\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 85, pp. 1--22.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2006 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2006/85\hfil A topology on inequalities] {A topology on inequalities} \author[A. M. D'Aristotile, A. Fiorenza\hfil EJDE-2006/85\hfilneg] {Anna Maria D'Aristotile, Alberto Fiorenza} % in alphabetical order \address{Anna Maria D'Aristotile Dipartimento di Costruzioni e Metodi Matematici in Architettura, Universit\`a di Napoli "Federico II", via Monteoliveto 3, 80134 Napoli, Italy} \email{daristot@unina.it} \address{Alberto Fiorenza \newline Dipartimento di Costruzioni e Metodi Matematici in Architettura, Universit\`a di Napoli "Federico II", via Monteoliveto 3, 80134 Napoli, Italy and Istituto per le Applicazioni del Calcolo "Mauro Picone", Consiglio Nazionale delle Ricerche, via Pietro Castellino 111, 80131 Napoli, Italy} \email{fiorenza@unina.it} \date{} \thanks{Submitted July 12, 2006. Published August 2, 2006.} \subjclass[2000]{46E99, 54C30, 54A20, 54B15} \keywords{Real analysis; topology; inequalities; homogeneous operators; \hfill\break\indent Banach spaces; Orlicz spaces; Sobolev spaces; norms; density} \begin{abstract} We consider sets of inequalities in Real Analysis and construct a topology such that inequalities usually called ``limit cases'' of certain sequences of inequalities are in fact limits - in the precise topological sense - of such sequences. To show the generality of the results, several examples are given for the notions introduced, and three main examples are considered: Sequences of inequalities relating real numbers, sequences of classical Hardy's inequalities, and sequences of embedding inequalities for fractional Sobolev spaces. All examples are considered along with their limit cases, and it is shown how they can be considered as sequences of one ``big'' space of inequalities. As a byproduct, we show how an abstract process to derive inequalities among homogeneous operators can be a tool for proving inequalities. Finally, we give some tools to compute limits of sequences of inequalities in the topology introduced, and we exhibit new applications. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{example}[theorem]{Example} \newtheorem{remark}[theorem]{Remark} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{definition}[theorem]{Definition} \section{Introduction} In Analysis it is frequent that authors consider inequalities that are limiting cases of sequences of inequalities, or, more generally, of a parametrized set of inequalities. The goal of this paper is to construct a topology such that inequalities usually called ``limit cases'' of certain sequences of inequalities are in fact limits - in the precise topological sense - of such sequences of inequalities. Such kind of problem can be studied from several points of view, because, for instance, it is possible - in a quite general, abstract setting - to speak about inequalities in ordered sets; moreover, even confining ourselves for instance to inequalities involving real functions, several notions of convergence can be considered. Our point of view has to be considered only as a first approach to the problem, which seems new. \section{The main question through examples} To give an idea of the general setting of our results, we will examine three examples of sequences of inequalities. The first one is the case of elementary numerical inequalities. The second one will be the classical integral inequality, known as \sl Hardy's inequality \rm and, finally, we conclude with a recent version of the Sobolev inequality for fractional Sobolev spaces. \subsection{Inequalities relating real numbers, part I} Let $(a_{n})$, $(b_{n})$ be sequences of positive real numbers such that $a_{n}\to a>0$, $b_{n}\to b>0$, and \begin{equation} a_{n}\le b_{n}\quad \forall n\in\mathbb{N} \label{numerireali} \end{equation} From elementary Analysis we know that one can ``pass to the limit'' in \eqref{numerireali}, obtaining \begin{equation} a\le b \label{limnumerireali} \end{equation} The question here is to identify the inequalities in \eqref{numerireali} with a sequence of ``Inequalities'' in a suitable topological space, let's call it ({$\mathcal{I}$},$\tau$), and to show that the limit of such sequence, in the space ({$\mathcal{I}$},$\tau$), is the ``Inequality'' identified with \eqref{limnumerireali}. \subsection{Hardy's inequality, part I} Let $p>1$, $f$ be a nonnegative (Lebesgue) measurable function on $(0,1)$. The classical Hardy's integral inequality states that (\cite{HLP}) \begin{equation} \int_{0}^{1} \Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)^{p}dx \le \Big( \frac{p}{p-1}\Big)^{p} \int_{0}^{1} f^{p}(x)dx \label{hardy} \end{equation} When $p\to 1+$ the constant (which is the best one such that \eqref{hardy} holds) $\big( \frac{p}{p-1}\big)^{p}$ blows up, and this leads immediately to conjecture that it cannot exist a constant $c>0$ such that the inequality $$ \int_{0}^{1} \Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx \le c \int_{0}^{1} f(x)dx \label{hardyp=1bad} $$ holds for any $f$. This conjecture is in fact true: it is sufficient to consider the sequence $f_n(x)=x^{-1+1/n}$ . Nevertheless, the ``limiting'' case of \eqref{hardy}, when $p\to 1+$, can be expressed through the norm of the Zygmund space $LlogL(0,1)$: \begin{equation} \int_{0}^{1} \Big(\frac{1}{x} \int_{0}^{x}f(t)dt\Big) dx \le C \| f \|_{LlogL(0,1)}, \label{hardyp=1good} \end{equation} where \begin{equation} \| f \|_{LlogL(0,1)}=\inf \big\{ \lambda>0:\int_0^1 | \frac{f(x)}{\lambda}| \log \big(e+ | \frac{f(x)}{\lambda}| \big)dx\leq 1\big\}. \label{zyg} \end{equation} For a recent digression on equivalent norms in $LlogL$ see e.g. \cite{FK}. As in the previous example, setting e.g. $p=1+1/n$ in \eqref{hardy}, it is natural to ask whether in some topological space it is really true that the sequence of Hardy's inequalities \sl converges \rm to the inequality \eqref{hardyp=1good}. \subsection{Sobolev inequalities for fractional Sobolev spaces, part I} Let $\Omega\subset\mathbb{R}^N$ $(N\ge 1)$ be a bounded smooth open set, and let $p\ge 1$. Consider the classical Sobolev space $W_0^{1,p}(\Omega)$, defined as the completion of $\mathcal{C}_0^\infty(\Omega)$ - the set of all functions defined on $\Omega$ which have derivatives of any order on $\Omega$, whose supports are compact sets - in the norm $\| \nabla f\|_{L^p(\Omega)}$. Let $00}T\big( \frac{f}{\lambda}\big) =T({\bf 0}):=m_T\in [0,+\infty[\quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) \label{inzero} \end{gather} For $T$, $S$ in $\mathcal{O}$ we shall often write $$ d=d(T,S),$$ instead of $$ Tf\le Sf \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) $$ We observe that the operators that we consider are very common in Analysis, because many known inequalities have both sides enjoying the properties listed above. Before listing some examples, let us recall some definitions which will be useful in the sequel. A function $A:[0,+\infty[\to[0,+\infty[$ is called \emph{N-function} if it is continuous, strictly increasing, convex and such that $$ \lim_{t\to0}\frac{A(t)}t=0,\quad \lim_{t\to\infty}\frac{A(t)}t=\infty. $$ Typical examples of N-functions are powers, with exponent greater than $1$. Starting from the notion of N-function it is possible to consider the norm $$ \|f\|_A=\| f \|_{L^A(0,1)}=\inf \big\{ \lambda>0:\int_0^1 A \big(\big| \frac{f(x)}{\lambda}\big| \big)dx\leq 1\big\} $$ which defines the \emph{Orlicz} space $L^A(0,1)$. If $A(t)=t\log (e+t)$, we get the norm considered in \eqref{zyg}; in this case the Orlicz space is called \emph{Zygmund} space. We refer to \cite{FK} for expressions for the norm in such spaces. For properties and further examples of N-functions and Orlicz spaces see e.g. \cite{A}. Let $f$ be a (Lebesgue) measurable function defined on $(0,1)$, a.e. finite, and for any Lebesgue measurable set $E\subset (0,1)$ let $|E|$ be its measure. The \emph{decreasing rearrangement} of $f$ is the function, denoted by $f^*$, defined by $$ f^*(t)=\inf\{ \lambda>0 : | \{ x\in (0,1) : |f(x)|>\lambda \} |\le t \} \quad t\in (0,1) $$ This definition is usually given for a much more general class of functions, but it is not in our purposes to give details here. For interested readers we refer to \cite{BS}. \begin{example} \label{exa4.1}\rm Let $X$ be a Banach space whose elements are measurable functions. Suppose that $\mathcal{C}_0^\infty(\Omega)\subset X$. An example of operator in $\mathcal{O}$ is $$ T_1f=\|f\|_X \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega). $$ \end{example} \begin{example} \label{exa4.2} \rm Let $p$ be a measurable function on $\Omega$, whose values are in $[1,\infty]$, and set $\Omega_\infty=\{ x\in\Omega : p(x)=\infty\}$. Then an example of operator in $\mathcal{O}$ is $$ T_2f=\inf\big\{ \lambda>0: \int_{\Omega\backslash\Omega_\infty}\big| \frac{f(x)}\lambda \big|^{p(x)}dx+\mathop{\rm ess\,sup}_{\Omega_\infty} \big| \frac{f(x)}\lambda \big|\le1 \big\} $$ This operator is a special case of the previous example, in fact it is the norm in the space $L^{p(\cdot)}(\Omega)$. For details see \cite{KR}. \end{example} \begin{example} \label{exa4.3}\rm Let $A:[0,+\infty[\to[0,+\infty[$ be an N-function. Then an example of operator in $\mathcal{O}$ is $$ T_3f=\int_\Omega A(f)dx \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega). $$ \end{example} \begin{example} \label{exa4.4} \rm Let $A_1, A_2, A_3:[0,+\infty[\to[0,+\infty[$ be continuous and strictly increasing functions such that $A_i(+\infty)=+\infty$, $i=1,2,3$, and let $w_1, w_2$ be nonnegative, locally integrable functions defined respectively in $\Omega\times\Omega$ and $\Omega$, such that $$ w_1(x, \cdot)\not\equiv 0\quad \forall x\in\Omega, \quad w_2>0 \quad\text{a.e. in }\Omega. $$ Then an example of operator in $\mathcal{O}$ is $$ T_4f=A_1\Big( \int_\Omega A_2\Big( \int_\Omega A_3(f(y))w_1(x,y)dy \Big)w_2(x)dx\Big) \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) $$ Next we consider the particular case: $\Omega=]0,1[\subset\mathbb{R}$, $A_1(t)=A_2(t)=A_3(t)=t$, $w_1(x,y)=\chi_{(0,x)}(y)$, $w_2(x)=1/x$, which gives the operator $$ T_5=\int_0^1 \Big( \frac1x \int_0^x f(t)dt\Big) dx \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) $$ \end{example} \begin{example} \label{exa4.5} \rm Let $A_1, A_2:[0,+\infty[\to[0,+\infty[$ be N-functions and let $X$ be a Banach Function Space. Then an example of operator in $\mathcal{O}$ is $$ T_6f=A_1( \|A_2(f)\|_X) \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega). $$ \end{example} \begin{example} \label{exa4.6}\rm Let us denote by $f^*$ the decreasing rearrangement of $f$, defined in the interval $]0, |\Omega|]$. An example of operator in $\mathcal{O}$ is $$ T_7f=f^*\big(\frac{|\mathop{\rm supp}(f)|}{2}\big)\quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) $$ where $|\mathop{\rm supp}(f)|$ denotes the Lebesgue measure of the support of $f$. \end{example} \begin{example} \label{exa4.7} \rm Let us denote by $f^*$ the decreasing rearrangement of $f$, defined in the interval $]0, |\Omega|]$, let $A_1$ be a strictly increasing, continuous function on $[0, |\Omega|]$ such that $A_1(0)=0$, $A_1(+\infty)=+\infty$, let $A_2:[0,+\infty[\to[0,+\infty[$ be an N-function, and let $w_1, w_2$ be positive, locally integrable functions defined in $]0,|\Omega|[$. Then an example of operator in $\mathcal{O}$ is $$ T_8f=\int_0^{|\Omega|}A_1\Big( \int_0^t A_2 (f^*(s))w_1(s)ds\Big)w_2(t)dt \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) $$ Operators of this type occur in Function Space Theory, see e.g. \cite{FKa}. \end{example} Let us now introduce a notion which will play a key role in the sequel. For each $T\in \mathcal{O}$ we define the \emph{associate family $\{T^{(\mu)}\}$ of homogeneous operators} as the class of operators $T^{(\mu)}: \mathcal{C}_{0,+}^\infty(\Omega) \to [0,+\infty[$ defined by $$ T^{(\mu)}f=\inf \big\{ \lambda>0: T\big( \frac{f}{\lambda}\big)\le \mu\big\}\quad\forall \mu>m_T\quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) $$ We remark that since $T\in \mathcal{O}$ the set $$ \big\{ \lambda>0: T\big( \frac{f}{\lambda}\big)\le \mu\big\} $$ is nonempty for all $f\in \mathcal{C}_{0,+}^\infty(\Omega)$ and all $\mu>m_T$, so that the definition of $T^{(\mu)}f$ is well posed for all $\mu>m_T$. The homogeneity of the operator $T^{(\mu)}$ is proved by the following result. \begin{proposition} If $T$ is admissible, then for all $\mu>m_T$ the operator $T^{(\mu)}$ is admissible and has the further property to be homogeneous (of degree $1$): $$ T^{(\mu)}(kf)=kT^{(\mu)}f\quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) \; \forall k\ge 0. $$ \label{homo} \end{proposition} \begin{proof} First we prove the homogeneity. If $k=0$ it is sufficient to see that \begin{equation} T^{(\mu)}{\bf 0}=0\quad \forall \mu>m_T, \label{T0} \end{equation} and this is trivial, because for all $\lambda>0$ it is $$ T\big( \frac{{\bf 0}}{\lambda}\big)=T{\bf 0}\le \mu\quad \forall \mu>m_T. $$ For $k>0$ we have \begin{align*} T^{(\mu)}(kf) &=\inf \big\{ \lambda>0: T\big(\frac{kf}{\lambda}\big)\le \mu\big\}\\ &=\inf \big\{ k\lambda>0: T\big( \frac{f}{\lambda}\big)\le \mu\big\} \\ &=k\inf \big\{ \lambda>0: T\big( \frac{f}{\lambda}\big)\le \mu\big\} \\ &=kT^{(\mu)}(f) \end{align*} Now we show that the operator $T^{(\mu)}$ is admissible. Property \eqref{cont} is trivial, because the homogeneity of $T^{(\mu)}$ implies that $$ \lambda\in [0,+\infty[ \to T^{(\mu)}(\lambda f)\in [0,+\infty[ $$ is linear. In order to show properties \eqref{incr} and \eqref{infinity}, because of the homogeneity of $T^{(\mu)}$, it is sufficient to see that $$ f\neq {\bf 0} \Longrightarrow T^{(\mu)}(f)>0 $$ Since $T$ is admissible, property \eqref{infinity} is true for $T$, therefore there exists $\lambda>0$ such that $ T\big( \frac{f}{\lambda}\big)>\mu$. The conclusion is that $T^{(\mu)}(f)>0$. Finally, we observe that both sides of \eqref{inzero} are equal to zero: the left hand side because of the homogeneity of $T^{(\mu)}$, the right hand side because of \eqref{T0}. \end{proof} \begin{remark} \rm If $T$ is homogeneous, then it is easy to show that $T^{(\mu)}=(1/\mu)T$ for all $\mu>0$. \label{Thomo} \end{remark} The main result of this Section is the following. \begin{theorem} Let $T,S\in \mathcal{O}$. The following equivalence holds: $$ Tf\le Sf\quad\forall f\in \mathcal{C}_{0,+}^\infty(\Omega) $$ if and only if $$ T^{(\mu)}f\le S^{(\mu)}f \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) \quad \forall \mu>\max(m_T, m_S) $$ \label{homogen} \end{theorem} \begin{proof} Let us assume first that \begin{equation} Tf\le Sf\quad\forall f\in \mathcal{C}_{0,+}^\infty(\Omega) \label{lhs} \end{equation} and let $\mu>\max(m_T, m_S)$. Fix $f\in \mathcal{C}_{0,+}^\infty(\Omega)$ and let $\lambda>0$ be such that $S\big(\frac{f}{\lambda}\big)\le \mu$. By \eqref{lhs} the number $\lambda$ is also such that $T\big(\frac{f}{\lambda}\big)\le \mu$; therefore, $$ \big\{ \lambda>0: T\big( \frac{f}{\lambda}\big)\le \mu\big\}\supseteq \big\{ \lambda>0: S\big( \frac{f}{\lambda}\big)\le \mu\big\} $$ from which the first part of the assertion follows. \par On the other hand, by contradiction, let us assume that there exists $ \overline{f}$ such that $$ T\overline{f}>S\overline{f} $$ We observe that $\overline{f}$ can be always chosen different from ${\bf 0}$. In fact, if $$ T{\bf 0}>S{\bf 0} $$ then, fixing any $f\neq {\bf 0}$, by property \eqref{cont} for $S$ in $\lambda=0$, there exists $\lambda$ sufficiently small such that $$ T{\bf 0}>S(\lambda f)>S{\bf 0} $$ and therefore, by property \eqref{incr} for $T$, $$ T(\lambda f)>T{\bf 0}>S(\lambda f). $$ Setting $\overline{f}=\lambda f$ we get the existence of $\mu>\max(m_T,m_S)$ such that \begin{equation} T\overline{f}>\mu>S\overline{f} \label{4contr} \end{equation} Now from the inequality $S\overline{f}<\mu$, applying property \eqref{cont} for $S$ in $\lambda=1$, we can consider $\epsilon>0$ such that $$ 1-\epsilon\in \{ \lambda>0: S\big( \frac{\overline{f}}{\lambda}\big)\le \mu\} $$ By our assumption $$ \inf \big\{ \lambda>0: T\big( \frac{\overline{f}}{\lambda}\big)\le \mu\big\}= T^{(\mu)}{\overline{f}}\le S^{(\mu)}{\overline{f}}\le 1-\epsilon $$ from which $T\overline{f}\le \mu$; this conclusion is in contrast with \eqref{4contr}. \end{proof} In the sequel we will use the first implication proved above, which, starting from a generic inequality, leads to a family of inequalities relating homogeneous operators. We will say that such family of inequalities is obtained \emph{homogenizing} the original inequality. \subsection*{Application 1.} Let $\varphi:[0,+\infty[\to[0,+\infty[$ be increasing and such that $\varphi(0)=0$, and suppose to know that \begin{equation} \varphi(\| f\|_X)\le \| f\|_Y \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega), \label{appl1} \end{equation} where $X$ and $Y$ are Banach Function Spaces (see \cite[Def 1.3 p. 3]{BS}), such that $\mathcal{C}_0^\infty(\Omega)$ functions are dense in $X$ and $Y$. Applying \cite[Theorem 1.8 p. 7]{BS}, one can deduce that $Y$ is continuously embedded into $X$; therefore, there exists a constant $k>0$ such that $$ \| f\|_X\le k \|f\|_Y $$ We observe that it is possible to get the same conclusion homogenizing the inequality \eqref{appl1}, getting also an estimate of the constant $k$. In fact, by Theorem \ref{homogen}, from \eqref{appl1} we get, for all $\mu>0$ and $f\in \mathcal{C}_{0,+}^\infty(\Omega)$, \begin{gather*} \inf \big\{ \lambda>0: \varphi\big( \big\|\frac{f}{\lambda}\big\|_X\big)\le \mu\big\} \le \inf \big\{ \lambda>0: \big\|\frac{f}{\lambda}\big\|_Y\le \mu\big\}, \\ \inf \big\{ \lambda>0: \frac1\lambda \|f\|_X\le \sup\{ \xi:\varphi(\xi)\le \mu\} \big\} \le \frac1\mu \|f\|_Y, \\ \|f\|_X\le \frac{\sup\{ \xi:\varphi(\xi)\le \mu\}}\mu \|f\|_Y\,. \end{gather*} In conclusion: $$ \|f\|_X\le \inf_{\mu>0} \frac{\sup\{ \xi:\varphi(\xi)\le \mu\}}\mu \|f\|_Y\quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) $$ and therefore the same inequality is true for all $f$, due to the assumed density of the $\mathcal{C}_0^\infty(\Omega)$ functions. \subsection*{Application 2.} Let $\Phi$ be an N-function. By the well-known Jensen inequality it is $$ \Phi\big( \int_0^1 f(x)dx\big)\le \int_0^1 \Phi(f)dx \quad\forall f\in \mathcal{C}_{0,+}^\infty(\Omega) $$ Let us homogenize this inequality. For all $\mu>0$ and $f\in \mathcal{C}_{0,+}^\infty(\Omega)$ we get \begin{gather*} \inf\{ \lambda>0: \Phi\big( \int_0^1 \frac{f(x)}\lambda dx\big) \le\mu \big\} \le \inf \big\{ \lambda>0: \int_0^1\Phi\big( \frac{f(x)}\lambda \big) dx \le\mu \big\}, \\ \inf \big\{ \lambda>0: \int_0^1 \frac{f(x)}\lambda dx\le\Phi^{-1}(\mu) \big\} \le \inf \big\{ \lambda>0: \int_0^1\big(\frac\Phi\mu\big)\big( \frac{f(x)}\lambda \big) dx\le1 \big\}, \\ \frac1{\Phi^{-1}(\mu)}\int_0^1 f(x)dx\le \|f\|_{\Phi/\mu}, \\ \int_0^1 f(x)dx\le \Phi^{-1}(\mu) \|f\|_{\Phi/\mu} \end{gather*} For $\mu=1$ such inequality reduces to $$ \int_0^1 f(x)dx\le \Phi^{-1}(1) \|f\|_{\Phi} $$ which is the inequality which shows that the Orlicz space $L^\Phi(0,1)$ is embedded in $L^1(0,1)$. Notice that the constant $\Phi^{-1}(1)$ on the right hand side is optimal (the inequality becomes equality for $f\equiv 1$). \subsection*{Application 3.} Let us consider - as usual, we will consider functions $f$ in $\mathcal{C}_{0,+}^\infty$ - a well known inequality by Hardy and Littlewood (see \cite[241 (i) page 169]{HLP}, or \cite[ vol.1, p.32, Theorem 13.15(iii)]{Z}): \begin{equation} \int_0^1 \Big( \frac1x \int_0^x f(t)dt\Big) dx \le c_1 \int_0^1 f(x)\log^+f(x)dx+c_2 \label{appl3} \end{equation} which is true for some $c_1, c_2>0$. Here $\log^+t=\max (\log t, 0)$. We now compute the family of the homogenized inequalities. Since the left hand side is already homogeneous, by Remark \ref{Thomo} it is immediate to compute the associated family of homogeneous operators. Let us consider the right hand side: $$ Tf=c_1 \int_0^1 f(x)\log^+f(x)dx+c_2=\int_0^1 [c_1f(x)\log^+f(x)+c_2] dx $$ Let $c_3>0$ be such that $$ c_1t\log^+t+c_2\ge c_3t\log(e+t)\quad\forall t>0\,. $$ For all $\mu>0$ we have, \begin{align*} T^{(\mu)}(f) &=\inf\big\{\lambda>0: T\big( \frac{f}{\lambda}\big)\le \mu\big\}\\ &=\inf\big\{\lambda>0: \int_0^1 \big[c_1\frac{f(x)}{\lambda}\log^+ \frac{f(x)}{\lambda}+c_2\big] dx\le \mu\big\}\\ &\ge \inf\big\{\lambda>0: \int_0^1 c_3\frac{f(x)}{\lambda}\log \big( e+\frac{f(x)}{\lambda}\big)dx\le \mu\big\} \\ &=\inf\big\{\lambda>0: \int_0^1 \frac{c_3}\mu\frac{f(x)}{\lambda}\log \big( e+\frac{f(x)}{\lambda}\big)dx\le 1\big\} \\ &\ge \min \big(1, \frac{c_3}\mu\big)\|f\|_{LlogL(0,1)} \end{align*} where $\|f\|_{LlogL(0,1)}$ is defined in \eqref{zyg}. Last inequality is easily obtained considering separately the cases $c_3\ge \mu$, $c_3<\mu$. On the other hand, for all $\mu>c_2$ we have \begin{align*} T^{(\mu)}(f) &=\inf\big\{\lambda>0: T\big( \frac{f}{\lambda}\big)\le \mu\big\} \\ &=\inf\big\{\lambda>0: \int_0^1 \big[c_1\frac{f(x)}{\lambda}\log^+ \frac{f(x)}{\lambda}+c_2\big] dx\le \mu\big\} \\ &=\inf\big\{\lambda>0: \int_0^1c_1\frac{f(x)}{\lambda}\log^+ \frac{f(x)}{\lambda} dx\le \mu-c_2\big\} \\ &\le \inf\big\{\lambda>0: \int_0^1 \frac{c_1}{\mu-c_2}\frac{f(x)}{\lambda} \log\big( e+\frac{f(x)}{\lambda}\big) dx\le 1\big\} \\ &\le \max \big(1, \frac{c_1}{\mu-c_2}\big)\|f\|_{LlogL(0,1)} \end{align*} In conclusion, the homogenized family of inequalities of \eqref{appl3} can be written as $$ \int_0^1 \big( \frac1x \int_0^x f(t)dt\big) dx\le C(\mu) \|f\|_{LlogL(0,1)} \quad \forall\mu>c_2. $$ We observe that such inequalities are of the type \eqref{hardyp=1good}. \medskip We conclude this Section answering to a natural question: What happens if we homogenize one of the inequalities obtained after a homogenization? Let $T,S\in \mathcal{O}$ and fix a number $\mu>\max(m_T, m_S)$. If $$ Tf\le Sf\quad\forall f\in \mathcal{C}_{0,+}^\infty(\Omega) $$ by Theorem \ref{homogen} we know that $$ T^{(\mu)}f\le S^{(\mu)}f \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) $$ Let us now apply again Theorem \ref{homogen} to this inequality, and consider $$ T^{(\mu)(\sigma)}f\le S^{(\mu)(\sigma)}f \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega)\quad \forall\sigma>0 $$ By Remark \ref{Thomo} we get $$ \frac1\sigma T^{(\mu)}f\le \frac1\sigma S^{(\mu)}f \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega)\quad\forall\sigma>0 $$ i.e. $$ T^{(\mu)}f\le S^{(\mu)}f \quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) $$ The conclusion is that for each fixed $\mu>0$ the inequality $T^{(\mu)}f\le S^{(\mu)}f $ has a trivial family of associated homogeneous inequalities, pairwise identical to the original one. \section{A topology on inequalities} Let us consider the set of inequalities $$ {\mathcal{I}}_{0}=\{ d(T,S) : T,S\in {\mathcal{O}}\} $$ We construct a topology in ${\mathcal{I}}_{0}$, taking inspiration from the classical procedure used to define the topology of pointwise convergence. Fix $d=d(T,S)\in {\mathcal{I}}_{0}$. For any $n\in \mathbb{N}$, for any finite subset $F\subset \mathcal{C}_{0,+}^\infty(\Omega)$, let us set $$ {\mathcal{U}}_{n,F}(d)=\{ d'=d'(T',S'): |T'f-Tf|<\frac1n \; \forall f\in F, \; |S'f-Sf|<\frac1n \; \forall f\in F\} $$ Let ${\mathcal{N}}(d)$ be the family of subsets of ${\mathcal{I}}_{0}$ whose elements contain some set of the type ${\mathcal{U}}_{n,F}(d)$: $$ A\in {\mathcal{N}}(d) \Leftrightarrow \exists n\in\mathbb{N} , \; \exists F\subset \mathcal{C}_{0,+}^\infty(\Omega) \text{ finite: }A \supseteq {\mathcal{U}}_{n,F}(d). $$ We will prove the following result. \begin{proposition} The family $\{ {\mathcal{N}}(d) \}_{d\in {\mathcal{I}}_{0}}$ satisfies the following Hausdorff's axioms: \begin{itemize} \item[(i)] $d\in A$ for all $A\in {\mathcal{N}}(d)$ \item[(ii)]$ A\in {\mathcal{N}}(d)$ and $B\in {\mathcal{N}}(d)$ implies $A\bigcap B\in {\mathcal{N}}(d)$ \item[(iii)] $A\in {\mathcal{N}}(d)$ and $B\supseteq A$ implies $B\in {\mathcal{N}}(d)$ \item[(iv)] for all $A\in {\mathcal{N}}(d)$ there exists $B\in {\mathcal{N}}(d)$ such that $A\in {\mathcal{N}}(d')$ for all $d'\in B$ \end{itemize} \end{proposition} After this proposition we know (see e.g. \cite[Theorem 3.2, p. 67]{D}) that there exists one and only one topology ${\tau}$ for ${\mathcal{I}}_{0}$ such that ${\mathcal{N}}(d)$ is, for any $d\in {\mathcal{I}}_{0}$, the set of the nbds of $d$. \begin{proof} (i) Let $A\in {\mathcal{N}}(d)$ and let ${\mathcal{U}}_{n,F}(d)\subseteq A$. Since $d\in {\mathcal{U}}_{n,F}(d)$, it is $d\in A$. \noindent (ii) Let ${\mathcal{U}}_{n,F}(d)\subseteq A$, ${\mathcal{U}}_{m,G}(d)\subseteq B$. Then ${\mathcal{U}}_{\max (n,m),F\bigcup G}(d)\subseteq A\bigcap B$. \noindent (iii) Let $A\in {\mathcal{N}}(d)$ and let ${\mathcal{U}}_{n,F}(d)\subseteq A$. Since $B\supseteq A$, then $B\supseteq {\mathcal{U}}_{n,F}(d)$; therefore $B\in {\mathcal{N}}(d)$ \noindent(iv) Let $A\in {\mathcal{N}}(d)$ and let ${\mathcal{U}}_{n,F}(d)\subseteq A$. We show that (iv) is true with $B={\mathcal{U}}_{n,F}(d)$: $$ d'\in {\mathcal{U}}_{n,F}(d)\quad \Rightarrow \quad A\in {\mathcal{N}}(d') $$ To this goal we need to find some ${\mathcal{U}}_{\nu,F}(d')$ such that ${\mathcal{U}}_{\nu,F}(d')\subset A$. Since $d'\in {\mathcal{U}}_{n,F}(d)$, \begin{gather*} |T'f-Tf|<\frac1n\quad\forall f\in F, \\ |S'f-Sf|<\frac1n\quad\forall f\in F\,. \end{gather*} Since $F$ is finite, we may consider $\nu\in\mathbb{N}$ such that $$ \frac1\nu < \min\big\{ \min_F \{ \frac1n - |T'f-Tf| \}, \min_F \{ \frac1n - |S'f-Sf| \} \big\}. $$ Let $d''\in {\mathcal{U}}_{\nu,F}(d')$. Then \begin{gather*} |T''f-T'f|<\frac1\nu\quad\forall f\in F,\\ |S''f-S'f|<\frac1\nu\quad\forall f\in F\,. \end{gather*} We have $$ |T''f-Tf|\le |T''f-T'f|+|T'f-Tf|<\frac1\nu + |T'f-Tf| <\frac1n \quad\forall f\in F $$ and similarly $|S''f-Sf|<1/n$ for all $f\in F$. Therefore, $d''\in {\mathcal{U}}_{n,F}(d)\subseteq A$. \end{proof} At this point we have a topology ${\tau}$ in ${\mathcal{I}}_{0}$. The notion of convergence in this topology is, as usual, $$ d_n\to d\text{ in } {\tau} \quad \Leftrightarrow\quad \forall {\mathcal{U}}_{\nu,F}(d)\; \exists \, n_0\in\mathbb{N}: d_n\in {\mathcal{U}}_{\nu,F}(d) \; \forall n>n_0 $$ Set $d_n=d_n(T_n,S_n)$, $d=d(T,S)$. It is readily seen that $d_n\to d$ if and only if \begin{equation} \forall \, f\in \mathcal{C}_{0,+}^\infty(\Omega) \quad T_nf\to Tf \quad\text{and}\quad S_nf\to Sf \label{converg} \end{equation} Now, we go back to the examples considered in Section 2, and we show how a suitable choice of the operators $T_n$, $S_n$, $T$, $S$ gives that the considered inequalities converge to their respective limits in the sense of \eqref{converg}. \subsection{Inequalities relating real numbers, part II} Let $(a_{n})$, $(b_{n})$ be sequences of positive real numbers such that $a_{n}\to a>0$, $b_{n}\to b>0$, and \begin{equation} a_{n}\le b_{n}\quad \forall n\in\mathbb{N} \label{numerireali2} \end{equation} We cannot consider the most natural operators $$ T_nf\equiv a_n\quad S_nf\equiv b_n\quad\forall f\in \mathcal{C}_{0,+}^\infty(0,1) $$ because they are not admissible (notice that property \eqref{incr} does not hold). Let us set $$ T_nf\equiv a_n \sup_{(0,1)}f(x)\quad S_nf\equiv b_n\sup_{(0,1)}f(x)\quad \forall f\in \mathcal{C}_{0,+}^\infty(0,1) $$ Observe that now the operators $T_n$, $S_n$ are admissible. The limit (in the sense of \eqref{converg}) of $$ T_{n}f \le S_{n}f\quad\forall f\in \mathcal{C}_{0,+}^\infty(0,1) $$ is $$ Tf \le Sf\quad\forall f\in \mathcal{C}_{0,+}^\infty(0,1), $$ where $$ Tf\equiv a \sup_{(0,1)}f(x)\quad Sf\equiv b\sup_{(0,1)}f(x) \quad\forall f\in \mathcal{C}_{0,+}^\infty(0,1). $$ \subsection{Hardy's inequality, part II}\label{har} We start with the homogenized version of inequality \eqref{hardy}: \begin{equation} \Big(\int_{0}^{1} \Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)^{p}dx\Big)^{1/p} \le \frac{p}{p-1} \Big(\int_{0}^{1} f^{p}(x)dx\Big)^{1/p} \label{hardy2} \end{equation} As a first step, fix $n\in\mathbb{N}$ and set $p=1+1/n$. Here, again, the most natural operators $T_n$, $S_n$, respectively equal to the left hand side and the right hand side of \eqref{hardy2} do not work, due to the blowup of the right hand side. In this case $T_n$ and $S_n$ would be admissible and it seems that there cannot be a better choice. The important consideration to be made at this point is to understand $which$ inequality we wish to use before passing to the limit. In fact the standard proof of Hardy's inequality leads to a better version of \eqref{hardy2}, containing one more term, which is usually dropped. Taking into consideration such term, we are able to prove that the limit of Hardy's inequality when $p\to 1+$ is inequality \eqref{hardyp=1good} (actually, even a better one). We stress that inequality \eqref{hardyp=1good} has an independent, classical, simple proof in \cite[vol.1, p.32, Theorem 13.15(iii)]{Z}. However, our main intention here is to prove a limiting process and to obtain, as a byproduct, a new tool for proving inequalities. We believe that the following procedure has an independent interest. For completeness, we start here with the simple proof of \eqref{hardy2} (as usual, we are assuming here to deal only with $\mathcal{C}_{0,+}^\infty$ functions, not identically zero). We have \begin{align*} \int_{0}^{1}\Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)^{p}dx &=\int_{0}^{1} \Big( \int_{0}^{x}f(t)dt \Big)^{p}x^{-p}dx \\ &=\int_{0}^{1} \Big( \int_{0}^{x}f(t)dt \Big)^{p}d \big(\frac{x^{1-p}}{1-p}\big) \\ &=\Big[-\frac{x^{1-p}\big( \int_{0}^{x}f(t)dt \big)^{p}}{p-1} \Big]_{x=0}^{x=1}- \int_0^1 \big(\frac{x^{1-p}}{1-p}\big) d\Big( \int_{0}^{x}f(t)dt \Big)^{p}\\ &=-\frac{\big( \int_{0}^{1}f(t)dt \big)^{p}}{p-1} +\frac{p}{p-1}\int_{0}^{1}\Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)^{p-1}f(x)dx \end{align*} Applying Holder's inequality, \begin{align*} &\int_{0}^{1}\Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)^{p}dx\\ &\le -\frac{\big( \int_{0}^{1}f(t)dt \big)^{p}}{p-1}+\frac{p}{p-1} \Big[\int_{0}^{1}\Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)^{p}dx\Big]^{1-1/p} \Big( \int_{0}^{1}f(t)^pdt \Big)^{1/p} \end{align*} At this point Hardy's inequality \eqref{hardy2} is readily obtained dropping the first added and multiplying each side by $\Big[\int_{0}^{1} \Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)^{p}dx\Big]^{1/p-1}$. We now $do$ $not$ drop the first added, which plays a key role when passing to the limit for $p\to 1+$, and we raise both sides to the power $1/p$, getting an inequality relating two homogeneous operators, which we call respectively $T_nf$ and $S_nf$. We now compute the limit of $T_nf$ and $S_nf$. It is immediate to see that the limit of $T_nf$ is exactly the left hand side of inequality \eqref{hardyp=1good}. As for $S_nf$, we have \begin{align*} (S_nf)^p &=-\frac{\big( \int_{0}^{1}f(t)dt \big)^{p}}{p-1} +\frac{p}{p-1}\Big[\int_{0}^{1} \Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)^{p} dx\Big]^{1-1/p}\Big( \int_{0}^{1}f(t)^pdt \Big)^{1/p} \\ &= \frac{\big( \int_{0}^{1}f(t)^pdt \big)^{1/p} -\big( \int_{0}^{1}f(t)dt \big)^{p}}{p-1} \\ &\quad +\frac{p\big[\int_{0}^{1} \big( \frac{1}{x} \int_{0}^{x}f(t)dt \big)^{p}dx \big]^{1-1/p}-1}{p-1}\Big( \int_{0}^{1}f(t)^pdt \Big)^{1/p} \end{align*} We can now compute the limit as $p\to 1+$, taking into account that the quotients are in fact difference-quotients (therefore it suffices to compute derivatives in $p$ and set $p=1$). We have: \begin{align*} &\lim_{p\to 1+}\frac{\big( \int_{0}^{1}f(t)^pdt \big)^{1/p} -\big( \int_{0}^{1}f(t)dt \big)^{p}}{p-1}\\ &= \int_0^1 f(x)\log f(x)dx-2\Big( \int_{0}^{1}f(t)dt \Big)\log \Big( \int_{0}^{1}f(t)dt \Big) \end{align*} and $$ \lim_{p\to 1+}\frac{p\Big[\int_{0}^{1} \Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)^{p}dx\Big]^{1-1/p}-1}{p-1} =1+\log\Big(\int_{0}^{1}\Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx\Big). $$ Hence the limit as $p\to 1+$ is the inequality \begin{align*} &\int_{0}^{1}\Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx \\ &\le \int_0^1 f(x)\log f(x)dx-2\Big( \int_{0}^{1}f(t)dt \Big) \log\Big( \int_{0}^{1}f(t)dt \Big)+\int_{0}^{1}f(t)dt \\ &\quad +\Big( \int_{0}^{1}f(t)dt \Big) \log\Big(\int_{0}^{1}\Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx\Big) \end{align*} We will prove that this inequality is finer than inequality \eqref{appl3}, which leads to inequality \eqref{hardyp=1good}(see Application 3 in Section 4). We consider two cases. \noindent First case: $\int_{0}^{1}\Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx\le1$ so that the last term of our inequality is nonpositive: $$ \Big( \int_{0}^{1}f(t)dt \Big)\log\Big(\int_{0}^{1} \Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx\Big)\le 0 $$ We can drop it and observe that since $$ \sup_{t>0} -2t\log t+t=M_1<\infty, $$ we get $$ \int_{0}^{1}\Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx \le \int_0^1 f(x)\log f(x)dx+M_1 $$ which is an inequality of the type \eqref{appl3}. \noindent Second case: $\int_{0}^{1} \Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx>1$. By Young's inequality $$ ab\le (a+\frac12)\log (1+2a)-a+\frac12(e^b-b-1)\quad\forall a,b\ge0 $$ and therefore, setting $$ a=\int_{0}^{1}f(t)dt \quad b=\log\Big(\int_{0}^{1} \Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx\Big) $$ the following inequality holds \begin{align*} &\Big( \int_{0}^{1}f(t)dt \Big)\log\Big(\int_{0}^{1} \Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx\Big) \\ &\le \Big(\int_{0}^{1}f(t)dt +\frac12\Big) \log \Big(1+2\int_{0}^{1}f(t)dt \Big) -\int_{0}^{1}f(t)dt +\frac12\int_{0}^{1} \Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx. \end{align*} Substituting this into our limit inequality we get \begin{align*} \int_{0}^{1}\Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx &\le \int_0^1 f(x)\log f(x)dx-2\Big( \int_{0}^{1}f(t)dt \Big) \log\Big(\int_{0}^{1}f(t)dt \Big)\\ &\quad +\int_{0}^{1}f(t)dt+ \Big(\int_{0}^{1}f(t)dt +\frac12\Big) \log \Big(1+2\int_{0}^{1}f(t)dt \Big)\\ &\quad -\int_{0}^{1}f(t)dt+\frac12\int_{0}^{1} \Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx \end{align*} i.e., \begin{align*} \int_{0}^{1}\Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx &\le 2\int_0^1 f(x)\log f(x)dx-4\Big(\int_{0}^{1}f(t)dt \Big) \log\Big(\int_{0}^{1}f(t)dt \Big) \\ &\quad +\Big(2\int_{0}^{1}f(t)dt +1\Big) \log \Big(1+2\int_{0}^{1}f(t)dt \Big). \end{align*} Finally, since $\sup_{t>0} -4t\log t+(2t+1)\log (1+2t)=M_2<\infty$, we get $$ \int_{0}^{1}\Big( \frac{1}{x} \int_{0}^{x}f(t)dt \Big)dx \le 2\int_0^1 f(x)\log f(x)dx+M_2 $$ which is exactly an inequality of the type \eqref{appl3}. \begin{remark} \label{remk5.2} \rm The passages to the limit in $p$, similar to that one made above, are made in \cite[ 6.8, p. 139]{HLP}, where a logarithm appears after the limiting process. Much more recently, a similar passage to the limit has been fruitful when studying maximal functions and related weight classes, see \cite{SW}. In both cases the limit of the Lebesgue quasinorm in $L^r$ has been studied when $r\to 0$. The appearance of the logarithm in a limit for $r\to 1$, like in our case, has been noted and used in \cite{MoS}. Finally, let us recall that the same procedure has been used to derive the $LlogL$ integrability of the Jacobian (see \cite[(8.44) p. 186)]{IM}. \end{remark} \subsection{Sobolev inequalities for fractional Sobolev spaces, part II} \label{sobo} The problem of the ``not natural'' blowup of the norm of $W_0^{s,p}$, $p2)$ in \eqref{sobembed2} and consider \begin{gather*} T_nf= \|f\|_{L^{nNp/(nN-p)}(\Omega)}, \\ S_nf=c_1(N)^{1/p}\frac{(1/n)^{1/p}}{(N-p/n)^{(p-1)/p}} \|f\|_{W_0^{1-1/n,p}(\Omega)}, \\ Tf= \|f\|_{L^{Np/(N-p)}(\Omega)}, \\ Sf=c_2(p,N) \|f\|_{W_0^{1,p}(\Omega)} \end{gather*} The limit (in the sense of \eqref{converg}) of \eqref{sobembed2} is \eqref{sobembed3}. \section{A topology on Inequalities} The topology on ${\mathcal{I}}_{0}$ is not satisfactory for our purposes, since, for instance, in ${\mathcal{I}}_{0}$ the inequalities \begin{gather*} Tf\le Sf\quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega), \\ 2Tf\le 2Sf\quad \forall f\in \mathcal{C}_{0,+}^\infty(\Omega) \end{gather*} are \emph{different} objects. We are going to build up an abstract setting in which we identify all equivalent inequalities, and we will consider a topology on these new objects. Of course the convergence proved in the previous Section will be preserved in this new setting. \subsection{Equivalence of inequalities} \begin{definition} \label{equineq} \rm Let $T_1,S_1,T_2,S_2\in \mathcal{O}$ and let the inequalities \begin{gather*} d_1:\quad T_1f\le S_1f\quad\forall f\in \mathcal{C}_{0,+}^\infty(\Omega), \\ d_2:\quad T_2f\le S_2f\quad\forall f\in \mathcal{C}_{0,+}^\infty(\Omega) \end{gather*} be given. We will say that $d_1$ is equivalent to $d_2$, and we will write $$ d_1\sim d_2 $$ if it is possible to deduce $d_2$ from $d_1$ or $d_1$ from $d_2$ by combining a finite number of the following two operations: \begin{itemize} \item There exists $W \in \mathcal{O}$ such that $T_2f=T_1f+Wf$ and $S_2f=S_1f+Wf$ for all $f\in \mathcal{C}_{0,+}^\infty(\Omega)$ \item There exists $\Phi$ nonnegative, strictly increasing function on $[0,\infty[$ such that $T_2f=\Phi(T_1f)$, $S_2f=\Phi(S_1f)$ for all $f\in \mathcal{C}_{0,+}^\infty(\Omega)$ \end{itemize} \label{equivalence} \end{definition} We immediately observe that such definition is well-posed, in fact it is trivial to prove the following \begin{proposition} \label{prop6.2} The notion of equivalence introduced in Definition \ref{equivalence} is reflexive, symmetric, transitive. \label{wellposed} \end{proposition} \subsection{The quotient space} Definition \ref{equivalence} leads naturally to consider classes of equivalent inequalities. Let us consider the set of inequalities $$ {\mathcal{I}}_{0}=\{ d(T,S) : T,S\in {\mathcal{O}}\} $$ and in such set we consider the classes of equivalence given by $\sim$: $$ {\mathcal{I}}=\frac{{\mathcal{I}}_{0}}{\sim} $$ An element of ${\mathcal{I}}$ will be denoted by $[d]$, which is the class of all inequalities $d_1\in {\mathcal{I}}_{0}$ equivalent to the inequality $d\in {\mathcal{I}}_{0}$: $$ [d]=\{ d_1\in {\mathcal{I}}_{0} : d\sim d_1\} $$ In order to remember that $[d]$ is not an inequality, but a class of inequalities, we will refer to it, in the sequel, as ``Inequality''. The notion of convergence of inequalities and the introduction of a topology for inequalities have much more sense when dealing with \sl Inequalities \rm rather than \sl inequalities. \rm We introduce in ${\mathcal{I}}$ the topology of the quotient space ${{\mathcal{I}}_{0}}/{\sim}$ (see e.g. \cite[p. 125]{D}): if we call $P$ the projection $$ P: {{\mathcal{I}}_{0}} \; \rightarrow\; {\mathcal{I}} =\frac{{\mathcal{I}}_{0}}{\sim} $$ then $$ {\mathcal{A}}\subseteq {\mathcal{I}} \text{ is open in } {\mathcal{I}} $$ if and only if $$ P^{-1}[{\mathcal{A}}]=\cup \{ A : A\in {\mathcal{A}} \} \text{ is open in } {{\mathcal{I}}_{0}}. $$ It is well-known that $$ d_n\to d \quad \text{in } {{\mathcal{I}}_{0}} \; \Rightarrow \; [d_n]\to [d] \quad\text{in }{\mathcal{I}} $$ therefore the convergence already shown in the previous Section still hold in $ {\mathcal{I}} $. We conclude writing explicitly, in terms of the admissible operators in $\mathcal{O}$, what does it mean that $[d_n]\to [d]$ in ${{\mathcal{I}}}$. The following notion of convergence represents the answer we wanted to find to our original question, settled in the Introduction. \begin{quote} Let $[d_n]$, $[d]$ in ${{\mathcal{I}}}$. Write $d=d(T,S)$, $d_n=d_n(T_n,S_n)$ $\forall\, n\in\mathbb{N}$. It is $[d_n]\to [d]$ in ${{\mathcal{I}}}$ if for any $d'=d'(T',S')$, $d'\sim d$, for any $n'=n'(T',S')\in\mathbb{N}$, for any $F'=F'(T',S')\subset \mathcal{C}_{0,+}^\infty(\Omega)$ finite, there exists $\nu\in\mathbb{N}$ such that for $n>\nu$ the following holds: $$ \forall\, d'_n(T'_n,S'_n)\sim d_n(T_n,S_n)\; \exists d'(T',S')\sim d(T,S) : d'_n(T'_n,S'_n)\in {\mathcal{U}}_{n',F'}(d'(T',S')) $$ \end{quote} \subsection{The three main examples} We now make a few comments on the examples discussed in Sections 5.1, 5.2, and 5.3. In the first case, the sequence of inequalities was a sequence of relations between norms, therefore, the step made in this last Section has no a relevant meaning. In the other two cases, we changed, for convenience, the sequences of inequalities: in the case of Hardy's inequality, we preferred to deal with \eqref{hardy2} rather than \eqref{hardy}; in the case of the Sobolev inequalities for fractional Sobolev spaces, we dealt with inequality \eqref{sobembed2}, and we used a relation of limit involving the right norms, up to the factor $(1-s)$. It is trivial that such transformations (to raise the inequalities to a certain power, and to multiply the inequalities by a constant) can be done, giving of course \emph{equivalent} inequalities. But without this last step (the construction of a topology on Inequalities, made in this Section 6), the trivial transformations would lead to $different$ inequalities, and this would be not natural for the problem we wished to study. \section{Computing limits} In this last Section we wish to provide some tools to compute explicitly the limits of some Inequalities. Some of them have been implicitly proved or used in the previous Sections, others come as a byproduct from Function Space Theory. We stress here that the novelty of the limits we are going to show is not in the difficulty of the computations, but in the new light given by our construction: more or less common ``passages to the limit'' are in fact concrete limits in a suitable topology. We will conclude the Section giving two applications, which show how the construction of the topology leads to the proof of new results. \subsection{Some basic tools} Given a sequence of true inequalities, it is evident that the explicit computation of the limit must be carried out by passing through equivalent inequalities (namely, the operations described in the Definition \ref{equineq}), and by computing the limits of the left hand side and the right hand side. Therefore the basic tools rely upon the study of sequences of admissible operators, rather than the inequalities themselves. Moreover, we observe that since the admissible operators have real values, the standard theorems on operation of limits (for instance, the limit of a sum, of a product, the composition with a continuous nonnegative real function) can be applied. We are going to show some first ``bricks'' that can be used in applications. For our purposes it will be sufficient to confine ourselves to functions defined in domains having measure $1$. The first tool, which has been already used in Sections \ref{har} and \ref{sobo}, is completely standard, and it can be found in \cite{HLP}, n. 194, p. 143. \begin{proposition} Let $0p_0$, is an admissible operator in the sense of Section \ref{homog}, by virtue of the classical H\"older's inequality. \begin{proposition} Let $1\le p_01$, and let $u\in \mathcal{C}_{0,+}^\infty(\Omega)$. Let $q,r,p$ be such that $$ 1\le r0$. Fix $$ 1\le r