Some Particular Norm in the Sobolev Space \(H^{1}[a,b]\)

  • Edward TutajEmail author
Open Access


This paper is a continuation of the recent paper of the author, where a certain reproducing kernel Hilbert space \(X_{\mathcal {S}}\) was constructed. The norm in \(X_{\mathcal {S}}\) is related to a certain generalized isoperimetric inequality in \({{\mathbb {R}}^2}\). In the present paper we give an alternative description of the space \(X_{\mathcal {S}}\), which appears to be a Sobolev space \(H^{1}[a,b]\) with some special norm.


Absolutely continuous functions Isoperimetric inequality Hilbert spaces Sobolev spaces Reproducing kernels 

Mathematics Subject Classification


1 Introduction

This paper is a continuation of the recent paper of the author, [8]. In this paper we consider a Sobolev space \(H^1[a,b]\). We recall some necessary definitions and properties of these spaces below, but at the beginning it is sufficient to know that the elements of \(H^1[a,b]\) are absolutely continuous functions with the derivatives in \(L_2[a,b]\). Then, if one will define an unitary structure in \(H^1[a,b]\), the formula for \(\langle f,g\rangle \) must necessarily contain the bilinear form (\(f,g \in H^1[a,b]\))
$$\begin{aligned} \int _a^b f'(x)\cdot g'(x)dx. \end{aligned}$$
Clearly the quadratic form generated by (1) is only a seminorm in \(H^1[a,b]\), hence to have a norm one must add ”something” to (1), or consider a smaller space. In (11) we present a typical example of this kind. In this paper we propose, a new formula to define an inner product in \(H^1[a,b]\) inspired by the very old idea of the isoperimetric inequality in the plane. The paper [8] contains the construction of a certain reproducing kernel Hilbert space \(X_\mathcal {S}\) (the necessary information concerning these class of spaces are to be found in the next section). Its kernel, given by the formula (7), is not very complicated. However the description of the function space \(X_\mathcal {S}\) is not satisfactory. The elements of this space are the equivalent classes of pairs of convex sets relative to some equivalence relation [formula (2)] and then it is not easy to look on the elements of \(X_\mathcal {S}\) as on the real function defined on some interval. But, on the other hand, we have a concrete positively defined function K [in our situation the function \(K(\varphi ,\psi )\) given by the formula (7)] so we may try to construct another, perhaps more convenient, model of the space \(\mathcal {H}(K)\) corresponding to the considered kernel. Such a situation occurs frequently in the theory of RKHS’s and has even his own name (the reconstruction problem, cf [4]). The aim of this paper is to give such a concrete description of the space \(\mathcal {H}(K)\) for the kernel given by (7). It will appear that in our case the space \(\mathcal {H}(K)\) is a Sobolev space \(H^1[a,b]\) equipped with some special norm. The construction of ”our” \(\mathcal {H}(K)\) is contained in the three theorems: Theorem 4, Theorem 5 and the Theorem 6.

In Theorem 4 we define a certain bilinear form \(\langle ,\rangle _i\) (14) and we prove that this form is positively defined. In the proof we use the Wirtinger inequality (3) in a general version for the derivatives from \(L_2[a,b]\). This variant of the Wirtinger inequality is to be found in [3].

In Theorem 5 we check the reproducing property of the given kernel (7). The proof is easy to understand, but may be is heavy to read because of lengthy calculations. The kernel (7), as far as we know, was not studied before.

In the last Theorem 6 we prove, that the constructed space is an RKHS. For this we must check, that the evaluation functionals are commonly bounded, and that the constructed space is complete. This type of arguments is typical in examples of Sobolev spaces \(H^1[a,b]\) [2]. Summarizing, the main result of this paper says, that the space \(H^{1}[-\frac{\pi }{2},\frac{\pi }{2}]\) equipped with the norm \(||\cdot ||_i\) is a reproducing kernel Hilbert space corresponding to the same kernel (7) as the space \(X_{\mathcal {S}}\) constructed in [8].

We present also one more construction of the considered space \(\mathcal {H}(K)\), called here the sequence model. This sequence model, if one looks from strictly theoretical point of view, does not bring new information, but gives tools for proofs and for numerical calculations. This model represents also a kind of presentation of the density of polygons in the space of convex sets.

From a number of known examples of the kernels in the space \(H^1[a,b]\), we present here one see formulas (12) and (13) in order to compare it with the considered here kernel (7).

This paper is organized as follows. In the second part we give a summary of the paper [8] necessary to understand the main result of the present paper. We recall also some information on the Sobolev of the type \(H^{1}[a,b]\) in the aim to see the ”particularity” of the norm we are going to construct. In the third section we define some special (particular) norm \(||\cdot ||_i\) in the space \(H^{1}[-\frac{\pi }{2},\frac{\pi }{2}]\). In the last section we present the construction of the sequence model.

2 About the Space of Generalized Convex Sets

In this subsection we recall the construction of the space \(X_\mathcal {S}\) from [8]. Let \(\mathcal {S}\) denote the cone of convex, compact and centrally symmetric sets in the plane \({\mathbb {R}}^2\). In this cone we consider the Minkowski addition and the scalar multiplication by non-negative real numbers. In the product \(\mathcal {S}\times \mathcal {S}\) we consider the equivalence relation
$$\begin{aligned} (U,V)\diamondsuit (P,Q)\Longleftrightarrow U+Q=V+P \end{aligned}$$
It is known, that the quotient \(\mathcal {S}\times \mathcal {S}/\diamondsuit \) is a real vector space (for details see e.g. [5]), and the space \(X_{\mathcal {S}}\) mentioned above is the completion of the quotient \(\mathcal {S}\times \mathcal {S}/\diamondsuit \) with respect to the so-called isoperimetric norm, which is constructed in a few steps. At first we prove, that the two dimensional Lebesgue measure \(m:\mathcal {S}\longrightarrow [0,\infty )\) can be extended in a unique way to the polynomial of the second degree \(m^{*}:\mathcal {S}\times \mathcal {S}/\diamondsuit \longrightarrow {\mathbb {R}}\) and the functional \(o:\mathcal {S}\longrightarrow \mathbb R\) where o(U) is a perimeter of U, can be extended to a linear functional on the whole \(\mathcal {S}\times \mathcal {S}/\diamondsuit \). Next we prove that for this extended measure \(m^{*}\) the generalized isoperimetric inequality holds, which means that for each \([U,V]\in \mathcal {S}\times \mathcal {S}/\diamondsuit \) we have
$$\begin{aligned} (o(U)-o(V))^2-4\pi \cdot m([U,V])\ge 0, \end{aligned}$$
where \(o(U)-o(V)\) is the perimeter of the pair [UV] and (here and in the sequel) we write m([UV]) instead of \(m^{*}([U,V])\). Using this fact it was proved, that the formula
$$\begin{aligned} ||[U,V]||^2_i=2\cdot (o(U)-o(V))^2-4\pi \cdot m([U,V]) \end{aligned}$$
is a unitary norm in \(\mathcal {S}\times \mathcal {S}/\diamondsuit \). Finally the space \(X_\mathcal {S}\) was defined as the completion of the unitary space \((\mathcal {S}\times \mathcal {S}/\diamondsuit ; ||\cdot ||_i)\).

2.1 About the RKHS

In the last section of [8] we prove that the space \(X_{\mathcal {S}}\) is a reproducing kernel Hilbert space (RKHS for short). Although the reproducing property was discovered by Zaremba more than hundred years ago (1906), the first systematic lecture is due to Aronszajn in 1950 (see [1]). The necessary information concerning RKHS are to be found in the book of Berlinet and Thomas-Agnant [2] or in the recent book of Szafraniec [6]. The RKHS are function spaces, so at first one must recognize the elements of \(\mathcal {S}\times \mathcal {S}/\diamondsuit \) as real functions on the interval \(\Delta =[-\frac{\pi }{2},\frac{\pi }{2}]\). This may be done as follows. We consider the functions
$$\begin{aligned} I_{\psi }:\left[ -\frac{\pi }{2},\frac{\pi }{2}\right] \ni \varphi \longrightarrow \sin |\varphi -\psi |, \end{aligned}$$
where \(\varphi ,\psi \in [-\frac{\pi }{2},\frac{\pi }{2}].\) These functions \(I_\psi \) are called diangles. Each diangle defines a line in \({\mathbb {R}}^2\) (\({\mathbb {R}} \cdot (\cos \psi ,\sin \psi ))\) which will be denoted also by \(I_\psi \) and for the set \(U\in \mathcal {S}\) the number \(\overline{U}(I_\psi )\) denotes the width of the set U with respect to the line \(I_\psi \). The correspondence
$$\begin{aligned}{}[U,V]\rightarrow f_{[U,V]} = \overline{U}(I_\varphi ) - \overline{V}(I_{\varphi }), \end{aligned}$$
allows us to see the equivalent classes [UV] as functions on the interval \(\Delta \) (the details in [8]). Now we are ready to formulate the main result from [8].

Theorem 1

The space \(X_\mathcal {S}\) with the norm (4) is a reproducing kernel Hilbert space with the kernel \(K:\Delta \times \Delta \longrightarrow {\mathbb {R}}\), where
$$\begin{aligned} K(\varphi ,\psi )= 2 - \frac{\pi }{2}\sin |\varphi -\psi |. \end{aligned}$$

The elements of \(X_\mathcal {S}\) are, roughly speaking, the ”differences” of convex sets (differences with respect to the Minkowski addition). This makes it possible to understand their geometrical character, but on the other hand, given a function \(f:\Delta \longrightarrow {\mathbb {R}}\) it is difficult to prove (or disprove), that \(f\in X_\mathcal {S}\). In this paper we will solve this problem. More exactly we will prove, that \(X_\mathcal {S}\) is a certain Sobolev space.

2.2 About the Sobolev spaces

We shall start by recalling some commonly known definitions and theorems concerning the Sobolev spaces.

Definition 2

Let \(\Delta =[a,b]\subset {\mathbb {R}}\) be a compact interval. A function \(f:\Delta \longrightarrow {\mathbb {R}}\) is said to be absolutely continuous on \(\Delta \) if for every \(\epsilon > 0\) there exists \(\delta >0\) such that
$$\begin{aligned} \sum _{k=1}^{p}|f(b_k)-f(a_k)|\le \epsilon \end{aligned}$$
for every finite number of nonoverlappig intervals \((a_k,b_k)\), \(k=1,2,\ldots ,p\) with \([a_k,b_k]\subset \Delta \) and
$$\begin{aligned} \sum _{k=1}^{p}(b_k-a_k)\le \delta . \end{aligned}$$
Let \(AC_1(\Delta )\) denote the vector space of all absolutely continuous functions on \(\Delta \). It is known, that each \(f\in AC_1(\Delta )\) is differentiable almost everywhere and its derivative \(f'\) is a Lebesgue integrable function (i.e. \(f'\in L_1(\Delta )\)). We will consider a subspace \(H^{1}(\Delta )\subset AC_1(\Delta )\) claiming that
$$\begin{aligned} f\in H^{1}(\Delta ) \Longleftrightarrow f\in AC_1(\Delta ) \wedge f'\in L_2(\Delta ). \end{aligned}$$
It is also known (see e.g. [4]), that one may consider an inner product \(\langle ,\rangle \) in \(H^{1}(\Delta )\) given by the formula (11) for \(f,g \in H^{1}(\Delta )\):
$$\begin{aligned} \langle f,g\rangle = \int _{a}^{b}(f\cdot g + f'\cdot g') \end{aligned}$$
which is well defined since \(f',g' \in L_2(\Delta )\). One may check, that \(H^{1}(\Delta )\), equipped with the above inner product (11), is a Sobolev space with the reproducing kernel K(xy) where
$$\begin{aligned} K(x,y)= & {} \frac{\cosh (x-a)\cosh (b-y)}{\sinh (b-a)},\, a\le x\le y\le b \end{aligned}$$
$$\begin{aligned} K(x,y)= & {} \frac{\cosh (x-a)\cosh (b-y)}{\sinh (b-a)},\, a \le y\le x \le b \end{aligned}$$
Some other examples of reproducing kernels in Sobolev spaces are to be found in [2] or in [4].

In the present paper we construct another inner product \(\langle ,\rangle _i\) defined in a subspace \(H^{1}_{0}(\Delta )\subset H^{1}(\Delta )\), related to another reproducing kernel \(K_i(x,y),\) which will be defined in the next section.

3 An Inner Product in the Sobolev Space \(H_0^{1}(\Delta )\)

Now we fix \(\Delta =[-\frac{\pi }{2},\frac{\pi }{2}]\) and we consider the subspace \(H^{1}_{0}\subset H^{1}(\Delta )\) claiming that \(f(-\frac{\pi }{2})=f(\frac{\pi }{2})\). This means, that \(H^1_0(\Delta )\) may be identified with the space of periodic functions with analogous properties (i.e. absolutely continuous and with the derivatives in \(L_2(\Delta )\). For \(f,g \in H_0^{1}(\Delta )\) we set
$$\begin{aligned} \langle f,g\rangle _i = \frac{1}{\pi ^2}\left[ 2\cdot \left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) \cdot \left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}} g\right) - \pi \cdot \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}} (f\cdot g - f'\cdot g')\right] . \end{aligned}$$
Here, and below we will often write \(\int _a^b f\) or \(\int _a^b f(x)\) instead of \(\int _a^b f(x)dx\). We will frequently also write \(\int f\) instead of \(\int _a^b f\) in the case when \(a= -\frac{\pi }{2}\) and \(b=\frac{\pi }{2}\).

It is clear, that the formula (14) is well defined (because of \(f',g' \in L_2(\Delta )\)) and defines a bilinear form in \(H_0^{1}(\Delta )\). The coefficient \(\frac{1}{\pi ^2}\) is clearly without importance and is chosen to have \(\langle \mathbf{1},\mathbf{1}\rangle _i = 1\) (where \(\mathbf{1}\) denotes the constant (and equal 1) function). Hence, to have a norm related to the form (14), it remains to show that \(\langle ,\rangle _i\) is positively defined. The proof of this positivity is similar in fact to the known proof of the isoperimetric inequality based on the Wirtinger inequality.

Let us start by recalling a variant of the Wirtinger inequality (for the proof and some other formulations see [3, 7]).

Theorem 3

Let f be a continuous and periodic function with the period \(\pi \). Let \(f'\in L_2(\Delta )\) and let
$$\begin{aligned} \overline{f} = \frac{1}{\pi }\left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) , \end{aligned}$$
$$\begin{aligned} \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f')^2 \ge \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f-\overline{f})^2, \end{aligned}$$
and equality holds if and only if \(f=a \cos (t) + b\sin (t)\).

Now we have the following

Theorem 4

Let \(f\in H_0^{1}(\Delta )\). Then
$$\begin{aligned} \langle f,f\rangle _i=2\left( \int _{-\pi /2}^{\pi /2}f\right) ^2 - \pi \int _{-\pi /2}^{\pi /2}((f)^2-(f')^2)\ge 0. \end{aligned}$$
and the equality holds only when \(f=0\). In other words, the form (14) is positively defined.


Let us set, as above
$$\begin{aligned} \overline{f} = \frac{1}{\pi }\left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) , \end{aligned}$$
and let
$$\begin{aligned} E(f)=\left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) ^2 - \pi \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}((f)^2-(f')^2) \end{aligned}$$
We want to show, that \(E(f)\ge 0\). Consider the difference
$$\begin{aligned} F(f)=\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f')^2 - \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f-\overline{f})^2. \end{aligned}$$
The Wirtinger inequality says precisely that \(F(f)\ge 0\). This implies that
$$\begin{aligned} E(f)=(\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f)^2 - \pi \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}((f)^2-(f')^2)\ge 0. \end{aligned}$$
$$\begin{aligned} 0\le F(f)= & {} \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f')^2 -\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f)^2 + 2 \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f \cdot \left( \frac{1}{\pi }\cdot \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) \\&\quad -\,\pi \cdot \left( \frac{1}{\pi ^2}\cdot \left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) ^2\right) \\= & {} \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f')^2 -\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f)^2 +\frac{1}{\pi }\left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) ^2, \end{aligned}$$
and this is equivalent to the desired inequality \(E(f)\ge 0\). Then \(\langle f,f\rangle _i\ge 0\).
If \(\langle f,f\rangle _i =0\) then
$$\begin{aligned} \langle f,f\rangle _i = \left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) ^2 + E(f) =0 \end{aligned}$$
and hence \(\overline{f}=0 \) and \(E(f)=0\). Thus
$$\begin{aligned} \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}((f)^2-(f')^2)=0 \end{aligned}$$
and by (17) \(F(f)=0\), which means that the equality holds in the Wirtinger inequality and additionally \(\overline{f}=0\). We know that in such a case \(f=a \cos (t) +b\sin (t)\) and moreover \(\overline{f}=0\). Since \(f(-\frac{\pi }{2})=f(\frac{\pi }{2})\) then \(b=0\) and since \(\overline{f}=0\) then \(a=0\). Hence \(f=0\).
In consequence we have proved that the \(H^1_{0}(\Delta )\) with the inner product (14) is an unitary space. The norm induced by (14) is
$$\begin{aligned} ||f||_i^2= & {} \frac{1}{\pi ^2}\cdot \left( 2\cdot \left( \int _{-\pi /2}^{\pi /2}f\right) ^2 -\pi \int _{-\pi /2}^{\pi /2}((f)^2-(f')^2) \right. \nonumber \\&\quad = \frac{1}{\pi ^2}(o^2(f) + E(f)), \end{aligned}$$
$$\begin{aligned} o^2(f)=\left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) ^2. \end{aligned}$$
This ends the proof of Theorem 4. \(\square \)

We will prove now the reproducing property.

Theorem 5

The function \(K:[-\frac{\pi }{2},\frac{\pi }{2}]\times [-\frac{\pi }{2},\frac{\pi }{2}]\longrightarrow {\mathbb {R}}\) defined by
$$\begin{aligned} K_i(x,y)= 2-\frac{\pi }{2}\cdot \sin |x-y|, \end{aligned}$$
is a reproducing kernel in the unitary space \(H^1_{0}(\Delta )\). This function will be called the isoperimetric kernel.


Let \(k_y(x)=2-\frac{\pi }{2}\cdot \sin |x-y|,\) (i.e. \(k_y= K_i(\cdot ,y)\) is a kernel function). We will check (this is the so-called reproduction property) that for each function \(f\in H^1_0(\Delta )\) and for each \(y\in [-\frac{\pi }{2},\frac{\pi }{2}]\) we have:
$$\begin{aligned} f(y)=\langle f,k_y\rangle _i \end{aligned}$$
Let us fix y and f as above. We must verify that
$$\begin{aligned} f(y) = \frac{1}{\pi ^2}\left( 2\cdot \left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) \left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}k_y\right) - \pi \cdot \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f\cdot k_y - f'\cdot k'_y)\right) . \end{aligned}$$
Let us denote
$$\begin{aligned} A= & {} 2\cdot \left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) \left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}k_y\right) , \\ B= & {} \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\cdot k_y, \\ C= & {} \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f'\cdot k'_y. \end{aligned}$$
We must check that
$$\begin{aligned} f(y)=\frac{1}{\pi ^2}(A-\pi B +\pi C). \end{aligned}$$
Let us recall, that in the calculations which will be done below, we will frequently write \(\int h\) instead of \(\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}h(x)dx.\)
We have
$$\begin{aligned} A= & {} 2\cdot \left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) \cdot \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left( 2-\frac{\pi }{2}\sin |x-y|\right) =2 \cdot \left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) \cdot (2\pi - \pi ) \nonumber \\= & {} 2\pi \cdot \left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\right) . \end{aligned}$$
Now we will compute B.
$$\begin{aligned} B= & {} \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}} f\cdot \left( 2-\frac{\pi }{2}\cdot \sin |x-y|\right) \nonumber \\= & {} 2\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}} f - \frac{\pi }{2}\cdot \int f\cdot \sin |x-y| = 2\int f - \frac{\pi }{2}\cdot \int f\cdot \sin |x-y| \nonumber \\= & {} 2\int f - \frac{\pi }{2}\cdot \left( \int _{-\frac{\pi }{2}}^{y}f\cdot \sin (y-x) + \int _{y}^{\frac{\pi }{2}}f\cdot \sin (x-y)\right) \end{aligned}$$
The case of C is more complicated. We have
$$\begin{aligned} C= & {} \int f'\cdot k'_y = \int f'\cdot \left( -\frac{\pi }{2}\cdot \sin |x-y|\right) \\= & {} \int _{-\frac{\pi }{2}}^{y}\left( -\frac{\pi }{2}\right) f' \cdot (-\cos (x-y))+\int ^{\frac{\pi }{2}}_{y}\left( -\frac{\pi }{2}\right) f'\cos (x-y) \\= & {} \frac{\pi }{2}\int _{-\frac{\pi }{2}}^{y} f'\cos (x-y) - \frac{\pi }{2}\int ^{\frac{\pi }{2}}_{y} f'\cos (x-y). \end{aligned}$$
Now we apply twice the integration by parts and we obtain that the above is equal
$$\begin{aligned}= & {} \frac{\pi }{2}\left[ f\cos (x-y)|_{-\frac{\pi }{2}}^y + \int _{-\frac{\pi }{2}}^{y}f\sin (x-y) - f\cos (x-y)|^{\frac{\pi }{2}}_y \right. \\&\quad \left. - \int ^{\frac{\pi }{2}}_{y}f\sin (x-y)\right] \\= & {} \frac{\pi }{2}\left[ 1\cdot f(y)-f\left( \frac{-\pi }{2}\right) \cos \left( -\frac{\pi }{2}-y\right) -f\left( \frac{\pi }{2}\right) \cos \left( \frac{\pi }{2}-y\right) + 1\cdot f(y) \right. \\&\quad \left. - \left( \int _{\frac{-\pi }{2}}^{y}f(-\sin (x-y)) + \int ^{\frac{\pi }{2}}_{y}f(-\sin (x-y))\right) \right] . \end{aligned}$$
Since \(f(-\frac{\pi }{2})=f(\frac{\pi }{2})\) then we obtain
$$\begin{aligned} \frac{\pi }{2} \left[ 2f(y)-f\left( \frac{\pi }{2}\right) \left( \cos \left( \frac{\pi }{2}+y\right) +\cos \left( \frac{\pi }{2}-y\right) \right) - \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\cdot \sin |x-y|\right] . \end{aligned}$$
But \(\cos (\frac{\pi }{2}+y)=-\sin (y)\) and \(\cos (\frac{\pi }{2}-y)=\sin (y)\) so finally we have
$$\begin{aligned} C=\frac{\pi }{2}\left[ 2f(y)-\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\cdot \sin |x-y|\right] . \end{aligned}$$
$$\begin{aligned} \langle f,k_y\rangle _i= & {} \frac{1}{\pi ^2}[A-\pi B+\pi C] \\= & {} \frac{1}{\pi ^2}\left[ 2\pi \left( \int f\right) -\pi \left( 2\left( \int f\right) -\frac{\pi }{2}f\cdot \sin |x-y|\right) \right. \\&\left. +\,\pi \frac{\pi }{2}\left( 2f(y)-\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\cdot \sin |x-y|\right) \right] \\= & {} \frac{1}{\pi ^2}\left[ 2\pi \left( \int f\right) -2\pi \left( \int f\right) \right. \\&\left. +\,\frac{\pi ^2}{2}\int f\sin |x-y| + \pi ^2 f(y) - \frac{\pi ^2}{2}\int f\sin |x-y|\right] \\= & {} \frac{1}{\pi ^2}\cdot {\pi ^2} f(y) = f(y). \end{aligned}$$
This ends the proof of the Theorem 5\(\square \)

It remains to prove now the main result of this section, which says that \(H^{1}_{0}(\Delta )\) with the norm (22) is an RKHS.

Theorem 6

With the notations as above, the space \(H_{0}(\Delta )\) is a reproducing kernel Hilbert space and its kernel is given by
$$\begin{aligned} K_i(x,y)=2-\frac{\pi }{2}\sin |x-y|. \end{aligned}$$


The argumentation is similar for example to that in [4]. First we check that the evaluation functionals \(e_x(f)=f(x)\) are bounded with respect to the isoperimetric norm. Since \(H_0(\Delta )=H_{00}+\mathbb {R}\cdot {\mathbf{1}}\) then it is sufficient to prove, that each \(e_x\) is bounded on the subspace \(H_{00}\subset H^1_{0}\), composed of those functions which vanish at 0, (i.e. we may additionally assume, that \(f(0)=0)\).

Let us suppose that \(f\in H^1_{0}(\Delta )\). Let \(x+h,x\in [-\frac{\pi }{2},\frac{\pi }{2}]\). By the reproducing property (just proved) we have:
$$\begin{aligned} f(x+h)-f(x) = \langle f,k_{x+h}\rangle _i - \langle f,k_x\rangle _i = \langle f,k_{x+h}-k_x\rangle _i, \end{aligned}$$
$$\begin{aligned} k_{x+h}(t)-k_x(t)= & {} \left( 2-\frac{\pi }{2}\sin |t-(x+h)|\right) -\left( 2-\frac{\pi }{2}\sin |t-x|\right) \\= & {} \frac{\pi }{2}(I_x(t)-I_{x+h}(t)), \end{aligned}$$
where \(I_{\psi }\) is a diangle (see (5), [8]). More exactly \(I_\psi \) is a function given by the formula \(I_\psi (\varphi )=\sin |\varphi -\psi |\). Hence we may write (using Schwarz inequality)
$$\begin{aligned} |f(x+h)-f(x)| = |\frac{\pi }{2}\langle f,I_x - I_{x +h}\rangle _i|\le \frac{\pi }{2}||f||_i\cdot ||I_x-I_{x+h}||_{i}. \end{aligned}$$
Now we may compute the isoperimetric norm of the difference of diangles \(I_x-I_{x+h}\). Namely, using the formula for the isoperimetric norm from [8] we have:
$$\begin{aligned} ||I_{x+h}-I_x||^{2}_{i}= & {} \frac{1}{4\pi ^2}\left[ 2(o(I_x)-o(I_{x+h}))^2 - 4\pi m ([I_{x+h},I_x])\right] \nonumber \\= & {} \frac{1}{4\pi ^2}\cdot 4\pi \cdot 4 \cdot 1\cdot 1 \cdot \sin |h| = \frac{4}{\pi }\sin |h|. \end{aligned}$$
However, if we want to have an independent proof of Theorem 6, we must compute the norm \(||I_{x+h}-I_x||^2\) using formula (17). A direct calculation of the definite integrals which are in (17) is perhaps not to difficult, but is rather time-absorbing. Thankfully, we have the reproducing property (just proved). Hence
$$\begin{aligned} ||I_\varphi - I_\psi ||^2= & {} \langle I_\varphi - I_\psi ,I_\varphi - I_\psi \rangle _i = \frac{2}{\pi }\langle I_\varphi - I_\varphi ,\frac{\pi }{2}I_\psi - 2, 2-\frac{\pi }{2}I_\psi \rangle _i \\= & {} \frac{2}{\pi }\langle I_\varphi - I_\psi ,-k_\psi +k\varphi \rangle _i \\= & {} \frac{2}{\pi }\left( \langle I_\varphi ,-k_\varphi \rangle _i + \langle I_\psi ,k_\varphi \rangle _i+ \langle I_\varphi ,k_\psi \rangle _i -\langle I_\psi ,k_\psi \rangle _i\right) \\= & {} \frac{2}{\pi }(-0 + I_\varphi (\psi ) + I_\psi (\varphi ) + 0) \\= & {} \frac{4}{\pi }\sin |\varphi -\psi |. \end{aligned}$$
Setting \(h=\varphi -\psi \) we obtain the desired equality. In consequence
$$\begin{aligned} ||I_{x+h}-I_x||_i \le \sqrt{\frac{4}{\pi }}\sqrt{|h|}, \end{aligned}$$
and hence
$$\begin{aligned} |f(x+h)-f(x)|\le \sqrt{\pi }\cdot \sqrt{|h|}(||f||_i). \end{aligned}$$
Setting \(x=0\) and taking f such that \(f(0)=0\) we obtain
$$\begin{aligned} |f(h)|\le \sqrt{\pi }\cdot {\sqrt{|h|}}(||f||_i). \end{aligned}$$
Hence all evaluation functionals \((e_t)_{t\in \Delta }\) are bounded. Moreover the norm of uniform convergence on \(\Delta \) is weaker than the convergence with respect to the isoperimetric norm.
To end the proof of Theorem 6 it remains to show, that \(H^{1}_{0}(\Delta )\) with the isoperimetric norm is complete. Take then a sequence \((f_n)\), which is a Cauchy sequence with respect to the isoperimetric norm. Hence
$$\begin{aligned} {\pi ^2}||f_n-f_m||^2 = o^2(f_n-f_m) + E(f_n-f_m). \end{aligned}$$
By our assumption both sequences \(o^2(f_n-f_m)\) and \( E(f_n-f_m)\) tend to 0 (when \(m,n \rightarrow \infty \)). But
$$\begin{aligned} E(f_n-f_m)= o^2(f_n-f_m) -\pi \left( \int (f_n-f_m)^2 - \int (f'_n-f'_m)^2\right) . \end{aligned}$$
Since \(o^2(f_n-f_m)\) tends to 0 then \((\int (f_n-f_m)^2 -\int (f'_n-f'_m)^2)\) tends to 0. But, by the remark made above, the sequence \((f_n)\) being a Cauchy sequence with respect to the isoperimetric norm is also a Cauchy sequence with respect to the norm of uniform convergence on the compact interval \([-\frac{\pi }{2}, \frac{\pi }{2}]\). Hence \(\int (f_n-f_m)^2\) tends to 0 and in consequence \(\int (f'_n-f'_m)^2)\) tends to 0. This means, in particular, that \((f'_n)\) is a Cauchy sequence in \(L^2[-\frac{\pi }{2}, \frac{\pi }{2}]\), then \(f'_n\) tends to a function h from \(L^2[-\frac{\pi }{2}, \frac{\pi }{2}]\) with respect to the \(L^2\) norm. But \(f_n\) tends uniformly to a continuous function f. Since
$$\begin{aligned} f(x) = \lim _{n}f_n(x) = \lim _n\int _{-\frac{\pi }{2}}^{x} f'_n(t)dt =\int _{-\frac{\pi }{2}}^{x} h(t)dt, \end{aligned}$$
then f is absolutely continuous and since \(f_n(-\frac{\pi }{2})=f_n(\frac{\pi }{2})\) then also \(f(-\frac{\pi }{2})=f(\frac{\pi }{2})\). Thus \(f\in H^1_{0}(\Delta )\) and \(||f_n-f||_i\) tends to 0. Hence \( H^1_{0}(\Delta )\) is complete with respect to the isoperimetric norm. \(\square \)

We will end this chapter by a number of remarks.

Remark 7

The last theorem asserts, that the space \(H_0(\Delta )\) with the norm
$$\begin{aligned} ||f||_i^2= & {} \frac{1}{\pi ^2}\cdot \left( 2\cdot \left( \int _{-\pi /2}^{\pi /2}f\right) ^2 -\pi \int _{-\pi /2}^{\pi /2}((f)^2-(f')^2))\right. \nonumber \\= & {} \frac{1}{\pi ^2}(o^2(f) + E(f)) \end{aligned}$$
is isometric to the space \(X_\mathcal {S}\) of generalized convex sets with the norm
$$\begin{aligned} ||[U,V]||^2 =\frac{1}{4\pi ^2}[2o^2([U,V])- 4\pi \cdot m([U,V]), \end{aligned}$$
where \(o([U,V]) = o(U)-o(V)\) is the generalized perimeter and
$$\begin{aligned} m([U,V])=2m(U)+2m(V)-m(U+V) \end{aligned}$$
is the generalized Lebesgue measure in the plane. This is true since both spaces have the same (isoperimetric) kernel. The details concerning the formulas (34) and (35) are described in [8]. The isometry between \(X_\mathcal {S}\) and \(H^1_0(\Delta )\) is established by the correspondence
$$\begin{aligned} \mathcal {S}\times \mathcal {S} \ni [U,V]\rightarrow f_{[U,V]}\in H^1_o(\Delta ) \end{aligned}$$
$$\begin{aligned} f_{[U,V})(\varphi ) = \overline{U}(I_\varphi ) - \overline{V}(I_{\varphi }), \end{aligned}$$
and where \(\overline{U}(I_\varphi )\) denotes the width of the set U with respect to the straight line generated by the diangle \(I_\varphi \), (see [8]).
This means, that the term \(\int f = \int f_{[U,V]}\) is (because of the isometry) the perimeter of [UV] (in particular of U) and the term
$$\begin{aligned} \int _{-\pi /2}^{\pi /2}(f_{[U,V]})^2-(f_{[U,V]}')^2= m([U,V]) \end{aligned}$$
equals to the measure of [UV]. This fact is known for a long time as the so-called Cauchy formula (true not only for centrally symmetric, but for all convex and compact sets). Let us notice that (6) may be considered as an extension of the Cauchy formulas for generalized measure.

Remark 8

Let us come back to the inequality (32) proved above, and saying that for each \(f\in H^1_0(\Delta )\), for each \(x\in \Delta \) and \(h\rangle 0\) the following inequality holds:
$$\begin{aligned} |f(x+h)-f(x)|\le \sqrt{\pi }\cdot ||f||_i\sqrt{h}. \end{aligned}$$
Setting \(x+h=x_2\) , \(x=x_1\) and \(\sqrt{\pi } ||f||_i = C(f)\) we obtain the evaluation
$$\begin{aligned} |f(x_2)-f(x_1)|\le C(f)\sqrt{|x_2-x_1|} = C(f)|x_2-x_1|^{\frac{1}{2}}. \end{aligned}$$
This means, that each function \(f\in H^1_0(\Delta )\) (each \([U,V]\in X_\mathcal {S}\) ) satisfies the Hölder condition with the exponent \(\frac{1}{2}\). Hence the function
$$\begin{aligned} u(x) = |x|^{\frac{1}{3}}\notin X_\mathcal {S} \end{aligned}$$
and the diangles are examples, that the exponent \(\frac{1}{2}\) cannot be improved. This is the known Hölder condition in Sobolev spaces and we see that it is not too difficult to prove, when we work in the representation \(H^1_0(\Delta )\). On the other hand, this is not easy to observe, if we work with generalized convex sets in \(X_\mathcal {S}\).

Remark 9

The third remark concerns the coefficient 2 in the formula
$$\begin{aligned} ||[U,V]||^2 =\frac{1}{4\pi ^2}[2o^2([U,V])- 4\pi \cdot m([U,V]), \end{aligned}$$
or in the formula (17). The generalized isoperimetric inequality
$$\begin{aligned} o^2([U,V])-4\pi m([U,V])\ge 0, \end{aligned}$$
or more exactly the inequality
$$\begin{aligned} o^2(x)-4\pi m(x)\ge 0 \end{aligned}$$
for \(x\in X_\mathcal {S}\) gives only a seminorm on \(X_\mathcal {S}\) with one-dimensional kernel \({{\mathbb {R}}}\cdot \mathbf{1}\). Thus as it was remarked in the Introduction, to have the ”true” norm one must add ”something” non-vanishing on the line \({{\mathbb {R}}}\cdot \mathbf{1}\). In this (and the previous [8] paper, this added term is the seminorm \(o^2([U,V])\) (more exactly \(o^2(x)\)—a square of the linear functional—for \(x\in X_\mathcal {S}\)). It is clear, that one may consider the family of norms of the form
$$\begin{aligned} ||[U,V]||^2 =\frac{1}{4\pi ^2}[\theta \cdot o^2([U,V])- 4\pi \cdot m([U,V]), \end{aligned}$$
where \(\theta > 1\).
The corresponding reproducing kernel has the following form:
$$\begin{aligned} K_\theta (\varphi ,\psi )= \theta -\frac{\pi }{2}\sin |\varphi -\psi |, \end{aligned}$$
where \(\theta \) is a greater than 1. For \(\theta =2\) we obtain the kernel \(K_i\). Let us observe additionally, that for \(\theta =1\) we obtain an interesting (since canonical) kernel
$$\begin{aligned} K_1(x,y) = 1-\frac{\pi }{2}\sin |x-y|, \end{aligned}$$
corresponding to the subspace of \(X_o\subset X_\mathcal {S}\), which is the kernel of the linear functional (perimeter) \(x\longrightarrow o(x)\).

4 A Sequence Model

The construction of the RKHS space \(H^1_0(\Delta )\) presented above is frequently called from space to kernel (see e.g. [6]). In this section we will construct the same function space—i.e. \(X_\mathcal {S}\)—but in a way, which is called from kernel to space. The aim for which we repeat this commonly known construction is that it gives a possibility to see a kind of finite dimensional version of the generalized isoperimetric inequality. As usual, in this model the space \(X_\mathcal {S}\) will appear to be the completion of a certain function space \(\mathcal {F}\) with respect to a suitable inner product (suitable norm), corresponding to a given isoperimetric kernel. The idea we will describe below in details, was used in [8] in many places, but in an implicit form.

4.1 Definition of a Certain Sequence Space

Let, as above \(\Delta = [-\pi /2,\pi /2]\) and let us consider the family of diangles, i.e. the family of functions \((I_\varphi )_{\varphi \in \Delta }\), where
$$\begin{aligned} I_{\varphi }:\Delta \ni \psi \longrightarrow I_{\varphi }(\psi )=\sin |\varphi -\psi |. \end{aligned}$$
Let \(\mathcal {F}\) denote the subspace of the vector space \({\mathbb R}^{\Delta }\), generated by the constant function \(\mathbf {1}\) and by the diangles i.e.
$$\begin{aligned} \mathcal {F}=\left\{ x_o \cdot \mathbf {1} + \sum _{i=1}^{k} x_i \cdot I_{\varphi _i} : x_i\in {\mathbb {R}}, k=1,2,...\right\} . \end{aligned}$$
Let us observe, that the space \(\mathcal {F}\) is exactly the linear space generated by the kernel functions \(k_\psi = 2\cdot \mathbf{{1}}-\frac{\pi }{2}I_\psi \) of the considered kernel \(K_i(\varphi ,\psi )= 2\cdot \mathbf{1}-\frac{\pi }{2}\sin |\varphi -\psi |.\) Now we will define a certain bilinear form \(\langle ,\rangle _i\) in \(\mathcal {F}\). Take two elements \(x,y \in \mathcal {F}\) where
$$\begin{aligned} x=x_o\cdot \mathbf {1} + \sum _{i=1}^{k} x_i \cdot I_{\varphi _i}, \end{aligned}$$
$$\begin{aligned} y=y_o\cdot \mathbf {1} + \sum _{j=1}^{m} y_j \cdot I_{\psi _j}. \end{aligned}$$
Since \(\langle ,\rangle _i\) is claimed to be bilinear, then
$$\begin{aligned}&\langle x,y\rangle _i=\langle x_o\cdot \mathbf {1} + \sum _{i=1}^{k} x_i \cdot I_{\varphi _i},y_o\cdot \mathbf {1} + \sum _{j=1}^{m} y_j \cdot I_{\psi _j}\rangle _i = x_o\cdot y_o \langle \mathbf {1},\mathbf {1}\rangle _i \\&\qquad + \sum _{j=1}^{m}x_oy_j\langle \mathbf {1},I_{\psi _j}\rangle _i + \sum _{i=1}^{k}\langle \mathbf {1},I_{\varphi _i}\rangle _i+ \sum _{i=1}^{k} \sum _{j=1}^{m} x_iy_j\langle I_{\varphi _i}, I_{\psi _j}\rangle _i. \end{aligned}$$
Setting \({\langle \mathbf 1,\mathbf 1 \rangle }_i=k_{oo}\), \(\langle \mathbf{1},I_\varphi \rangle _i = k_{o,\varphi }\), \(\langle I_\varphi ,I_\psi \rangle _i=k_{\varphi ,\psi }\) where \(k_{oo}, k_{o,\varphi }, k_{\varphi ,\psi }\) are arbitrarily chosen real numbers, one always obtains a bilinear form, but not necessarily positively defined. Let us set
$$\begin{aligned} k_{oo}= & {} \langle \mathbf {1},\mathbf {1}\rangle _i=1, \end{aligned}$$
$$\begin{aligned} k_{o,\varphi }= & {} \langle \mathbf {1},I_{\varphi }\rangle _i=\frac{2}{\pi }, \end{aligned}$$
$$\begin{aligned} k_{\varphi ,\psi }=\langle I_{\varphi },I_{\psi }\rangle _i=\frac{4}{{\pi }^2}\left( 2-\frac{\pi }{2}\cdot \sin |\varphi _i - \psi _j|\right) . \end{aligned}$$
Now we set

Definition 10

For the vectors \(x,y \in \mathcal {F}\) defined as above we set
$$\begin{aligned} \langle x,y\rangle _i= & {} x_oy_o + \frac{2}{\pi }\sum _{i=1}^{k}x_iy_o + \frac{2}{\pi }\sum _{j=1}^{m}x_oy_j \nonumber \\&+ \frac{4}{\pi ^2} \sum _{i=1}^{k} \sum _{j=1}^{m}\left( 2-\frac{\pi }{2}\cdot \sin |\varphi _i - \psi _j|\right) x_iy_j. \end{aligned}$$

The coefficients in the formula (42) are chosen in such a way, that one may easily check directly the reproducing property of the form \(\langle ,\rangle _i\). In other words it is easy to verify, that \(\mathcal {F}\) equipped with the scalar product given by the (42) is identical (isometric) with the subspace of the space \(X_{\mathcal {S}}\) spanned by the unit disc and all diangles. In consequence \(\langle ,\rangle _i\) given by (42) is really an inner product (is positively defined) and the completion of \(\mathcal {F}\) equals \(X_{\mathcal {S}}\).

Setting in (42) \(x=y\) we obtain the following formula for the (isoperimetric) norm in \(\mathcal {F}\). Namely
$$\begin{aligned} ||x||_i^2 = x_o^2 + \frac{4}{\pi }\sum _{i=0}^{k}x_ox_i + \frac{4}{\pi ^2}\sum _{i=1}^{k}\sum _{j=1}^{k}\left( 2-\frac{\pi }{2} \sin |\varphi _i-\varphi _j|\right) x_ix_j. \end{aligned}$$
This formula may be rewritten in the form
$$\begin{aligned} ||x||_i^2= & {} \frac{4}{\pi ^2}\cdot \left[ \left( \frac{\pi }{2}x_o+\sum _{i=1}^{k}x_i\right) ^2 - \left( \sum _{i=1}^{k}x_i\right) ^2 \right. \nonumber \\&\left. +\sum _{i=1}^{k} \sum _{j=1}^{k}\left( 2-\frac{\pi }{2}\sin |\varphi _i-\varphi _j|\right) x_ix_j\right] , \end{aligned}$$
or in the form
$$\begin{aligned} ||x_||_i^2= & {} \frac{4}{\pi ^2}\cdot \left[ \left( \frac{\pi }{2}x_o+\sum _{i=1}^{k}x_i\right) ^2 - \left( \sum _{i=1}^{k}x_i\right) ^2+ 2\sum _{i,j=1}^{k}x_ix_j \right. \nonumber \\&\left. \left. - \frac{\pi }{2}\sum _{i=1}^{k}\sum _{j=1}^{k}\sin |\varphi _i-\varphi _j|\right) x_ix_j\right] . \end{aligned}$$
Now, since \(||x||^2\ge 0\) then we obtain the following inequality:
$$\begin{aligned}&\left( \frac{\pi }{2}x_o+\sum _{i=1}^{k}x_i\right) ^2 + 2\sum _{i,j=1}^{k}x_ix_j\ge \left( \sum _{i=1}^{k}x_i\right) ^2 \nonumber \\&\qquad + \frac{\pi }{2}\sum _{i=1}^{k}\sum _{j=1}^{k}\sin |\varphi _i-\varphi _j| x_ix_j. \end{aligned}$$
This inequality may be considered as the isoperimetric inequality for sequences. If we put \(x_0=0\) we obtain the form
$$\begin{aligned} 2\sum _{i,j=1}^{k}x_ix_j\ge \frac{\pi }{2}\sum _{i=1}^{k}\sum _{j=1}^{k} (\sin |\varphi _i-\varphi _j|)x_ix_j, \end{aligned}$$
or, finally, the form
$$\begin{aligned} \sum _{i=1}^{k}\sum _{j=1}^{k}(\sin |\varphi _i-\varphi _j|)x_ix_j\le \frac{4}{\pi }\left( \sum _{i,j=1}^{k}x_ix_j\right) ,(=\frac{4}{\pi }(\sum _{i=1}^{k}x_i)^2). \end{aligned}$$
Let us remark, that in our interpretation, the sum of diangles \(W=\sum _{i=1}^{k}x_iI_{\varphi _i}\) represents (for non-negatives \(x_i\)) a polygon with sides \(I_{\varphi _i}\). Since the perimeter of a diangle is \(o(I_\varphi )= 4x_i\) then the number \((\sum _{i=1}^{k}x_i)^2\) equals \(\frac{1}{16}o^2(W)\), so, for polygon W the last inequality gives
$$\begin{aligned} \sum _{i=1}^{k} \sum _{j=1}^{k}\sin |\varphi _i-\varphi _j| x_ix_j\le \frac{1}{4\pi } o^2(W). \end{aligned}$$
It is also not hard to check, that the area of the polygon \(W=\sum _{i=1}^{k}x_iI_{\varphi _i}\) is given by
$$\begin{aligned} area(W) = m(W) = \sum _{i=1}^{k}\sum _{j=1}^{k}\sin |\varphi _i-\varphi _j| x_ix_j. \end{aligned}$$
Hence the classical isoperimetric inequality for sequences says simply, that the isoperimetric inequality is valid for polygons (with positive sides). Explicitely (for W as above)
$$\begin{aligned} \sum _{i=1}^{k}\sum _{j=1}^{k}\sin |\varphi _i-\varphi _j| x_ix_j = m(W)\le \frac{1}{4\pi }o^2(W)= \frac{4}{\pi }\left( \sum _{i=1}^{k}x_i\right) ^2. \end{aligned}$$
Let us remark, that as we have proved, the last inequality is valid for all and not only for non-negative sequences \((x_i)\).



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Authors and Affiliations

  1. 1.Jagiellonian UniversityKrakówPoland
  2. 2.State Higher Vocational School in TarnowTarnówPoland

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