Abstract
In the paper there are two kinds of functional symmetry considered, i.e., (1)-power symmetry and \((-1)\)-power symmetry, in a rich subfamily \(\mathcal {B}\) of the family of all complex valued functions on a symmetric set \(\mathcal {G}\subset \mathbb {C}\). The first one defines the values \(f(-z)\) as identical with f(z) and the other as the inverse of the f(z) in \(\mathcal {G}\). The announced notions allow a unique decomposition of functions \(f\in \mathcal {B}\) into a product of two factors \(f_{1},f_{-1}\) having the (1)- and the \((-1)\)-power symmetry property, respectively. In the paper there are also given examples of such partitions \(f=f_{1}\cdot f_{-1}\), for various \(f\in \mathcal {B},\) a solution of the problem of invariance of the above functional symmetries with respect to some one- and two-argument function operations and two applications of the mentioned function decomposition.
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1 Introduction
It is known that for every complex valued function f on a symmetric set \(\mathcal {G}\) \(\subset \mathbb {C}\) \((-1\mathcal {G}=\mathcal {G})\), there exists exactly one function \(f_{E}\) from the class \(\mathcal {F}_{E}(\mathcal {G})\) of even functions and exactly one function \(f_{O}\) from the class \(\mathcal {F} _{O}(\mathcal {G})\) of odd functions on G, such that
Moreover,
In view of the uniqueness of this partition, the functions \(f_{E}\) and \(f_{O} \) are called the even and odd part of f, respectively.
This partition has interesting applications, see for instance the proof of the Borsuk sandwich theorem 1.4.1. in the paper [1] and see also [5]. Of course, not every complex function of a symmetric set \(\mathcal {G}\) \(\subset \mathbb {C}\) is the product of an even and an odd function. For example no non-zero even function can be represented by the product of an even and an odd function. We will show that a function decomposition into the product of two functional factors is possible, if we define other special kinds of functional symmetry and a family of selected functions \(f:\mathcal {G\rightarrow }\mathbb {C}\).
2 Definitions and basic properties
Let \(\mathcal {G}\) \(\subset \mathbb {C}\) be a symmetric set such that \(0\in \mathcal {G}\). Then, by \(\mathcal {B}\) let us denote the set of all continuous functions \(f:\mathcal {G}\rightarrow \mathbb {C}\) normalized by \(f(0)=1\) and with \({\text {Re}}f(z)>0,z\in \mathcal {G}\). Of course, then \(f(z)\ne 0\) for \(z\in \mathcal {G}\).
Now we give the definition of the mentioned functional symmetries as well as the proposed names and designations.
Definition 1
The function \(f\in \mathcal {B}\) will be called \(\left( 1\right) \)-power symmetric and \(\left( -1\right) \)-power symmetric, if it satisfies the following conditions
respectively.
For \(j=1,-1\) the family of all (j)-power symmetric functions \(f\in \mathcal {B}\) will be denoted by \(\mathcal {B}_{j}\). Obviously, the constant function \(f=1\) belongs to both families \(\mathcal {B}_{1}\),\(\mathcal {B}_{-1}\) and there does not exist any other such function. Let us show that the families \(\mathcal {B}_{j},j=1,-1\) are non-trivial.
Example
1. Let the symmetric set \(\mathcal {G}\) be the horizontal strip
Then the function \(f(z)=exp(z),z\in \mathcal {G},\) belongs to \(\mathcal {B}_{1}.\)
Indeed, the function f belongs to \(\mathcal {B},\) because f is continuous, \(f(0)=1\) and \({\text {Re}}\exp (z)=e^{x}\cos y>0\) for \(z\in \mathcal {G}.\) On the other hand
Hence \(f\in \mathcal {B}_{-1}.\) Of course, \(f\notin \mathcal {B}_{-1},\) if we replace the horizontal strip \(\mathcal {G}\) by the plane \(\mathbb {C}\).
2. Let the symmetric set \(\mathcal {G}\) be the vertical strip
Then the function \(f(z)=\cos \left( z\right) ,z\in \mathcal {G}\) belongs to \(\mathcal {B}_{1}.\) Of course
\(f\in \mathcal {B},\) because f is continuous, normalized by \(f(0)=1\) and for \(z\in \mathcal {G}\) satisfies the condition
For the relation \(f\in \mathcal {B}_{1}\) it suffices to observe that \(\cos \left( -z\right) =\cos \left( z\right) ,z\in \mathcal {G}\).
Note that \(f\notin \mathcal {B}_{1}\) if we replace the set \(\mathcal {G}\) by the plane \(\mathbb {C}\).
3. Similarly, the function \(f(z)=\cosh \left( z\right) =\frac{e^{z}+e^{-z}}{2},\) reduced to the above horizontal strip \(\mathcal {G}\), belongs to the family \(\mathcal {B}_{1}.\)
4. Let us observe, that the functions \(f(z)=\sin \left( z\right) \) and \(f(z)=\sinh \left( z\right) \) reduced to a symmetric set do not belong to \(\mathcal {B}_{1}\) or \(\mathcal {B}_{-1},\) because \(f(0)\ne 1\).
5. Let the symmetric set \(\mathcal {G}\) be the open unit disc \(\mathcal {U}=\left\{ z\in \mathbb {C}:\left| z\right| <1\right\} .\) Then the function
belongs to \(\mathcal {B}_{1}.\)
Indeed, the function f belongs to \(\mathcal {B},\) because f is continuous, \(f(0)=1\) and \({\text {Re}}f(z)>\frac{1}{2}.\) Now it suffices to observe that
From now, we will always take the unit disc \(\mathcal {U}\) as the symmetric set \(\mathcal {G}\).
In the paper, for a precise definition of the mentioned decomposition of a function into the product of a (1)-power symmetric and a \((-1)\)-power symmetric function, we use the following very useful properties of the continuous unique main branch of the square root
(the symbol \(\sqrt{\cdot }\) in the whole paper means the above main branch of the square root).
Lemma 2.1
Let
For \(z,w\in \) \(\pi ^{+}\) and the above main branch of the square root the following equalities hold:
Note that Lemma 2.1. is a direct consequence of the following result [6, Chap. 5] for the continuous unique main branch of the logarithm
Proposition 1
For \(z,w\in \) \(\pi ^{+}\) and the above main branch of the logarithm, the following equalities hold:
Now, we give two theorems about the invariance of (j)-power symmetry \(j=1,-1\) with respect some functional operations.
We start with the case of operations of one function argument. Of course if \(f\in \mathcal {B}\), then the power \(f^{0}=1\) belongs to both families \(\mathcal {B}_{1},\mathcal {B}_{-1}.\)
On the other hand we have the following.
Theorem 2.1
Let n be an integer, \(f\in \mathcal {B}\), its power \(f^{n}\) belongs to \(\mathcal {B}\) and \(\sqrt{\cdot }\) be the main branch of the square root. Then the following implications are true:
Proof
(i) We take an \(n\in \mathbb {Z}\backslash \{0\}\), then for \(f\in \mathcal {B}_{1}\) and \(z\in \mathcal {U}\) we have:
This gives the thesis \(\left( i\right) \). Similarly, for \(n\in \mathbb {Z}\backslash \{0\}\), \(f\in \mathcal {B}_{-1}\) and \(z\in \mathcal {U}\) we have
This gives the thesis \(\left( ii\right) .\)
(iii) Let \(f\in \mathcal {B}_{1}\) (of course \(\sqrt{f}\in \mathcal {B)}\). Then from the definition of the (1)-power symmetric function, we have for \(z\in \mathcal {U}\):
hence \(\sqrt{f}\in \mathcal {B}_{1}.\)
(iv) Now let \(f\in \mathcal {B}_{-1}\), then from the definition of the \((-1) \)-power symmetric function and Lemma 2.1. we have successively for \(z\in \mathcal {U}\):
which proves that \(\sqrt{f}\in \mathcal {B}_{-1}\). \(\square \)
The next theorem concerns the invariance of the (j)-power symmetry \(j=1,-1\) with respect to some functional operations of two function arguments.
Theorem 2.2
Let \(f,g\in \mathcal {B},f\cdot g\in \) \(\mathcal {B}\) and \(\frac{f}{g}\in \) \(\mathcal {B}\). Then the following implications are true:
Proof
(i) Let f, g \(\in \mathcal {B}_{1}\). Using the definition of the (1)-power symmetric function we show (i) as follows:
Thus \(f\cdot g\) is (1)-power symmetric.
(ii) Let \(f,g\in \) \(\mathcal {B}_{-1}\), then using the definition of the \((-1\))-power symmetric function we show similarly:
Hence \(f\cdot g\) is \((-1)\)-power symmetric.
For the proof of parts (iii) and (iv) it suffices to save \(\frac{f}{g}=f\cdot \frac{1}{g}\) and use the above proved theses (i), (ii) and Theorem 2.1. (i), (ii), for \(n=-1.\) \(\square \)
Now, we give the main theorem of the paper. It concerns the decomposition of a function into (1)- and \((-1)\)-power symmetric factors.
Theorem 2.3
Let \(f\in \mathcal {B}\). Then:
-
(i)
The functions
$$\begin{aligned} f_{1}(z)&=\sqrt{f(z)\cdot f(-z)},z\in \mathcal {U},\\ f_{-1}(z)&=\sqrt{f(z)\cdot [f(-z)]^{-1}},\,z\in \mathcal {U}, \end{aligned}$$with the main branch of the square root, are (1)- and \((-1)\)-power symmetric, respectively.
-
(ii)
The above function f can be represented as the product of the functions \(f_{1}\) and \(\ f_{-1}\) , i.e.
$$\begin{aligned} f(z)=f_{1}(z)\cdot f_{-1}(z),\qquad z\in \mathcal {U}. \end{aligned}$$ -
(iii)
This decomposition of a function f is unique. It means that, if there exist g \(\in \mathcal {B}_{1}\) and \(h\in \) \(\mathcal {B} _{-1}\) such that, \(f(z)=g(z)\cdot h(z)\) for \(z\in \mathcal {U},\) then \(f_{1}=g\) and \(f_{-1}=h.\)
Proof
Firstly, we show that the functions \(f_{1},f_{-1}\) are well defined. Indeed, since \(Re\,f(-z)>0,\) for \(z\in U,\) we also have \(Re\,\frac{1}{f(-z)}>0,z\in U.\) Then
and
Then the functions are well defined, because \(\sqrt{\cdot }\) means the main branch of the square root.
Now we show that \(f_{1}\) is (1)-power symmetric and \(f_{-1}\) is \((-1)\)-power symmetric. Let us note that \(f_{1}(0)=1=f_{-1}(0)\) and \(f_{1},f_{-1} :\mathcal {U}\rightarrow \pi ^{+}.\) Obviously the functions \(f_{1}\) and \(f_{-1}\) are continuous, because f is continuous, the product and quotient of continuous functions are continuous, the main branch of the square root is a continuous. From the form of \(f_{1},\) we have for \(z\in \mathcal {U}\):
Thus \(f_{1}\) is (1)-power symmetric. From the definition of \(f_{-1}\) and Lemma 2.1, we show that \(f_{-1}\) is \((-1)\)-power symmetric. This follows from the following equalities for \(z\in \mathcal {U}\):
This completes the proof of (i).
Next we show (ii). In view of thesis (i) and Lemma 2.1, we can write for \(z\in \mathcal {U}\) the product \(f_{1}\cdot \) \(f_{-1}\) of the functions \(f_{1},\) \(f_{-1},\) step by step:
Now we will prove (iii), i.e., that the decomposition
is unique.
Suppose that: \(f(z)=g(z)\cdot h(z),z\in \mathcal {U},\) where \(g\in \mathcal {B}_{1}\), \(h\in \mathcal {B}_{-1}\). Then
Thus, using the fact that \(g\in \mathcal {B}_{1},\) \(h\in \mathcal {B}_{-1}\) and Lemma 2.1, we get for \(z\in \mathcal {U}:\)
Similarly for \(z\in \mathcal {U}\):
Using once again Lemma 2.1 and the fact that \(g\in \mathcal {B}_{1} ,h\in \mathcal {B}_{-1},\) we get for \(z\in \mathcal {U}:\)
This gives (iii) and the proof is complete. \(\square \)
By the uniqueness of the above decomposition, we will call the function \(f_{1}\) and \(f_{-1}\) the (1)- and \((-1)\)-power symmetric parts of the function f.
Now, we turn to examples involving the form of the decomposition of various functions \(f\in \mathcal {B}\) into (1)-power symmetric \(f_{1}\) and \((-1)\)-power symmetric \(f_{-1}\) factors. We will focus on examples of decompositions of functions from Caratheodory’s class of holomorphic functions \(f:\mathcal {U}\rightarrow \mathbb {C},f(0)=1\) with \({\text {Re}}f\left( z\right) >0,z\in \mathcal {U}.\) Caratheodory’s class of functions plays an important role in geometric function theory (see for instance [2]).
Example
1. Let us consider the function
The function f belongs to \(\mathcal {B}\), because it is continuous, f\((0)=1\) and \({f(U)=\pi ^{+}}\). Putting \(-z\) in place of z, we obtain:
Thus f is a \((-1)\)-power symmetric function. Hence, by the uniqueness of the decomposition of the function f in Theorem 2.3., we have \(f_{_{1}}(z)=1,\) and \(\ \ f_{_{-1}}\left( z\right) =f(z),\) for \(z\in \mathcal {U}\) .
2. Let us consider the function
We see that f is a superposition of the previous function with the square function \(w(z)=z^{2},z\in \mathcal {U}\), hence f \(\in \)\(\mathcal {B}\). Putting \(\left( -z\right) \) in place of\(\ \ z\), we obtain that \(f\left( -z\right) =f(z)\) for \(z\in \mathcal {U}.\) Thus f is a (1)-power symmetric function. Hence, by the uniqueness of the decomposition of the function f in Theorem 2.3, we obtain that \(f_{1}(z)=f\left( z\right) \) and \(f_{-1}(z)=1.\)
3. Let us consider the function
Of course, f is continuous, normalized by f\((0)=1\) and \({f(U)}\subset \pi ^{+}\). Then the function f belongs to the family \(\mathcal {B}.\) Putting \(\left( -z\right) \) in place of\(\ z,\) we obtain:
We can see that \(f\left( -z\right) \ne f(z)\) and \(f\left( -z\right) \ne \frac{1}{f(z)}.\) Thus f is neither (1)-power symmetric nor \((-1)\)-power symmetric. Therefore we cannot use the previous method of decomposition. Below we determine directly the power symmetric parts \(f_{1}\) and \(f_{-1}\) of the function f:
3 Applications
Now we show the potential possibility of using (1)- and \((-1)\)- power symmetric functions to solve certain functional equations.
Let us therefore consider the equations:
where \(F\in \mathcal {B}\) is a given function and \(f\in \mathcal {B}\) is an unknown function.
These equations belong to the collection of the following type of functional equations
where G, g are given functions and \(f\in \mathcal {B}\) is the wanted function (see Kuczma’s monograph [4]). This follows from the observation that Eq. (1) has the form
where \(g(z)=-z\), \(F\in \) \(\mathcal {B}\) is a fixed function and Eq. (2) has the form
respectively.
Let us add that Eq. (1) corresponds to the Schröder-type equation, i.e., equation [4] of the form
where \(s\in \mathbb {C}\) and \(g(z)=-z\).
We will use Theorem 2.3. to solve Eqs. (1) and (2). We will prove the following:
Theorem 3.1
Equation (1) has a solution \(f\in \mathcal {B},\) if and only if \(F\in \) \(\mathcal {B}_{1}.\)
Proof
“\(\Longrightarrow \)” Let \(f\in \mathcal {B}\) be a solution of Eq. (1). Then by replacing z by \(\left( -z\right) ,\) we obtain:
From the above and from (1) we can see that
Hence, \(F\in \) \(\mathcal {B}_{1}.\)
“\(\Longleftarrow \)” Let \(F\in \) \(\mathcal {B}_{1}.\) Then, applying decomposition (ii) from Theorem 2.3. to Eq. (1), we have for \(z\in \mathcal {U}\):
Next, using part \(\left( i\right) \) of Theorem 2.3., we have further:
Therefore:
(Let us recall that \(\sqrt{\cdot }\) means the unique continuous main branch \(\sqrt{\cdot }:\mathbb {C}\ \backslash (-\infty ,0]\rightarrow \) \(\mathbb {C} ,\sqrt{1}=1,\) of the square root).
We check that the function \(f(z)=\sqrt{F(z)}\cdot g(z)\), \(z\in \mathcal {U}\), with a \((-1)\)-power symmetric function g, is a solution of Eq. (1). Indeed, since \(F\in \) \(\mathcal {B}_{1}\) and \(g\in \mathcal {B}_{-1}\), we obtain for \(z\in \mathcal {U}\):
This ends the proof. \(\square \)
Corollary
If \(F\in \) \(\mathcal {B}_{1},\) then every solution f of Eq. (1) has the form:
with the main branch of the function \(\sqrt{\cdot }\) and with a \((-1)\)-power symmetric function g.
Theorem 3.2
Equation (2) has a solution \(f\in \mathcal {B},\) if and only if \(F\in \) \(\mathcal {B}_{-1}.\)
Proof
“\(\Longrightarrow ''\)Let \(f\in \mathcal {B}\) be a solution of Eq. (2). Then, putting \(\left( -z\right) \) in place of z, we obtain:
Multiplying this equation and Eq. (1) we obtain for \(z\in \mathcal {U}\):
Hence, we get that
Thus \(F\in \) \(\mathcal {B}_{-1}.\)
“\(\Longleftarrow ''\) Let \(F\in \) \(\mathcal {B}_{-1}.\) Then applying decomposition (ii) from Theorem 2.3 to Eq. (2), we have for \(z\in \mathcal {U}\):
Next, using part \(\left( i\right) \) of Theorem 2.3. we have further:
Hence
where \(\sqrt{\cdot }\) means the main branch of the square root.
Now, we check that the function
with a \((-1)\)-power symmetric function g, is a solution of Eq. (2). To this aim, we observe that
Since \(F\in \) \(\mathcal {B}_{-1}\) and \(g\in \) \(\mathcal {B}_{1,}\)using Lemma 2.1. we obtain for \(z\in \mathcal {U}\)
This ends the proof. \(\square \)
Corollary
If \(F\in \) \(\mathcal {B}_{-1}\) then every solution of Eq. (2) has the form:
with the main branch of the square root and a (1)-power symmetric function g.
Now we will show the possibility of using (j)-power-symmetric functions, \(j=1,-1\), to investigate the properties of functions defined by geometrical conditions.
We start with the definition of starlikeness of order \(\alpha \in \left[ 0,1\right) .\)
Definition 2
We say that a holomorphic function \(g:\mathcal {U}\rightarrow \mathbb {C}\) , \(g\left( 0\right) =0,\) \(g^{\prime }\left( 0\right) =1\) belongs to the family \(\mathcal {S}^{*}\left( \alpha \right) ,\alpha \in \left[ 0,1\right) ,\) of starlike mappings of order \(\alpha ,\) if
The following property of \(g\in \mathcal {S}^{*}\left( \alpha \right) \) is known (see [2, thm 2.3.4]).
Lemma 3.1
A function g belongs to \(\mathcal {S}^{*}\left( \alpha \right) ,\) if only if
Here we use the continuous unique main branch \(\left( \cdot \right) ^{\frac{1}{1-\alpha }}:\mathbb {C}\ \backslash (-\infty ,0]\rightarrow \mathbb {C},\) \((1)^{^{\frac{1}{1-\alpha }}}=1,\) of the power function.
The mentioned application of functional power symmetries relates to a family \(K_{\mathcal {S}}\) (see [3]).
Definition 3
The holomorphic function \(f:\mathcal {U}\rightarrow \mathbb {C},f\left( 0\right) =0,f^{\prime }\left( 0\right) =1,\) belongs to the family \(\mathcal {K}_{\mathcal {S}}\) if there exists a function \(g\in \mathcal {S}^{*}\left( \frac{1}{2}\right) \) such that
We prove the following theorem
Theorem 3.3
For every \(f\in \mathcal {K}_{\mathcal {S}},\) there exsists \(g\in \mathcal {S} ^{*}\left( \frac{1}{2}\right) ,\) such that
where the function \(\left( \frac{g(z)}{z}\right) _{1}\), is the \(\left( 1\right) \)-power symmetric part of the function \(\frac{g(z)}{z}.\)
Proof
Since \(g\in \mathcal {S}^{*}\left( \frac{1}{2}\right) ,\) we have from Lemma 3.1 that the function
belongs to the family \(\mathcal {S}^{*}\) of starlike univalent functions. Hence (see for instance [2, Thm. 2.3.5]) the function
with the main branch of the square root has positive real part in \(\mathcal {U}\) and also
Consequently, the function \(\frac{g(z)}{z}\) belongs to the family \(\mathcal {B}\) and we can compute its \(\left( 1\right) \)-power symmetric part. Using the definition of the \(\left( 1\right) \)-power symmetric part of functions, we have:
Hence
Inserting this equality into the defining inequality (1) we obtain the thesis of Theorem 3.3. \(\square \)
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Bartoszek, B., Długosz, R. On a multiplicative distribution of functions in complex plane. Aequat. Math. 97, 425–437 (2023). https://doi.org/10.1007/s00010-022-00903-4
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DOI: https://doi.org/10.1007/s00010-022-00903-4
Keywords
- Functions decomposition onto symmetric factors
- Power-symmetric functions
- Functions with positive real part
- Starlike functions
- Functional equations