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<h1>Sandwich theorems for radiant functions</h1>
<p class="authors">
<span class="author">Costică Mustăţa\(^{*}\)</span>
</p>
<p class="date">December 9, 2014.</p>
</div>
<p>\(^{*}\)<i class="it">Tiberiu Popoviciu</i> Institute of Numerical Analysis, Romanian Academy, P.O. Box 68-1, Cluj-Napoca, Romania, e-mail: <span class="tt">cmustata2001@yahoo.com, cmustata@ictp.acad.ro</span> Dedicated to prof. I. Păvăloiu on the occasion of his 75th anniversary </p>

<div class="abstract"><p> We show that between two graphs, one of a radiant function and the other of a coradiant, both defined on a real interval containing \(0\), there exists at least one line which separates the graphs. The conditions for the uniqueness of a separating linear function are also established. </p>
<p><b class="bf">MSC.</b> 46A22, 26A16, 47N10. </p>
<p><b class="bf">Keywords.</b> Sandwich theorems, radiant functions, coradiant functions. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p>The paper is concerned with the existence and uniqueness of a linear function whose graph separates the graphs of two real-valued functions defined on an interval in \(\mathbb {R}\) containing zero (i.e. a radiant subset of \(\mathbb {R}\)), one of which being radiant and the other one coradiant. We show that, under some conditions, this sandwich-type problem has at least one solution. The uniqueness of the solution is also discussed. As application, one gives a sufficient of Hyers-Ulam type stability conditions for positively homogeneous functions. </p>
<p>Sandwich theorems for diverse classes of real-valued functions (monotonic, convex, quasiconvex) were considered in <span class="cite">
	[
	<a href="#BaronMatkovskiNikodem" >1</a>
	]
</span>, <span class="cite">
	[
	<a href="#ForgRobNicodemPales" >2</a>
	]
</span>, <span class="cite">
	[
	<a href="#NicodemWasowicz" >9</a>
	]
</span>, <span class="cite">
	[
	<a href="#SuzukiKuroiwa" >13</a>
	]
</span> and for more general functions in <span class="cite">
	[
	<a href="#FuchssteinerLusky" >3</a>
	]
</span>, <span class="cite">
	[
	<a href="#Simons" >12</a>
	]
</span>, <span class="cite">
	[
	<a href="#Szaz" >14</a>
	]
</span>, etc. </p>
<p>Let \(I\) be an interval in \(\mathbb {R}\) containing \(0.\) Then \(I\) is a radiant set, i.e. for every \(x\in I\) and \(\lambda \in \lbrack 0,1]\) it follows \(\lambda x\in I.\) </p>
<p>A function \(f:I\rightarrow \mathbb {R}\) is called a <i class="it">radiant</i> function if</p>
<div class="equation" id="f.1">
<p>
  <div class="equation_content">
    \begin{equation}  f(\lambda x)\leq \lambda f(x), \label{f.1}\end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> for all \(x\in I\) and \(\lambda \in \lbrack 0,1].\) </p>
<p>A function \(g:I\rightarrow \mathbb {R}\) is called a <i class="it">coradiant</i> function if</p>
<div class="equation" id="f.2">
<p>
  <div class="equation_content">
    \begin{equation}  g(\lambda x)\geq \lambda g(x), \label{f.2}\end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> for all \(x\in I\) and \(\lambda \in \lbrack 0,1].\) </p>
<p>Obviously, a function \(f\) is radiant iff \(-f\) is coradiant. Every radiant function \(f\) verifies the inequality \(f(0)\leq 0\) and every coradiant function \(g\) verifies <br />\(g(0)\geq 0.\) </p>
<p>A function \(h:I\rightarrow \mathbb {R}\) is called a Lipschitz function if there exists a constant \(K\geq 0\) (depending on \(f\) and \(I\)), such that</p>
<div class="equation" id="f.3">
<p>
  <div class="equation_content">
    \begin{equation}  \left\vert h(x)-h(y)\right\vert \leq K\left\vert x-y\right\vert ,\label{f.3}\end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p> for all \(x,y\in I.\) </p>
<p>The constant \(K\) is called a Lipschitz constant (for \(f\)) and the smallest Lipschitz constant is given by following expression:</p>
<div class="equation" id="f.4">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert f\right\Vert _{I}=\sup \left\{  \frac{\left\vert f(x)-f(y)\right\vert }{\left\vert x-y\right\vert }:x\neq y,\  x,y\in I\right\} . \label{f.4}\end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p> Denote by \(Lip_{0}I\) the real linear space of Lipschitz functions on \(I\) vanishing at \(0,\) i.e.</p>
<div class="equation" id="f.5">
<p>
  <div class="equation_content">
    \begin{equation}  Lip_{0}I:=\{ f:I\rightarrow \mathbb {R},\  f(0)=0\  \text{and\  }\left\Vert f\right\Vert _{I}<\infty \} . \label{f.5}\end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p> The functional \(\left\Vert \  \right\Vert _{I}:Lip_{0}I\rightarrow \lbrack 0,\infty )\) is a norm (called the Lipschitz norm) and \(Lip_{0}I\) is a Banach space with respect to this norm <span class="cite">
	[
	<a href="#Johnson" >4</a>
	]
</span>. </p>
<p>Denote</p>
<div class="equation" id="f.6">
<p>
  <div class="equation_content">
    \begin{equation}  r{\text-}Lip_{0}I:=\{ f\in Lip_{0}I,\  f\  radiant\} , \label{f.6}\end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<div class="equation" id="f.7">
<p>
  <div class="equation_content">
    \begin{equation}  cr{\text-}Lip_{0}I:=\{ g\in Lip_{0}I,\  g\  coradiant\}  \label{f.7}\end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p> and</p>
<div class="equation" id="f.8">
<p>
  <div class="equation_content">
    \begin{equation}  o{\text-}Lip_{0}I:=\{ h\in Lip_{0}I:h(\lambda x)=\lambda h(x),\  \lambda \in \lbrack 0,1],\  x\in I\} . \label{f.8}\end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p> Then \(r\)-\(Lip_{0}I\) and \(cr\)-\(Lip_{0}I\) are convex cones in the linear space \(Lip_{0}I\) and \(o\)-\(Lip_{0}I\) is a subspace of \(Lip_{0}I.\) Also</p>
<div class="equation" id="f.9">
<p>
  <div class="equation_content">
    \begin{equation}  o{\text-}Lip_{0}I=(r{\text-}Lip_{0})\cap (cr{\text-}Lip_{0}I). \label{f.9}\end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<h1 id="a0000000003">2 Extensions preserving radiantness of a Lipschitz functions</h1>
<p>The following extension result for real valued Lipschitz functions defined on a subset of a metric space was given by McShane <span class="cite">
	[
	<a href="#McShane" >5</a>
	]
</span>: </p>
<p><div class="theorem_thmwrapper " id="Th.A">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \((X,d)\) be a metric space, \(Y\) a nonvoid subset of \(X\) and \(f:Y\rightarrow \mathbb {R}\) be a Lipschitz function having the Lipschitz constant \(K(f)\)(on \(Y\)). Then there exists a Lipschitz function \(F:X\rightarrow \mathbb {R}\) such that</p>
<div class="equation" id="f.10">
<p>
  <div class="equation_content">
    \begin{equation}  \left. F\right\vert _{Y}=f\  \  \  \text{and \  \  }K(F)=K(f). \label{f.10}\end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>

  </div>
</div> </p>
<p>Such a function \(F\) is called a Lipschitz extension of \(f,\) preserving the Lipschitz constant. In the proof of this theorem one shows that the following two functions</p>
<div class="equation" id="f.11">
<p>
  <div class="equation_content">
    \begin{equation}  F(f)(x):=\inf _{y\in Y}\{ f(y)+K(f)d(x,y)\} ,\  x\in X, \label{f.11}\end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div>
<p> and</p>
<div class="equation" id="f.12">
<p>
  <div class="equation_content">
    \begin{equation}  G(f)(x):=\inf _{y\in Y}\{ f(y)-K(f)d(x,y)\} ,\  x\in X \label{f.12}\end{equation}
  </div>
  <span class="equation_label">12</span>
</p>
</div>
<p> are Lipschitz extensions of \(f,\) preserving the constant \(K(f),\) and every other extension \(H\) verifies the inequalities:</p>
<div class="equation" id="f.13">
<p>
  <div class="equation_content">
    \begin{equation}  F(x)\geq H(x)\geq G(x),\  x\in X \label{f.13}\end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p> (see also <span class="cite">
	[
	<a href="#Mustata2" >7</a>
	]
</span>, <span class="cite">
	[
	<a href="#Mustata3" >8</a>
	]
</span> for a more general situation). </p>
<p>In the framework considered above we obtain the following result: </p>
<p><div class="theorem_thmwrapper " id="Th.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(I\subset \mathbb {R}\) such that \(0\in I\) and \(f:I\rightarrow \mathbb {R}\). </p>
<ol class="enumerate">
  <li><p>If \(f\in r\)-\(Lip_{0}I\) then the greatest extension of \(f,\) namely</p>
<div class="equation" id="f.14">
<p>
  <div class="equation_content">
    \begin{equation}  F(f)(x):=\inf _{y\in I}\{ f(y)+\left\Vert f\right\Vert _{I}\left\vert x-y\right\vert \} ,\  x\in \mathbb {R} \label{f.14}\end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<p> is a radiant function on \(\mathbb {R},\) </p>
</li>
  <li><p>If \(g\in cr\)-\(Lip_{0}I,\) then the smallest extension of \(g,\) namely</p>
<div class="equation" id="f.15">
<p>
  <div class="equation_content">
    \begin{equation}  G(g)(x):=\sup _{y\in I}\{ g(y)-\left\Vert g\right\Vert _{I}\left\vert x-y\right\vert \} ,\  x\in \mathbb {R} \label{f.15}\end{equation}
  </div>
  <span class="equation_label">15</span>
</p>
</div>
<p> is a coradiant function on \(\mathbb {R}.\) </p>
</li>
  <li><p>If \(h\in o\)-\(Lip_{0}I\) then the extension \(F(h)\) defined by <span class="rm">(<a href="#f.14">14</a>)</span> is in the cone \(r\)-\(Lip_{0}\mathbb {R},\) and \(G(h)\) defined by <span class="rm">(<a href="#f.15">15</a>)</span> is in the cone \(cr\)-\(Lip_{0}\mathbb {R}.\) </p>
</li>
  <li><p>If \(f\in Lip_{0}\mathbb {R}\) and there exists \(I\subset \mathbb {R}\) with \(0\in I\) such that \(\left\Vert \left. f\right\vert _{I}\right\Vert _{I}=\left\Vert f\right\Vert _{R}\) and \(f|_{I}\in o\)-\(Lip_{0}I,\) then</p>
<div class="equation" id="f.16">
<p>
  <div class="equation_content">
    \begin{equation}  G(\left. f\right\vert _{I})(x)\leq f(x)\leq F(\left. f\right\vert _{I})(x),\  x\in \mathbb {R}. \label{f.16}\end{equation}
  </div>
  <span class="equation_label">16</span>
</p>
</div>
</li>
</ol>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000004">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> The proof is similar to that in <span class="cite">
	[
	<a href="#Mustata3" >8</a>
	]
</span>. For the sake of completeness we sketch the proof. </p>
<p>a) Let \(f\in r\)-\(Lip_{0}I.\) Then, for all \(\lambda \in \lbrack 0,1]\) and \(y\in I,\) </p>
<div class="displaymath" id="a0000000005">
  \begin{align*}  F(f)(\lambda x)&  =\inf \limits _{y\in I}\{ f(y)+\left\Vert f\right\Vert _{I}\left\vert \lambda x-y\right\vert \} \\ &  \leq \inf _{y\in I}\{ f(\lambda y)+\left\Vert f\right\Vert _{I}\left\vert \lambda x-\lambda y\right\vert \} \\ &  \leq \inf _{y\in I}\{ \lambda f(y)+\lambda \left\Vert f\right\Vert _{I}\left\vert x-y\right\vert \} \\ &  =\lambda \inf _{y\in I}\{ f(y)+\left\Vert f\right\Vert _{I}\left\vert x-y\right\vert \} \\ & =\lambda F(f)(x), \end{align*}
</div>
<p> for every \(x\in \mathbb {R}\). It follows</p>
<div class="displaymath" id="a0000000006">
  \[  F(f)(\lambda x)\leq \lambda F(f)(x),  \]
</div>
<p> for all \(\lambda \in \lbrack 0,1]\) and all \(x\in \mathbb {R}\). </p>
<p>b) Let \(g\in cr\)-\(Lip_{0}I\) and \(x\in \mathbb {R}\). Then for all \(\lambda \in \lbrack 0,1]\) and \(y\in I,\)</p>
<div class="displaymath" id="a0000000007">
  \begin{align*}  G(g)(\lambda x) &  =\sup _{y\in I}\{ g(y)-\left\Vert g\right\Vert _{I}\left\vert \lambda x-y\right\vert \} \\ &  \geq \sup _{y\in I}\{ g(\lambda y)-\left\Vert g\right\Vert _{I}\left\vert \lambda x-\lambda y\right\vert \} \\ &  \geq \lambda \sup _{y\in I}\{ g(y)-\left\Vert g\right\Vert _{I}\left\vert x-y\right\vert \} \\ &  =\lambda G(g)(x) \end{align*}
</div>
<p> Consequently \(G(g)(\lambda x)\geq \lambda G(g)(x)\) for all \(\lambda \in \lbrack 0,1]\) and all \(x\in \mathbb {R}.\) </p>
<p>c) If \(h\in o\)-\(Lip_{0}I=(r\)-\(Lip_{0}I)\cap (cr\)-\(Lip_{0}I),\) then the assertions from c) follow from a) and b). </p>
<p>d) Let \(f\in Lip_{0}\mathbb {R}\) and suppose that there exists \(I\subset \mathbb {R}\) with \(0\in I\) such that \(\left. f\right\vert _{I}\in o\)-\(Lip_{0}I\) and \(\left\Vert f\right\vert _{I}=\left\Vert f\right\Vert _{R}.\) Then \(f\) is an extension of \(\left. f\right\vert _{I}\) preserving the smallest Lipschitz constant \(\left\Vert \left. f\right\vert _{I}\right\Vert \) and the assertion follows by (<a href="#f.13">13</a>) and c). <div class="proof_wrapper" id="a0000000008">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<h1 id="a0000000009">3 Sandwich theorems</h1>
<p>The assertion d) in Theorem <a href="#Th.1">2</a> suggests the following problem: Let \(f,\  g\) be two functions, on an interval \(I\) \((0\in I),\) \(f\) radiant (coradiant) and \(g\) coradiant (radiant), and \(f(x)\leq g(x)\) for all \(x\in I.\) Is there a function \(h:I\rightarrow \mathbb {R}\) verifying \(h(\lambda x)=\lambda h(x),\) for all \(x\in I\) and all \(\lambda \in \lbrack 0,1],\) and such that \(f(x)\leq h(x)\leq g(x),\) \(x\in I?\) Problems of such type were considered for example in <span class="cite">
	[
	<a href="#BaronMatkovskiNikodem" >1</a>
	]
</span>, <span class="cite">
	[
	<a href="#ForgRobNicodemPales" >2</a>
	]
</span>, <span class="cite">
	[
	<a href="#NicodemWasowicz" >9</a>
	]
</span> and, more general in <span class="cite">
	[
	<a href="#FuchssteinerLusky" >3</a>
	]
</span>, <span class="cite">
	[
	<a href="#Simons" >12</a>
	]
</span>, <span class="cite">
	[
	<a href="#Szaz" >14</a>
	]
</span>, etc. The studies in this direction are motivated by applications to the theory of optimization and numerical analysis. </p>
<p>We show that the answer is affirmative in both cases. </p>
<p>Firstly, we need the following Lemma. </p>
<p><div class="lemma_thmwrapper " id="Lemma 2">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">3</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(f,g:[0,a]\rightarrow \mathbb {R},\) \(f\) radiant and \(g\) coradiant such that \(f(x)\leq g(x),\) \(x\in \lbrack 0,a].\) Then the following assertions hold: </p>
<ol class="enumerate">
  <li><p>The function \(p_{f}:(0,a]\rightarrow \mathbb {R}\) defined by</p>
<div class="equation" id="f.17">
<p>
  <div class="equation_content">
    \begin{equation}  p_{f}(x)=\frac{f(x)}{x},\  \label{f.17}\end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p> is nondecreasing on \((0,a],\) and the function \(p_{g}:(0,a]\rightarrow \mathbb {R}\) defined by</p>
<div class="equation" id="f.18">
<p>
  <div class="equation_content">
    \begin{equation}  p_{g}(x)=\frac{g(x)}{x}, \label{f.18}\end{equation}
  </div>
  <span class="equation_label">18</span>
</p>
</div>
<p> is nonincreasing on \((0,a];\) </p>
</li>
  <li><p>For every \(x_{0}\in (0,a],\)</p>
<div class="equation" id="f.19">
<p>
  <div class="equation_content">
    \begin{equation}  \frac{f(x)}{x}\leq \frac{f(x_{0})}{x_{0}}\leq \frac{g(x_{0})}{x_{0}}\leq \frac{g(x)}{x},\  x\in (0,x_{0}]. \label{f.19}\end{equation}
  </div>
  <span class="equation_label">19</span>
</p>
</div>
</li>
</ol>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000010">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> a) Let \(f:[0,a]\rightarrow \mathbb {R},\) \(f\) radiant. For every \(x\in (0,a]\) let \(\lambda \in \lbrack 0,1]\) be such that \(x=\lambda a.\) Then</p>
<div class="displaymath" id="a0000000011">
  \[  p_{f}(x)=\frac{f(x)}{x}=\frac{f(\lambda a)}{\lambda a}\leq \frac{\lambda f(a)}{\lambda a}=\frac{f(a)}{a}.  \]
</div>
<p> Let \(x_{1},x_{2}\in (0,a],\) \(x_{1}\leq x_{2}.\) Then there exists \(\lambda \in (0,1]\) such that \(x_{1}=\lambda x_{2},\) so that</p>
<div class="displaymath" id="a0000000012">
  \[  p_{f}(x_{1})=\frac{f(x_{1})}{x_{1}}=\frac{f(\lambda x_{2})}{\lambda x_{2}}\leq \frac{\lambda f(x_{2})}{\lambda x_{2}}=\frac{f(x_{2})}{x_{2}}=p_{f}(x_{2}).  \]
</div>
<p> Also</p>
<div class="displaymath" id="a0000000013">
  \[  p_{g}(x_{1})=\frac{g(x_{1})}{x_{1}}=\frac{g(\lambda x_{2})}{\lambda x_{2}}\geq \frac{\lambda g(x_{2})}{\lambda x_{2}}=\frac{g(x_{2})}{x_{2}}=p_{g}(x_{2}).  \]
</div>
<p> Consequently \(p_{f}\) is nondecreasing and \(p_{g}\) is nonincreasing on \((0,a].\) </p>
<p>b) Taking into account the hypothesis and the conclusions in a), the inequalities (<a href="#f.19">19</a>) follow. <div class="proof_wrapper" id="a0000000014">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="Remark 3">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">4</span>
  </div>
  <div class="remark_thmcontent">
  <p>The inequalities (<a href="#f.17">17</a>), (<a href="#f.18">18</a>) and (<a href="#f.19">19</a>) imply that every line passing through the points (0,0) and \((x,f(x))\) is above the graph of \(f\) on the interval \([0,x],\) and every line passing through the points (0,0) and \((x,g(x))\) is bellow the graph of \(g\) on \([0,x].\) Also, the line passing through (0,0) and \((x,g(x))\) is above the line passing through (0,0) and \((x,f(x)).\)<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We obtain the main result of the paper. </p>
<p><div class="theorem_thmwrapper " id="Th.4">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f,g:[0,a]\rightarrow \mathbb {R},\) \(f\) radiant and \(g\) coradiant, and \(f(x)\leq g(x),\) for all \(x\in \lbrack 0,a].\) Then there exists at least a function \(h:[0,a]\rightarrow \mathbb {R}\) having the form \(h(x)=\alpha x,\) \(\alpha \in \mathbb {R}\) and such that</p>
<div class="equation" id="f.20">
<p>
  <div class="equation_content">
    \begin{equation}  f(x)\leq h(x)\leq g(x),\  x\in \lbrack 0,a]. \label{f.20}\end{equation}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p> If there exists \(x_{0}\in (0,a]\) such that \(f(x_{0})=g(x_{0})\) then the unique function \(h:[0,a]\rightarrow \mathbb {R}\) verifying <span class="rm">(<a href="#f.20">20</a>)</span> is \(h(x)=\frac{f(a)}{a}x.\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000015">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By Lemma <a href="#Lemma 2">3</a>, taking into account (<a href="#f.19">19</a>), one obtains</p>
<div class="displaymath" id="a0000000016">
  \[  \frac{f(x)}{x}\leq \frac{f(a)}{a}\leq \frac{g(a)}{a}\leq \frac{g(x)}{x},x\in (0,a].  \]
</div>
<p> Then, for \(\alpha \in \lbrack \frac{f(a)}{a},\frac{g(a)}{a}]\) the function \(h(x)=\alpha x,\) \(x\in \lbrack 0,a]\) verifies the properties from the conclusions of first part of theorem. </p>
<p>Now, if \(f(x_{0})=g(x_{0})\) at \(x_{0}\in (0,a]\) one obtains</p>
<div class="displaymath" id="a0000000017">
  \[  \frac{f(x_{0})}{x_{0}}\leq \frac{f(a)}{a}\leq \frac{g(a)}{a}\leq \frac{g(x_{0})}{x_{0}} \]
</div>
<p> and then \(f(a)=g(a).\) For \(x\in \lbrack x_{0},a]\) one obtains</p>
<div class="displaymath" id="a0000000018">
  \[  \frac{f(x)}{x}=\frac{g(x)}{x} \]
</div>
<p> i.e \(f(x)=g(x)=\frac{f(a)}{a}x.\) Then \(h(x)=\frac{f(a)}{a}x\) verifies \(f(x)\leq h(x)\leq g(x),x\in \lbrack 0,a],\) and the second part of the theorem follows. <div class="proof_wrapper" id="a0000000019">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>In the case \(I=[0,\infty )\) one obtains. </p>
<p><div class="theorem_thmwrapper " id="Th.5">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">6</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f,g:[0,\infty )\rightarrow \mathbb {R},\) \(f\) radiant and \(g\) coradiant, and \(f(x)\leq g(x)\) for all \(x\in \lbrack 0,\infty ).\) Then there exists at least a function \(h:[0,\infty )\rightarrow \mathbb {R}\), of the form \(h(x)=\alpha x,\) \(\alpha \in \mathbb {R}\) fixed such that \(f(x)\leq h(x)\leq g(x),\) \(x\in \lbrack 0,\infty ).\) </p>
<p>If there exists \(x_{0}{\gt}0\) such that \(f(x_{0})=g(x_{0})\) then the unique function \(h:[0,\infty )\) \(\rightarrow \mathbb {R}\) satisfying \(f(x)\leq h(x)\leq g(x)\) is \(h(x)=\frac{f(x_{0})}{x_{0}}x.\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000020">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By Lemma <a href="#Lemma 2">3</a></p>
<div class="displaymath" id="a0000000021">
  \[  \frac{f(x)}{x}\leq \frac{f(a)}{a}\leq \frac{g(a)}{a}\leq \frac{g(x)}{x},x\in (0,a],  \]
</div>
<p> for every \(a\in (0,\infty ),\) </p>
<p>Then</p>
<div class="displaymath" id="a0000000022">
  \[  \sup _{0{\lt}x\leq a}\frac{f(x)}{x}\leq \inf _{0{\lt}x\leq a}\frac{g(x)}{x},  \]
</div>
<p> and</p>
<div class="displaymath" id="a0000000023">
  \[  \lim _{a\rightarrow \infty }\sup _{0{\lt}x\leq a}\frac{f(x)}{x}\leq \lim _{a\rightarrow \infty }\inf _{0{\lt}x\leq a}\frac{g(x)}{x}.  \]
</div>
<p> By considering \(\alpha \in \lbrack \lim _{a\rightarrow \infty }\sup _{0{\lt}x\leq a}\frac{f(x)}{x},\lim _{a\rightarrow \infty }\inf _{0{\lt}x\leq a}\frac{g(x)}{x}]\) one obtains</p>
<div class="displaymath" id="a0000000024">
  \[  \frac{f(x)}{x}\leq \alpha \leq \frac{g(x)}{x},x\in (0,\infty ),  \]
</div>
<p> and consequently</p>
<div class="displaymath" id="a0000000025">
  \[  f(x)\leq \alpha x\leq g(x),\  x\in \lbrack 0,\infty ).  \]
</div>
<p> Then \(h(x)=\alpha x,x\in \lbrack 0,\infty ),\) satisfies the first conclusion of the theorem. </p>
<p>Let \(x_{0}{\gt}0\) such that \(f(x_{0})=g(x_{0}).\) For every \(x{\gt}x_{0}\) we have</p>
<div class="displaymath" id="a0000000026">
  \[  \frac{f(x_{0})}{x_{0}}\leq \frac{f(x)}{x}\leq \frac{g(x)}{x}\leq \frac{g(x_{0})}{x_{0}} \]
</div>
<p> It follows \(f(x)=g(x)=\frac{f(x_{0})}{x_{0}}x,\) for all \(x{\gt}x_{0}\) and consequently, for \(x\in \lbrack 0,\infty )\) the function \(h(x)=\frac{f(x_{0})}{x_{0}}x\) verifies</p>
<div class="displaymath" id="a0000000027">
  \[  f(x)\leq h(x)\leq g(x).  \]
</div>
<p> <div class="proof_wrapper" id="a0000000028">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>Now we consider \(I=[a,0]\). <div class="theorem_thmwrapper " id="Th.6">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">7</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f,g:[a,0]\) \(\rightarrow \mathbb {R}\) \((a{\lt}0),\) \(f\) radiant and \(g\) coradiant, and \(f(x)\leq g(x),\) for all \(x\in \lbrack a,0].\) Then there exists at least a function \(h:[a,0]\rightarrow \mathbb {R}\) having the form \(h(x)=\alpha x,\) \(\alpha \in \mathbb {R}\) fixed, and verifying the inequalities:</p>
<div class="equation" id="f.21">
<p>
  <div class="equation_content">
    \begin{equation}  f(x)\leq h(x)\leq g(x),\  x\in \lbrack a,0].\label{f.21}\end{equation}
  </div>
  <span class="equation_label">21</span>
</p>
</div>
<p>If there exists \(x_{0}\in \lbrack a,0)\) such that \(f(x_{0})=g(x_{0})\) then the only function \(h:[a,0]\rightarrow \mathbb {R}\) such that \(f(x)\leq h(x)\leq g(x)\) is \(h(x)=\frac{f(a)}{a}x.\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000029">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Observe that for all \(x\in \lbrack a,0]\) one obtains \(f(x)\leq \frac{f(a)}{a}x.\) Indeed, if \(x\in \lbrack a,0),\) there exists \(\lambda \in \lbrack 0,1]\) such that \(x=\lambda a.\) Then</p>
<div class="displaymath" id="a0000000030">
  \[  f(x)=f(\lambda a)\leq \lambda f(a)=\frac{f(a)}{a}(\lambda a)=\frac{f(a)}{a}x.  \]
</div>
<p> Analogously,</p>
<div class="displaymath" id="a0000000031">
  \[  g(x)=g(\lambda a)\geq \lambda g(a)=\frac{g(a)}{a}(\lambda a)=\frac{g(a)}{a}x.  \]
</div>
<p> But \(g(a)\geq f(a)\) implies \(\frac{g(a)}{a}\leq \frac{f(a)}{a},\) and for \(\alpha \in \left[ \frac{g(a)}{a},\frac{f(a)}{a}\right] \) it follows</p>
<div class="displaymath" id="a0000000032">
  \[  \frac{g(a)}{a}x\geq \alpha x\geq \frac{f(a)}{a}x,\  x\in \lbrack a,0]  \]
</div>
<p> and, consequently \(f(x)\leq \alpha x\leq g(x),\) \(x\in \lbrack 0,a].\) </p>
<p>Now, let \(x_{0}\in \lbrack a,0)\) such that \(f(x_{0})=g(x_{0}).\) </p>
<p>Then</p>
<div class="displaymath" id="a0000000033">
  \[  \frac{f(a)}{a}\leq \frac{f(x_{0})}{x_{0}}=\frac{g(x_{0})}{x_{0}}\leq \frac{g(a)}{a}.  \]
</div>
<p> We are lead to \(f(a)\geq g(a),\) and because of hypothesis \(f(a)\leq g(a)\) it follows \(\frac{f(a)}{a}=\frac{g(a)}{a}.\) </p>
<p>Then \(h(x)=\frac{f(a)}{a}x\) is the only function having the graph between the graphs of \(f\) and \(g\) on \([a,0].\) <div class="proof_wrapper" id="a0000000034">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>Suppose now that \(f\) is radiant, \(g\) is coradiant and \(f(x)\geq g(x)\) for all \(x\in \lbrack 0,a].\) Because \(f(0)\leq 0\) and \(g(0)\geq 0\) it follows \(f(0)=g(0)=0.\) </p>
<p>The following theorem holds. </p>
<p><div class="theorem_thmwrapper " id="Th.7">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">8</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f,g:[0,a]\rightarrow \mathbb {R},\) \(f\) radiant, \(g\) coradiant and \(f(x)\geq g(x),\) \(x\in \lbrack 0,a].\) Then there exists at least a function \(h:[0,a]\rightarrow \mathbb {R}\) having the form \(h(x)=\alpha x,\) \(\alpha \in \mathbb {R}\) fixed, and such that</p>
<div class="equation" id="f.22">
<p>
  <div class="equation_content">
    \begin{equation}  f(x)\geq h(x)\geq g(x),\  x\in \lbrack 0,a]. \label{f.22}\end{equation}
  </div>
  <span class="equation_label">22</span>
</p>
</div>
<p> If there exists \(x_{0}\in (0,a]\) such that \(f(x_{0})=g(x_{0})\) then \(f(x)\) \(=g(x)\) for every \(x\in \lbrack 0,x_{0}]\) and the only function \(h:[0,a\} \rightarrow \mathbb {R}\) verifying <span class="rm">(<a href="#f.22">22</a>)</span> is \(h(x)=\frac{f(x_{0})}{x_{0}}x\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000035">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Let \(x_{0}\in (0,a]\) be such that \(f(x_{0})=g(x_{0}).\) If \(x\in (0,x_{0}],\) by Lemma <a href="#Lemma 2">3</a> one obtains</p>
<div class="displaymath" id="a0000000036">
  \[  \frac{g(x)}{x}\geq \frac{g(x_{0})}{x_{0}}=\frac{f(x_{0})}{x_{0}}\geq \frac{f(x)}{x}.  \]
</div>
<p> It follows \(g(x)\geq f(x),\) \(x\in \lbrack 0,x_{0}]\) and because \(f(x)\geq g(x)\) (by hypothesis) one obtains \(f(x)=g(x)\) for every \(x\in \lbrack 0,x_{0}].\) </p>
<p>The line \(h(x)=\frac{f(x_{0})}{x_{0}}x\  (=\frac{g(x_{0})}{x_{0}}x)\) is between the graphs of \(f\) and \(g\) over the interval \([0,x_{0}].\) </p>
<p>For \(x{\gt}x_{0},\) \(x\leq a\)</p>
<div class="displaymath" id="a0000000037">
  \[  \frac{f(x_{0})}{x_{0}}\leq \frac{f(x)}{x}\  \text{and }\frac{f(x_{0})}{x_{0}}\leq \frac{g(x)}{x} \]
</div>
<p> and consequently the function \(h(x)=\frac{f(x_{0})}{x_{0}}x\), \(x\in \lbrack 0,a]\) verifies</p>
<div class="displaymath" id="a0000000038">
  \[  f(x)\geq h(x)\geq g(x),\  x\in \lbrack 0,a].  \]
</div>
<p> Consider now the remaining case \(f(x){\gt}g(x),\) \(x\in (0,a].\) Because</p>
<div class="displaymath" id="a0000000039">
  \[  \frac{g(a)}{a}\leq \frac{g(x)}{x}{\lt}\frac{f(x)}{x}\leq \frac{f(a)}{a},\  x\in (0,a]  \]
</div>
<p> it follows that \(\inf \left\{  \frac{f(x)}{x}:x\in (0,a]\right\}  \) and \(\sup \left\{  \frac{g(x)}{x}:x\in (0,a]\right\}  \) are finite, and</p>
<div class="displaymath" id="a0000000040">
  \[  \sup _{0{\lt}x\leq a}\frac{g(x)}{x}\leq \inf _{0{\lt}x\leq a}\frac{f(x)}{x}.  \]
</div>
<p> By considering \(\alpha \in \Big[ \sup \limits _{0{\lt}x\leq a}\frac{g(x)}{x},\inf \limits _{0{\lt}x\leq a}\frac{f(x)}{x}\Big] ,\) the line \(h(x)=\alpha x,\) \(x\in \lbrack 0,a]\) lies between the graphs of \(f\) and \(g,\) i.e.</p>
<div class="displaymath" id="a0000000041">
  \[  f(x)\geq h(x)\geq g(x),\  x\in \lbrack 0,a].  \]
</div>
<p> <div class="proof_wrapper" id="a0000000042">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="Remark 8">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">9</span>
  </div>
  <div class="remark_thmcontent">
  <p> The result in Theorem <a href="#Th.7">8</a> is valid also if \(f,g:(-\infty ,0]\rightarrow \mathbb {R}\). In this case there exists \(\alpha \in \lbrack \lim \limits _{a\rightarrow -\infty }\! \! \sup \nolimits _{a{\lt}x\leq 0}\frac{g(x)}{x},\lim \limits _{a\rightarrow -\infty }\! \! \inf \nolimits _{a{\lt}x\leq 0}\frac{f(x)}{x}]\) such that the function \(h(x)=\alpha x\) satisfies the inequalities</p>
<div class="displaymath" id="a0000000043">
  \[  f(x)\leq h(x)\leq g(x),\  x\in (-\infty ,0].\hfil \qed  \]
</div>

  </div>
</div> </p>
<p>Now consider \(I=[a,b]\) where \(a{\lt}0{\lt}b,\) or \(I=\mathbb {R}\). </p>
<p>By the above results it follows: </p>
<p><div class="corollary_thmwrapper " id="Corollary 9">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">10</span>
  </div>
  <div class="corollary_thmcontent">
  
<ol class="enumerate">
  <li><p>Let \(f,g:[a,b]\rightarrow \mathbb {R},\) \(a{\lt}0{\lt}b,\) \(f\) radiant, \(g\) coradiant and such that \(f(x)\leq g(x),\) \(x\in \lbrack a,b].\) Then there exists \(h:[a,b]\rightarrow \mathbb {R},\)</p>
<div class="displaymath" id="a0000000044">
  \[  h(x)=\left\{  \begin{array}[c]{c}\alpha x,\  x\in \lbrack a,0],\  \alpha \in \left[ \frac{g(a)}{a},\frac{f(a)}{a}\right] ,\smallskip \\ \beta x,\  x\in (0,b],\  \beta \in \left[ \frac{f(b)}{b},\frac{g(b)}{b}\right] . \end{array} \right.  \]
</div>
<p> such that \(f(x)\leq h(x)\leq g(x),\) \(x\in \lbrack a,b].\) </p>
</li>
  <li><p>Let \(f,g:\mathbb {R\rightarrow R},\  f\) radiant, \(g\) coradiant, \(f(x)\leq g(x),\) for all \(x\in \mathbb {R}\). Then there exists \(h:\mathbb {R\rightarrow R}\) of the form</p>
<div class="displaymath" id="a0000000045">
  \[  h(x)=\left\{  \begin{array}[c]{c}\alpha x,\  x\in (-\infty ,0],\\ \beta x,\  x\in (0,\infty ), \end{array} \right.  \]
</div>
<p> where</p>
<div class="displaymath" id="a0000000046">
  \[  \alpha \in \left[ \lim _{a\rightarrow -\infty }\sup \nolimits _{x\geq a}\frac{g(x)}{x},\lim _{a\rightarrow -\infty }\inf \nolimits _{x\geq a}\frac{f(x)}{x}\right]  \]
</div>
<p> and</p>
<div class="displaymath" id="a0000000047">
  \[  \beta \in \left[ \lim _{a\rightarrow \infty }\sup \nolimits _{x\leq a}\frac{f(x)}{x},\lim _{a\rightarrow \infty }\inf \nolimits _{x\leq a}\frac{g(x)}{x}\right],  \]
</div>
<p> verifying the inequalities:</p>
<div class="displaymath" id="a0000000048">
  \[  f(x)\leq h(x)\leq g(x),\  x\in \mathbb {R}.  \]
</div>
</li>
</ol>

  </div>
</div> </p>
<p>A similar result is valid if \(f(x)\geq g(x),\) \(f\) radiant and \(g\) coradiant on \([a,b],\) \(a{\lt}0{\lt}b,\) respectively on \(\mathbb {R}.\) </p>
<p>In Corollary <a href="#Corollary 11">11</a> a), if \(\left[ \frac{g(a)}{a},\frac{f(a)}{a}\right] \cap \left[ \frac{f(b)}{b},\frac{g(b)}{b}\right] \neq \emptyset \) and \(\mu \) is a number from this set, then \(h(x)=\mu x,\) \(x\in [a,b]\) verify</p>
<div class="displaymath" id="a0000000049">
  \[  f(x)\leq \mu x\leq g(x),\  x\in [a,b].  \]
</div>
<p> A similar result follows in the case b). </p>
<p>If \(f:I\rightarrow \mathbb {R}\) \(\  (0\in I),\) and \(f\) is a convex (concave) function on \(I\), \(f(0)=0\), then \(f\) is radiant (coradiant). The above results may be enounced for convex and concave function defined on \(I\), vanishing at zero. </p>
<p><div class="examples_thmwrapper " id="a0000000050">
  <div class="examples_thmheading">
    <span class="examples_thmcaption">
    Examples
    </span>
  </div>
  <div class="examples_thmcontent">
  <p>1\(^{0}\) Let \(f_{m},g:[0,2]\rightarrow \mathbb {R}\) be the functions defined by</p>
<div class="displaymath" id="a0000000051">
  \begin{align*} & f_{m}(x) = \begin{cases}  mx^{3},&  x\in \lbrack 0,1]\\ mx,&  x\in (1,2] \end{cases} ,\  m\in \mathbb {R}, \\ & g(x)= \begin{cases}  -x^{2}+2,&  x\in \lbrack 0,1],\\ x,&  x\in (1,2]. \end{cases}\end{align*}
</div>
<p> Then, for \(m{\gt}0\) the function \(f_{m}\) is radiant, \(g\) is coradiant and \(f_{m}(x)\leq g(x),\) \(x\in \lbrack 0,2].\) Every function \(h:[0,2]\rightarrow \mathbb {R},\) \(h(x)=\alpha x,\) where \(\alpha \in \lbrack m,1]\) verifies</p>
<div class="displaymath" id="a0000000052">
  \[  f_{m}(x)\leq h(x)\leq g(x),\  x\in \lbrack 0,2]  \]
</div>
<p> Also, for \(m=1,\) \(f_{1}(x)\leq g(x),\  x\in \lbrack 0,2]\) and because \(f_{1}(1)=g(1)=1\) one obtains that \(h(x)=x\) is the unique function verifying \(f_{1}(x)\leq h(x)\leq g(x),\) \(x\in \lbrack 0,2].\) Consequently Theorem <a href="#Th.5">6</a> is fulfilled. </p>
<p>2\(^{0}\) Let \(a{\gt}1,\) \(m\in (0,1]\) and let \(f_{m},\) \(g_{a}:[0,\infty )\rightarrow \mathbb {R},\) be the functions:</p>
<div class="displaymath" id="a0000000053">
  \[  f_{m}(x)= \begin{cases}  \frac{mx^{3}}{a^{2}},&  x\in \lbrack 0,a],\\ mx,&  x\in (a+\infty ), \end{cases}  \]
</div>
<div class="displaymath" id="a0000000054">
  \[  g_{a}(x)=\begin{cases}  -x^{2}+(1+a)x,&  x\in \lbrack 0,a]\\ x,&  x\in (a,+\  \infty ) \end{cases}  \]
</div>
<p> Then \(f_{m}(x)\leq g_{a}(x),\) for \(x\in \lbrack 0,\infty ),\) \(f_{m}\) is radiant and \(g_{a}\) is coradiant. </p>
<p>For every \(\alpha \in \lbrack m,1]\) one obtains</p>
<div class="displaymath" id="a0000000055">
  \[  f_{m}(x)\leq h(x)=\alpha x\leq g(x),\  x\in \lbrack 0,\infty )  \]
</div>
<p> For \(m=1,\  f_{1}(a)=g_{a}(a)=a\) and consequently \(h(x)=x\) is the only function verifying \(f_{1}(x)\leq h(x)=x\leq g(x),\) \(x\in \lbrack 0,\infty ).\) Theorem <a href="#Th.6">7</a> is fulfilled. </p>
<p>3\(^{0}\) Let \(f,g:[-2,0]\rightarrow \mathbb {R}\), \(f(x)=x^{2}+x\) and \(g(x)=-x^{2}-4x.\) Then \(f\) is radiant, \(g\) is coradiant and \(f(x)\leq g(x),\) \(x\in \lbrack -2,0].\) Every line \(h(x)=\alpha x,\) where \(\alpha \in \lbrack -2,-1]\) has the graph between the graphs of \(f\) and \(g.\) </p>
<p>Now let \(f,g:[-3,0]\rightarrow \mathbb {R}\) be the functions</p>
<div class="displaymath" id="a0000000056">
  \begin{align*} & f(x) = \begin{cases}  \frac{-x}{2},&  x\in \lbrack -3,-1]\\ x^{2}+\frac{x}{2},&  x\in (-1,0], \end{cases}\\ & g(x) = \begin{cases}  \frac{-x}{2},&  x\in \lbrack -3,-1]\\ -4x^{2}\frac{9}{2}x,&  x\in (-1,0]. \end{cases}\end{align*}
</div>
<p> Then \(f\) is radiant, \(g\) is coradiant and \(f(x)\leq g(x),\  x\in \lbrack -3,0].\) </p>
<p>The only function \(h\) such that \(f(x)\leq h(x)\leq g(x),\) \(x\in \lbrack -3,0]\) is \(h(x)=-\frac{1}{2}x.\) Thus Theorem <a href="#Th.7">8</a> is fulfilled. </p>
<p>4\(^{0}\) Let \(f_{m},g:[0,a]\rightarrow \mathbb {R},\) \((a{\gt}0,\) \(m\geq 0)\  \)be the functions</p>
<div class="displaymath" id="a0000000057">
  \begin{align*}  f_{m}(x) &  =x^{2}+mx,\  \\ g(x) &  =-x^{3}. \end{align*}
</div>
<p> Then \(f_{m}\  \)is radiant, \(g\) is coradiant and \(f(x)\geq g(x),\  x\in \lbrack 0,a].\) The function \(h(x)=\alpha x,\) where \(\alpha \in \lbrack 0,m]\) is such that</p>
<div class="displaymath" id="a0000000058">
  \[  f(x)\geq h(x)\geq g(x),\  x\in \lbrack 0,a].  \]
</div>
<p> For \(m=0,\) the function \(h(x)=0\) verify</p>
<div class="displaymath" id="a0000000059">
  \[  f(x)\geq h(x)\geq g(x),\  x\in \lbrack 0,a].  \]
</div>
<p>5\(^{0}\) Let \(f,g:[0,2]\rightarrow \mathbb {R}\) be given by</p>
<div class="displaymath" id="a0000000060">
  \begin{align*} &  f(x) =\begin{cases}  x,&  x\in \lbrack 0,1],\\ 2(x-1),&  x\in (1,2]. \end{cases}\\ & g(x) =\begin{cases}  x,&  x\in \lbrack 0,1]\\ -3(x-1),&  x\in (1,2]. \end{cases}\end{align*}
</div>
<p> Then \(f\) is radiant, \(g\) is coradiant, \(f(1)=g(1)\) and \(f(x)\geq g(x),\) \(x\in \lbrack 0,2].\) The function \(h(x)=x,\) \(x\in \lbrack 0,2]\) is the only function having the graph between graphs of \(f\) and \(g.\) Thus Theorem <a href="#Th.7">8</a> is fulfilled. </p>

  </div>
</div> </p>
<h1 id="a0000000061">4 Applications</h1>
<p>Let \(I=[0,a]\) and let \(f:I\rightarrow \mathbb {R}\). The function \(f\) is called \(\varepsilon \)-positively homogeneous if \(\left\vert f(\lambda x)-\lambda f(x)\right\vert {\lt} \varepsilon \), for all \(x \in [0,a]\) and \(\lambda \in [0,1]\). </p>
<p>The function \(f:[0,a]\rightarrow \mathbb {R}\) is both radiant and coradiant iff \(f\) is positively homogeneous, i.e., \(f(\lambda x)=\lambda f(x)\), for all \(x \in [0,a]\) and \(\lambda \in [0,1]\). </p>
<p>The above results gives sufficient stability conditions of Hyers-Ulam type for positively homogeneous functions. </p>
<p><div class="corollary_thmwrapper " id="Corollary 11">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">11</span>
  </div>
  <div class="corollary_thmcontent">
  <p> Let \(f:[0,a]\rightarrow \mathbb {R}\) such that \(f\) is radiant or coradiant. Let \(\varepsilon {\gt}0\) be a real number. In order to obtain</p>
<div class="displaymath" id="a0000000062">
  \[  \left\vert f(\lambda x)-\lambda f(x)\right\vert {\lt}\varepsilon  \]
</div>
<p> for all \(x\in \lbrack 0,a]\) and \(\lambda \in \lbrack 0,1]\) it is sufficient that</p>
<div class="displaymath" id="a0000000063">
  \[  \left\vert f(x)-\frac{f(a)}{a}x\right\vert {\lt}\varepsilon ,\  x\in \lbrack 0,a].  \]
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000064">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Let \(f:[0,a]\rightarrow \mathbb {R}\) be radiant. By Lemma <a href="#Lemma 2">3</a> and Theorem <a href="#Th.5">6</a> it follows \(f(x)\leq \frac{f(a)}{a}x,\) for all \(x\in \lbrack 0,a].\) </p>
<p>Then</p>
<div class="displaymath" id="a0000000065">
  \begin{align*}  0 &  \leq \lambda f(x)-f(\lambda x)\leq \lambda \frac{f(a)}{a}x-f(\lambda x)=\\ &  =\frac{f(a)}{a}(\lambda x)-f(\lambda x)=\\ &  =\left\vert \frac{f(a)}{a}(\lambda x)-f(\lambda x)\right\vert {\lt}\varepsilon . \end{align*}
</div>
<p> Consequently</p>
<div class="displaymath" id="a0000000066">
  \[  \lambda f(x)-f(\lambda x)=\left\vert f(\lambda x)-\lambda f(x)\right\vert {\lt}\varepsilon .  \]
</div>
<p> Now, if \(f\) is coradiant, then by Theorem <a href="#Th.5">6</a> it follows \(f(x)\geq \frac{f(a)}{a}x\) and</p>
<div class="displaymath" id="a0000000067">
  \begin{align*}  0 &  \leq f(\lambda x)-\lambda f(x)\leq f(\lambda x)-\lambda \frac{f(a)}{a}x\\ &  =f(\lambda x)-\frac{f(a)}{a}\lambda x=\left\vert f(\lambda x)-\frac{f(a)}{a}\lambda x\right\vert {\lt}\varepsilon . \end{align*}
</div>
<p> Consequently</p>
<div class="displaymath" id="a0000000068">
  \[  f(\lambda x)-\lambda f(x)=\left\vert f(\lambda x)-\lambda f(x)\right\vert {\lt}\varepsilon  \]
</div>
<p> if</p>
<div class="displaymath" id="a0000000069">
  \[  \left\vert f(x)-\frac{f(a)}{a}x\right\vert {\lt}\varepsilon .  \]
</div>
<p> <div class="proof_wrapper" id="a0000000070">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>By Theorem <a href="#Th.5">6</a>, it follows that if \(f:[0,a]\rightarrow \mathbb {R}\) is positively homogeneous, \(x_0 \in (0,a]\) and \(f(x_0)\) is given, then the function is exactly \(f(x)=\tfrac {f(x_0)}{x_0}x,\  x \in [0,a]\). </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
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</dd>
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</dd>
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  <dd><p><a href ="http://dx.doi.org/10.2307/1995044"> <i class="sc">J.A. Johnson</i>, <i class="itshape">Banach spaces of Lipschitz functions and vector-valued Lipschitz functions</i>, Trans. Amer. Math. Soc., <b class="bfseries">148</b>(1970), pp.&#160;147–169. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
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  <dd><p><a href ="http://dx.doi.org/10.1090/S0002-9904-1934-05978-0"> <i class="sc">E.J. McShane</i>, <i class="itshape">Extension of range of functions</i>, Bull. Amer. Math. Soc., <b class="bfseries">40</b> (1934), pp.&#160;837–842. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
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</dd>
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</a> </p>
</dd>
  <dt><a name="Mustata3">8</a></dt>
  <dd><p><a href ="http://dx.doi.org/10.1007/s00025-011-0206-x"> <i class="sc">C. Mustăţa</i>, <i class="itshape">On the extensions preserving the shape of semi-Hölder function</i>, Results. Math., <b class="bfseries">63</b> (2013), pp.&#160;425–433. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
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  <dd><p><a href ="http://dx.doi.org/10.1007/BF01827935"> <i class="sc">K. Nikodem</i> and <i class="sc">S. Wasowicz</i>, <i class="itshape">A sandwich theorem and Hyers-Ulam stability of affine functions</i>, Aequationes Math., <b class="bfseries">49</b> (1995), pp.&#160;160–164. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
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  <dd><p><i class="sc">A.M. Rubinov</i>, <i class="itshape">Abstract Convexity and Global Optimization</i>, Kluwer Academic Publisher, Boston-Dordrecht-London, 2000. </p>
</dd>
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  <dd><p><i class="sc">A.M. Rubinov</i> and <i class="sc">A.P. Shveidel</i>, <i class="itshape">Radiant and star-shaped functions</i>, Pacific Journal of Optimization, <b class="bfseries">3</b> (2007) no. 1, pp.&#160;193–212. </p>
</dd>
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</dd>
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  <dd><p><a href ="http://dx.doi.org/10.1016/j.jmaa.2011.01.061"> <i class="sc">S. Suzuki</i> and <i class="sc">D. Kuroiwa</i>, <i class="itshape">Sandwich theorem for quasiconvex functions and its applications</i>, J. Math. Anal. Appl., <b class="bfseries">379</b> (2011), pp.&#160;649–655. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
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</dd>
</dl>


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