \alpha , \beta )\| \leq 1
2(n+ 1)!
\bigl[
\| a - \alpha \| n+1 + \| \beta - a\| n+1
\bigr]
\times
\times \mathrm{m}\mathrm{a}\mathrm{x}
\Biggl\{
\mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| , \mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| \Biggr\} .
We have the following theorem.
Theorem 5. Let \scrB be a unital Banach algebra, a \in \scrB , G be a convex domain of \BbbC with
\sigma (a) \subset G and \alpha , \beta \in G with \alpha \not = \beta . If f : G \rightarrow \BbbC is analytic on G, then, for n \geq 1, we have
the representations (3.7) and (3.8) and the remainders Ln(a, \alpha , \beta ) and Pn(a, \alpha , \beta ) satisfy the norm
inequalities
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TWO POINTS AND nTH DERIVATIVES NORM INEQUALITIES FOR ANALYTIC FUNCTIONS . . . 1101
\bigm\| \bigm\| Ln(a, \alpha , \beta )
\bigm\| \bigm\| \leq 1
n!| \beta - \alpha |
\bigm\| \bigm\| (\beta - a)(a - \alpha )
\bigm\| \bigm\| \times
\times
\left[ \| a - \alpha \| n
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)
\bigl(
(1 - s)\alpha + sa
\bigr) \bigm\| \bigm\| \bigm\| (1 - s)nds+
+\| \beta - a\| n
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)
\bigl(
(1 - s)a+ s\beta
\bigr) \bigm\| \bigm\| \bigm\| sn ds
\right] \leq
\leq 1
n!| \beta - \alpha |
\bigm\| \bigm\| (\beta - a)(a - \alpha )
\bigm\| \bigm\| \times
\times
\left[
\| a - \alpha \| n
\left\{
1
n+ 1
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| ,
1
(nq + 1)1/q
\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,
\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| ds
+
+\| \beta - a\| n
\left\{
1
n+ 1
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| ,
1
(nq + 1)1/q
\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| ds
\right]
(4.3)
and
\bigm\| \bigm\| Pn(a, \alpha , \beta )
\bigm\| \bigm\| \leq 1
n!| \beta - \alpha |
\left[ \| a - \alpha \| n+2
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)((1 - s)\alpha + sa)
\bigm\| \bigm\| \bigm\| (1 - s)nds+
+\| \beta - a\| n+2
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)
\bigl(
(1 - s)a+ s\beta
\bigr) \bigm\| \bigm\| \bigm\| sn ds
\right] \leq
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
1102 S. S. DRAGOMIR
\leq 1
n!| \beta - \alpha |
\left[
\| a - \alpha \| n+2
\left\{
1
n+ 1
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| ,
1
(nq + 1)1/q
\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| ds
+
+ \| \beta - a\| n+2
\left\{
1
n+ 1
\mathrm{s}\mathrm{u}\mathrm{p}s\in [0,1]
\bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| ,
1
(nq + 1)1/q
\biggl( \int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| pds\biggr) 1/p
for p, q > 1, where
1
p
+
1
q
= 1,\int 1
0
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| ds
\right]
. (4.4)
Proof. From Theorem 3 we have\bigm\| \bigm\| Ln(a, \alpha , \beta )
\bigm\| \bigm\| \leq 1
n!| \beta - \alpha |
\bigm\| \bigm\| (\beta - a)(a - \alpha )
\bigm\| \bigm\| \times
\times
\left[ \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| (a - \alpha )n
1\int
0
f (n+1)((1 - s)\alpha + sa)(1 - s)nds
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| +
+
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| (\beta - a)n
1\int
0
f (n+1)
\bigl(
(1 - s)a+ s\beta
\bigr)
sn ds
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\|
\right] \leq
\leq 1
n!| \beta - \alpha |
\bigm\| \bigm\| (\beta - a)(a - \alpha )
\bigm\| \bigm\| \left[ \| (a - \alpha )n\|
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\|
1\int
0
f (n+1)((1 - s)\alpha + sa)(1 - s)nds
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| +
+\| (\beta - a)n\|
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\|
1\int
0
f (n+1)
\bigl(
(1 - s)a+ s\beta
\bigr)
sn ds
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\|
\right] \leq
\leq 1
n!| \beta - \alpha |
\bigm\| \bigm\| (\beta - a)(a - \alpha )
\bigm\| \bigm\| \left[ \| a - \alpha \| n
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)
\bigl(
(1 - s)\alpha + sa
\bigr) \bigm\| \bigm\| \bigm\| (1 - s)nds+
+\| \beta - a\| n
1\int
0
\bigm\| \bigm\| \bigm\| f (n+1)
\bigl(
(1 - s)a+ s\beta
\bigr) \bigm\| \bigm\| \bigm\| sn ds
\right] =: B,
which proves the first inequality in (4.3). The second part follows by Hölder’s integral inequality.
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TWO POINTS AND nTH DERIVATIVES NORM INEQUALITIES FOR ANALYTIC FUNCTIONS . . . 1103
The inequality (4.4) can be proved in a similar way.
Theorem 5 is proved.
Remark 4. In the case n = 0 we get
f(a) =
1
\beta - \alpha
[f(\alpha )(\beta - a) + f(\beta )(a - \alpha )] + L(a, \alpha , \beta ),
where \bigm\| \bigm\| L(a, \alpha , \beta )\bigm\| \bigm\| \leq 1
| \beta - \alpha |
\bigm\| \bigm\| (\beta - a)(a - \alpha )
\bigm\| \bigm\| \times
\times
1\int
0
\bigm\| \bigm\| f \prime ((1 - s)\alpha + sa)ds - f \prime (sa+ (1 - s)\beta )
\bigm\| \bigm\| ds (4.5)
and
f(a) =
1
\beta - \alpha
\bigl[
f(\alpha )(a - \alpha ) + f(\beta )(\beta - a)
\bigr]
+ P (a, \alpha , \beta ),
where \bigm\| \bigm\| P (a, \alpha , \beta )
\bigm\| \bigm\| \leq 1
| \beta - \alpha |
\times
\times
\left[ \| a - \alpha \| 2
1\int
0
\bigm\| \bigm\| f \prime ((1 - s)\alpha + sa)
\bigm\| \bigm\| ds+ \| \beta - a\| 2
1\int
0
\bigm\| \bigm\| f \prime \bigl( (1 - s)a+ s\beta
\bigr) \bigm\| \bigm\| ds
\right] .
Moreover, if there exist La > 0 such that\bigm\| \bigm\| f \prime ((1 - s)\alpha + sa)ds - f \prime (sa+ (1 - s)\beta )
\bigm\| \bigm\| \leq (1 - \alpha )La| \alpha - \beta |
for all s \in [0, 1], then by (4.5) we get\bigm\| \bigm\| L(a, \alpha , \beta )\bigm\| \bigm\| \leq 1
2
La
\bigm\| \bigm\| (\beta - a)(a - \alpha )
\bigm\| \bigm\| .
5. Examples for exponential function. Let \scrB be a unital Banach algebra, a \in \scrB . Consider the
exponential function f(z) = \mathrm{e}\mathrm{x}\mathrm{p}(z), z \in \BbbC and put
Ea,z := \mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \mathrm{e}\mathrm{x}\mathrm{p}[(1 - s)a+ sz]
\bigm\| \bigm\| < \infty , n \geq 0.
Observe that
\mathrm{e}\mathrm{x}\mathrm{p}((1 - t)\lambda + ta) = \mathrm{e}\mathrm{x}\mathrm{p}[(1 - t)\lambda ] \mathrm{e}\mathrm{x}\mathrm{p}(ta),
which gives \bigm\| \bigm\| \mathrm{e}\mathrm{x}\mathrm{p}((1 - t)\lambda + ta)
\bigm\| \bigm\| = | \mathrm{e}\mathrm{x}\mathrm{p}[(1 - t)\lambda ]| \| \mathrm{e}\mathrm{x}\mathrm{p}(ta)\| =
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1104 S. S. DRAGOMIR
= \mathrm{e}\mathrm{x}\mathrm{p}[(1 - t)\mathrm{R}\mathrm{e}\lambda ]\| \mathrm{e}\mathrm{x}\mathrm{p}(ta)\| \leq
\leq \mathrm{e}\mathrm{x}\mathrm{p}[(1 - t)\mathrm{R}\mathrm{e}\lambda ] \mathrm{e}\mathrm{x}\mathrm{p}(t\| a\| ) = \mathrm{e}\mathrm{x}\mathrm{p}
\bigl[
(1 - t)\mathrm{R}\mathrm{e}\lambda + t\| a\|
\bigr]
\leq
\leq \mathrm{e}\mathrm{x}\mathrm{p}(\mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{R}\mathrm{e}\lambda , \| a\| \} )
for any t \in [0, 1], \lambda \in \BbbC .
Therefore,
Ea,z \leq \mathrm{e}\mathrm{x}\mathrm{p}
\bigl(
\mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{R}\mathrm{e} z, \| a\| \}
\bigr)
.
Let \scrB be a unital Banach algebra, a \in \scrB , G be a convex domain of \BbbC with \sigma (a) \subset G and \alpha ,
\beta \in G. If f : G \rightarrow \BbbC is analytic on G, then by the inequality (4.1) we have
\| Tn(a, \alpha , \beta )\| \leq 1
2(n+ 1)!
\| a - \alpha \| n+1 \mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| +
+
1
2(n+ 1)!
\| \beta - a\| n+1 \mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| . (5.1)
If we apply the inequality (5.1) for the exponential function, then we get the norm inequality\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \mathrm{e}\mathrm{x}\mathrm{p} a - \mathrm{e}\mathrm{x}\mathrm{p}\alpha + \mathrm{e}\mathrm{x}\mathrm{p}\beta
2
- 1
2
n\sum
k=1
1
k!
\Bigl[
\mathrm{e}\mathrm{x}\mathrm{p}(\alpha )(a - \alpha )k + ( - 1)k \mathrm{e}\mathrm{x}\mathrm{p}(\beta )(\beta - a)k
\Bigr] \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \leq
\leq 1
2(n+ 1)!
\Bigl[
\| a - \alpha \| n+1 \mathrm{e}\mathrm{x}\mathrm{p}(\mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{R}\mathrm{e}\alpha , \| a\| \} ) +
+ \| \beta - a\| n+1 \mathrm{e}\mathrm{x}\mathrm{p}
\bigl(
\mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{R}\mathrm{e}\beta , \| a\| \}
\bigr) \Bigr]
\leq
\leq 1
2(n+ 1)!
\bigl[
\| a - \alpha \| n+1 + \| \beta - a\| n+1
\bigr]
\mathrm{e}\mathrm{x}\mathrm{p}(\mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{R}\mathrm{e}\alpha ,\mathrm{R}\mathrm{e}\beta , \| a\| \} ).
By using the inequality (4.3), we have, for \alpha \not = \beta , that\bigm\| \bigm\| Ln(a, \alpha , \beta )
\bigm\| \bigm\| \leq 1
(n+ 1)!| \beta - \alpha |
\bigm\| \bigm\| (\beta - a)(a - \alpha )
\bigm\| \bigm\| \times
\times
\Biggl[
\| a - \alpha \| n \mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| +
+\| \beta - a\| n \mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| \Biggr] . (5.2)
If we apply the inequality (5.2) for the exponential function, we obtain\bigm\| \bigm\| \bigm\| \bigm\| \mathrm{e}\mathrm{x}\mathrm{p} a - 1
\beta - \alpha
[\mathrm{e}\mathrm{x}\mathrm{p}(\alpha )(\beta - a) + \mathrm{e}\mathrm{x}\mathrm{p}(\beta )(a - \alpha )] -
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TWO POINTS AND nTH DERIVATIVES NORM INEQUALITIES FOR ANALYTIC FUNCTIONS . . . 1105
- (\beta - a)(a - \alpha )
\beta - \alpha
n\sum
k=1
1
k!
\Bigl\{
\mathrm{e}\mathrm{x}\mathrm{p}(\alpha )(a - \alpha )k - 1 + ( - 1)k \mathrm{e}\mathrm{x}\mathrm{p}(\beta )(\beta - a)k - 1
\Bigr\} \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \leq
\leq 1
(n+ 1)!| \beta - \alpha |
\bigm\| \bigm\| (\beta - a)(a - \alpha )
\bigm\| \bigm\| \times
\times
\bigl[
\| a - \alpha \| n \mathrm{e}\mathrm{x}\mathrm{p}(\mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{R}\mathrm{e}\alpha , \| a\| \} ) + \| \beta - a\| n \mathrm{e}\mathrm{x}\mathrm{p}
\bigl(
\mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{R}\mathrm{e}\beta , \| a\| \}
\bigr) \bigr]
\leq
\leq 1
(n+ 1)!| \beta - \alpha |
\bigm\| \bigm\| (\beta - a)(a - \alpha )
\bigm\| \bigm\| \times
\times
\bigl[
\| a - \alpha \| n + \| \beta - a\| n
\bigr]
\mathrm{e}\mathrm{x}\mathrm{p}(\mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{R}\mathrm{e}\alpha ,\mathrm{R}\mathrm{e}\beta , \| a\| \} ).
By using the inequality (4.4), we get\bigm\| \bigm\| Pn(a, \alpha , \beta )
\bigm\| \bigm\| \leq 1
(n+ 1)!| \beta - \alpha |
\times
\times
\biggl[
\| a - \alpha \| n+2 \mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)\alpha + sa]
\bigm\| \bigm\| \bigm\| +
+ \| \beta - a\| n+2 \mathrm{s}\mathrm{u}\mathrm{p}
s\in [0,1]
\bigm\| \bigm\| \bigm\| f (n+1)[(1 - s)a+ s\beta ]
\bigm\| \bigm\| \bigm\| \biggr] (5.3)
for \alpha \not = \beta .
By writing this inequality (5.3) for the exponential function, we have\bigm\| \bigm\| \bigm\| \bigm\| \mathrm{e}\mathrm{x}\mathrm{p}(a) - 1
\beta - \alpha
n\sum
k=0
1
k!
\Bigl\{
\mathrm{e}\mathrm{x}\mathrm{p}(\alpha )(a - \alpha )k+1 + ( - 1)k \mathrm{e}\mathrm{x}\mathrm{p}(\beta )(\beta - a)k+1
\Bigr\} \bigm\| \bigm\| \bigm\| \bigm\| \leq
\leq 1
(n+ 1)!| \beta - \alpha |
\Bigl[
\| a - \alpha \| n+2 \mathrm{e}\mathrm{x}\mathrm{p}(\mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{R}\mathrm{e}\alpha , \| a\| \} ) +
+ \| \beta - a\| n+2 \mathrm{e}\mathrm{x}\mathrm{p}
\bigl(
\mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{R}\mathrm{e}\beta , \| a\| \}
\bigr) \Bigr]
\leq
\leq 1
(n+ 1)!| \beta - \alpha |
\bigl[
\| a - \alpha \| n+2 + \| \beta - a\| n+2
\bigr]
\mathrm{e}\mathrm{x}\mathrm{p}(\mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{R}\mathrm{e}\alpha ,\mathrm{R}\mathrm{e}\beta , \| a\| \} ).
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Received 11.05.20
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 8
|
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| resource_txt_mv | umjimathkievua/be/86cf6bbe16370e90b47ba3a92421d5be.pdf |
| spelling | umjimathkievua-article-61162022-10-24T17:55:16Z Two points and $n$ th derivatives norm inequalities for analytic functions in Banach algebras Two points and $n$ th derivatives norm inequalities for analytic functions in Banach algebras Dragomir, S. S. Dragomir, Silvestru Sever Dragomir, S. S. Banach algebras, Ostrowski inequality, Norm inequalities, Analytic functional calculus. UDC 517.5 Let $\mathcal{B}$ be a unital Banach algebra, let $a \in \mathcal{B},$ $G$ be a convex domain of $\mathbb{C}$ with $\sigma (a) \subset G,$ let $\alpha, \beta \in G,$ and let $f \colon G \rightarrow \mathbb{C}$ be analytic on $G.$By using the analytic functional calculus, we obtain among others the following result:\begin{gather*}\left\| f(a) - \frac{1}{2}\sum_{k=0}^{n}\frac{1}{k!}\left[f^{(k) }(\alpha) (a-\alpha)^{k}+(-1)^{k}f^{(k) }(\beta) (\beta-a)^{k}\right] \right\|\leq\\\leq \frac{1}{2(n+1) !}\left[\|a-\alpha\|^{n+1}+\|\beta - a\|^{n+1}\right]\times\\\times \max \left\{\sup_{s\in [0,1] }\left\| f^{(n+1) }[(1-s) \alpha+sa] \right\|,\sup_{s\in [0,1] }\left\| f^{(n+1) }[(1-s) a+s\beta] \right\| \right\}.\end{gather*}Some examples for the exponential function on Banach algebras are also given. УДК 517.5нерiвностi для норми двох точок i $n$ -ї похiдної для аналiтичних функцiй у банахових алгебрах Нехай $\mathcal{B}$ — унiтальна алгебра Банаха, $a \in \mathcal{B},$ $G$ — опукла область в $\mathbb{C}$ з $\sigma (a) \subset G,$, $\alpha, \beta \in G,$, а $f \colon G \rightarrow \mathbb{C}$є аналiтичною на $G$. Використовуючи аналiтичне функцiональне числення, ми отримуємо серед iнших такий результат:\begin{gather*}\left\| f(a) - \frac{1}{2}\sum_{k=0}^{n}\frac{1}{k!}\left[f^{(k) }(\alpha) (a-\alpha)^{k}+(-1)^{k}f^{(k) }(\beta) (\beta-a)^{k}\right] \right\|\leq\\\leq \frac{1}{2(n+1) !}\left[\|a-\alpha\|^{n+1}+\|\beta - a\|^{n+1}\right]\times\\\times \max \left\{\sup_{s\in [0,1] }\left\| f^{(n+1) }[(1-s) \alpha+sa] \right\|,\sup_{s\in [0,1] }\left\| f^{(n+1) }[(1-s) a+s\beta] \right\| \right\}.\end{gather*}Наведено також деякi приклади для експоненцiальних функцiй на алгебрах Банаха Institute of Mathematics, NAS of Ukraine 2022-10-04 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/6116 10.37863/umzh.v74i8.6116 Ukrains’kyi Matematychnyi Zhurnal; Vol. 74 No. 8 (2022); 1086 - 1106 Український математичний журнал; Том 74 № 8 (2022); 1086 - 1106 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/6116/9288 Copyright (c) 2022 Silvestru Sever Dragomir |
| spellingShingle | Dragomir, S. S. Dragomir, Silvestru Sever Dragomir, S. S. Two points and $n$ th derivatives norm inequalities for analytic functions in Banach algebras |
| title | Two points and $n$ th derivatives norm inequalities for analytic functions in Banach algebras |
| title_alt | Two points and $n$ th derivatives norm inequalities for analytic functions in Banach algebras |
| title_full | Two points and $n$ th derivatives norm inequalities for analytic functions in Banach algebras |
| title_fullStr | Two points and $n$ th derivatives norm inequalities for analytic functions in Banach algebras |
| title_full_unstemmed | Two points and $n$ th derivatives norm inequalities for analytic functions in Banach algebras |
| title_short | Two points and $n$ th derivatives norm inequalities for analytic functions in Banach algebras |
| title_sort | two points and $n$ th derivatives norm inequalities for analytic functions in banach algebras |
| topic_facet | Banach algebras Ostrowski inequality Norm inequalities Analytic functional calculus. |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/6116 |
| work_keys_str_mv | AT dragomirss twopointsandnthderivativesnorminequalitiesforanalyticfunctionsinbanachalgebras AT dragomirsilvestrusever twopointsandnthderivativesnorminequalitiesforanalyticfunctionsinbanachalgebras AT dragomirss twopointsandnthderivativesnorminequalitiesforanalyticfunctionsinbanachalgebras |