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.
Authors
Costică Mustăța
Tiberiu Popoviciu Institute of Numerical Analysis, Romania
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Paper (preprint) in HTML form
2015-Mustata-SANDWICH THEOREMS FOR RADIANT FUNCTIONS-JNAAT
SANDWICH THEOREMS FOR RADIANT FUNCTIONS
COSTICĂ MUSTĂŢA*Dedicated to prof. I. Păvăloiu on the occasion of his 75th anniversary
Abstract
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.
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 R\mathbb{R} containing zero (i.e. a radiant subset of R\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.
Sandwich theorems for diverse classes of real-valued functions (monotonic, convex, quasiconvex) were considered in [1], [2], [9], [13] and for more general functions in [3], [12], [14, etc.
Let II be an interval in R\mathbb{R} containing 0 . Then II is a radiant set, i.e. for every x in Ix \in I and lambda in[0,1]\lambda \in[0,1] it follows lambda x in I\lambda x \in I.
A function f:I rarrRf: I \rightarrow \mathbb{R} is called a radiant function if
for all x in Ix \in I and lambda in[0,1]\lambda \in[0,1].
Obviously, a function ff is radiant iff -f-f is coradiant. Every radiant function ff verifies the inequality f(0) <= 0f(0) \leq 0 and every coradiant function gg verifies g(0) >= 0g(0) \geq 0.
A function h:I rarrRh: I \rightarrow \mathbb{R} is called a Lipschitz function if there exists a constant K >= 0K \geq 0 (depending on ff and II ), such that
for all x,y in Ix, y \in I.
The constant KK is called a Lipschitz constant (for ff ) and the smallest Lipschitz constant is given by following expression:
{:(4)||f||_(I)=s u p{(|f(x)-f(y)|)/(|x-y|):x!=y,x,y in I}:}\begin{equation*}
\|f\|_{I}=\sup \left\{\frac{|f(x)-f(y)|}{|x-y|}: x \neq y, x, y \in I\right\} \tag{4}
\end{equation*}
Denote by Lip_(0)IL i p_{0} I the real linear space of Lipschitz functions on II vanishing at 0 , i.e.
{:(5)Lip_(0)I:={f:I rarrR,f(0)=0" and "||f||_(I) < oo}.:}\begin{equation*}
\operatorname{Lip}_{0} I:=\left\{f: I \rightarrow \mathbb{R}, f(0)=0 \text { and }\|f\|_{I}<\infty\right\} . \tag{5}
\end{equation*}
The functional ||||_(I):Lip_(0)I rarr[0,oo):}\left\|\|_{I}: \operatorname{Lip}_{0} I \rightarrow[0, \infty)\right. is a norm (called the Lipschitz norm) and Lip_(0)I\operatorname{Lip}_{0} I is a Banach space with respect to this norm [4].
Denote
{:[(6)r-Lip_(0)I:={f in Lip_(0)I,f" radiant "}","],[(7)cr-Lip_(0)I:={g in Lip_(0)I," g coradiant "}]:}\begin{align*}
r-L i p_{0} I & :=\left\{f \in L i p_{0} I, f \text { radiant }\right\}, \tag{6}\\
c r-L i p_{0} I & :=\left\{g \in L i p_{0} I, \text { g coradiant }\right\} \tag{7}
\end{align*}
Then r-r- Lip _(0)I_{0} I and cr-c r- Lip _(0)I_{0} I are convex cones in the linear space Lip _(0)I_{0} I and o-Lip_(0)Io-L i p_{0} I is a subspace of Lip_(0)IL i p_{0} I. Also
{:(9)o-Lip_(0)I=(r-Lip_(0))nn(cr-Lip_(0)I).:}\begin{equation*}
o-L i p_{0} I=\left(r-L i p_{0}\right) \cap\left(c r-L i p_{0} I\right) . \tag{9}
\end{equation*}
2. EXTENSIONS PRESERVING RADIANTNESS OF A LIPSCHITZ FUNCTIONS
The following extension result for real valued Lipschitz functions defined on a subset of a metric space was given by McShane [5]:
Theorem 1. Let ( X,dX, d ) be a metric space, YY a nonvoid subset of XX and f:Y rarrRf: Y \rightarrow \mathbb{R} be a Lipschitz function having the Lipschitz constant K(f)K(f) (on YY ). Then there exists a Lipschitz function F:X rarrRF: X \rightarrow \mathbb{R} such that
{:(10)F|_(Y)=f quad" and "quad K(F)=K(f).:}\begin{equation*}
\left.F\right|_{Y}=f \quad \text { and } \quad K(F)=K(f) . \tag{10}
\end{equation*}
Such a function FF is called a Lipschitz extension of ff, preserving the Lipschitz constant. In the proof of this theorem one shows that the following two functions
{:(11)F(f)(x):=i n f_(y in Y){f(y)+K(f)d(x","y)}","x in X:}\begin{equation*}
F(f)(x):=\inf _{y \in Y}\{f(y)+K(f) d(x, y)\}, x \in X \tag{11}
\end{equation*}
and
{:(12)G(f)(x):=i n f_(y in Y){f(y)-K(f)d(x","y)}","x in X:}\begin{equation*}
G(f)(x):=\inf _{y \in Y}\{f(y)-K(f) d(x, y)\}, x \in X \tag{12}
\end{equation*}
are Lipschitz extensions of ff, preserving the constant K(f)K(f), and every other extension HH verifies the inequalities:
{:(13)F(x) >= H(x) >= G(x)","x in X:}\begin{equation*}
F(x) \geq H(x) \geq G(x), x \in X \tag{13}
\end{equation*}
(see also [7], 8] for a more general situation).
In the framework considered above we obtain the following result:
Theorem 2. Let I subRI \subset \mathbb{R} such that 0in I0 \in I and f:I rarrRf: I \rightarrow \mathbb{R}.
a) If f in rf \in r-Lip _(0)I{ }_{0} I then the greatest extension of ff, namely
{:(14)F(f)(x):=i n f_(y in I){f(y)+||f||_(I)|x-y|}","x inR:}\begin{equation*}
F(f)(x):=\inf _{y \in I}\left\{f(y)+\|f\|_{I}|x-y|\right\}, x \in \mathbb{R} \tag{14}
\end{equation*}
is a radiant function on R\mathbb{R},
b) If g in crg \in c r-Lip _(0)I{ }_{0} I, then the smallest extension of gg, namely
{:(15)G(g)(x):=s u p_(y in I){g(y)-||g||_(I)|x-y|}","x inR:}\begin{equation*}
G(g)(x):=\sup _{y \in I}\left\{g(y)-\|g\|_{I}|x-y|\right\}, x \in \mathbb{R} \tag{15}
\end{equation*}
is a coradiant function on R\mathbb{R}.
c) If h in o-h \in o- Lip _(0)I_{0} I then the extension F(h)F(h) defined by (14) is in the cone rr-Lip _(0)R{ }_{0} \mathbb{R}, and G(h)G(h) defined by (15) is in the cone cr-Lip _(0)R{ }_{0} \mathbb{R}.
d) If f inf \in Lip _(0)R_{0} \mathbb{R} and there exists I subRI \subset \mathbb{R} with 0in I0 \in I such that ||f|_(I)||_(I)=||f||_(R)\left\|\left.f\right|_{I}\right\|_{I}= \|f\|_{R} and f|_(I)in o-\left.f\right|_{I} \in o- Lip _(0)I_{0} I, then
Proof. The proof is similar to that in [8. For the sake of completeness we sketch the proof.
a) Let f in rf \in r-Lip _(0)I{ }_{0} I. Then, for all lambda in[0,1]\lambda \in[0,1] and y in Iy \in I,
{:[F(f)(lambda x)=i n f_(y in I){f(y)+||f||_(I)|lambda x-y|}],[ <= i n f_(y in I){f(lambda y)+||f||_(I)|lambda x-lambda y|}],[ <= i n f_(y in I){lambda f(y)+lambda||f||_(I)|x-y|}],[=lambdai n f_(y in I){f(y)+||f||_(I)|x-y|}],[=lambda F(f)(x)]:}\begin{aligned}
F(f)(\lambda x) & =\inf _{y \in I}\left\{f(y)+\|f\|_{I}|\lambda x-y|\right\} \\
& \leq \inf _{y \in I}\left\{f(\lambda y)+\|f\|_{I}|\lambda x-\lambda y|\right\} \\
& \leq \inf _{y \in I}\left\{\lambda f(y)+\lambda\|f\|_{I}|x-y|\right\} \\
& =\lambda \inf _{y \in I}\left\{f(y)+\|f\|_{I}|x-y|\right\} \\
& =\lambda F(f)(x)
\end{aligned}
for all lambda in[0,1]\lambda \in[0,1] and all x inRx \in \mathbb{R}.
b) Let g in crg \in c r-Lip _(0)I{ }_{0} I and x inRx \in \mathbb{R}. Then for all lambda in[0,1]\lambda \in[0,1] and y in Iy \in I,
{:[G(g)(lambda x)=s u p_(y in I){g(y)-||g||_(I)|lambda x-y|}],[ >= s u p_(y in I){g(lambda y)-||g||_(I)|lambda x-lambda y|}],[ >= lambdas u p_(y in I){g(y)-||g||_(I)|x-y|}],[=lambda G(g)(x)]:}\begin{aligned}
G(g)(\lambda x) & =\sup _{y \in I}\left\{g(y)-\|g\|_{I}|\lambda x-y|\right\} \\
& \geq \sup _{y \in I}\left\{g(\lambda y)-\|g\|_{I}|\lambda x-\lambda y|\right\} \\
& \geq \lambda \sup _{y \in I}\left\{g(y)-\|g\|_{I}|x-y|\right\} \\
& =\lambda G(g)(x)
\end{aligned}
Consequently G(g)(lambda x) >= lambda G(g)(x)G(g)(\lambda x) \geq \lambda G(g)(x) for all lambda in[0,1]\lambda \in[0,1] and all x inRx \in \mathbb{R}.
c) If h in o-Lip_(0)I=(r-Lip_(0)I)nn(cr-Lip_(0)I)h \in o-\operatorname{Lip}_{0} I=\left(r-\operatorname{Lip}_{0} I\right) \cap\left(c r-\operatorname{Lip}_{0} I\right), then the assertions from c) follow from a) and b).
d) Let f inLip_(0)Rf \in \operatorname{Lip}_{0} \mathbb{R} and suppose that there exists I subRI \subset \mathbb{R} with 0in I0 \in I such that f|_(I)in o-\left.f\right|_{I} \in o- Lip _(0)I_{0} I and ||f|_(I)=||f||_(R)\left\|\left.f\right|_{I}=\right\| f \|_{R}. Then ff is an extension of f|_(I)\left.f\right|_{I} preserving the smallest Lipschitz constant ||f|_(I)||\left\|\left.f\right|_{I}\right\| and the assertion follows by (13) and c).
3. SANDWICH THEOREMS
The assertion d) in Theorem 2 suggests the following problem: Let f,gf, g be two functions, on an interval I(0in I),fI(0 \in I), f radiant (coradiant) and gg coradiant (radiant), and f(x) <= g(x)f(x) \leq g(x) for all x in Ix \in I. Is there a function h:I rarrRh: I \rightarrow \mathbb{R} verifying h(lambda x)=lambda h(x)h(\lambda x)=\lambda h(x), for all x in Ix \in I and all lambda in[0,1]\lambda \in[0,1], and such that f(x) <= h(x) <= g(x),x in If(x) \leq h(x) \leq g(x), x \in I ? Problems of such type were considered for example in [1], [2], [9] and, more general in [3], [12], [14], etc. The studies in this direction are motivated by applications to the theory of optimization and numerical analysis.
We show that the answer is affirmative in both cases.
Firstly, we need the following Lemma.
Lemma 3. Let f,g:[0,a]rarrR,ff, g:[0, a] \rightarrow \mathbb{R}, f radiant and gg coradiant such that f(x) <= g(x),x in[0,a]f(x) \leq g(x), x \in[0, a]. Then the following assertions hold:
a) The function p_(f):(0,a]rarrRp_{f}:(0, a] \rightarrow \mathbb{R} defined by
Proof. a) Let f:[0,a]rarrR,ff:[0, a] \rightarrow \mathbb{R}, f radiant. For every x in(0,a]x \in(0, a] let lambda in[0,1]\lambda \in[0,1] be such that x=lambda ax=\lambda a. Then
Let x_(1),x_(2)in(0,a],x_(1) <= x_(2)x_{1}, x_{2} \in(0, a], x_{1} \leq x_{2}. Then there exists lambda in(0,1]\lambda \in(0,1] such that x_(1)=lambdax_(2)x_{1}=\lambda x_{2}, so that
Consequently p_(f)p_{f} is nondecreasing and p_(g)p_{g} is nonincreasing on ( 0,a0, a ].
b) Taking into account the hypothesis and the conclusions in a), the inequalities (19) follow.
Remark 4. The inequalities (17), (18) and (19) imply that every line passing through the points ( 0,0 ) and ( x,f(x)x, f(x) ) is above the graph of ff on the interval [ 0,x0, x ], and every line passing through the points ( 0,0 ) and ( x,g(x)x, g(x) ) is bellow the graph of gg on [0,x][0, x]. Also, the line passing through ( 0,0 ) and (x,g(x))(x, g(x)) is above the line passing through (0,0)(0,0) and (x,f(x))(x, f(x)).
We obtain the main result of the paper.
Theorem 5. Let f,g:[0,a]rarrR,ff, g:[0, a] \rightarrow \mathbb{R}, f radiant and gg coradiant, and f(x) <= g(x)f(x) \leq g(x), for all x in[0,a]x \in[0, a]. Then there exists at least a function h:[0,a]rarrRh:[0, a] \rightarrow \mathbb{R} having the form h(x)=alpha x,alpha inRh(x)=\alpha x, \alpha \in \mathbb{R} and such that
If there exists x_(0)in(0,a]x_{0} \in(0, a] such that f(x_(0))=g(x_(0))f\left(x_{0}\right)=g\left(x_{0}\right) then the unique function h:[0,a]rarrRh:[0, a] \rightarrow \mathbb{R} verifying 20 is h(x)=(f(a))/(a)xh(x)=\frac{f(a)}{a} x.
Proof. By Lemma 3, taking into account (19), one obtains
Then, for alpha in[(f(a))/(a),(g(a))/(a)]\alpha \in\left[\frac{f(a)}{a}, \frac{g(a)}{a}\right] the function h(x)=alpha x,x in[0,a]h(x)=\alpha x, x \in[0, a] verifies the properties from the conclusions of first part of theorem.
Now, if f(x_(0))=g(x_(0))f\left(x_{0}\right)=g\left(x_{0}\right) at x_(0)in(0,a]x_{0} \in(0, a] one obtains
i.e f(x)=g(x)=(f(a))/(a)xf(x)=g(x)=\frac{f(a)}{a} x. Then h(x)=(f(a))/(a)xh(x)=\frac{f(a)}{a} x verifies f(x) <= h(x) <= g(x),x in[0,a]f(x) \leq h(x) \leq g(x), x \in [0, a], and the second part of the theorem follows.
In the case I=[0,oo)I=[0, \infty) one obtains.
Theorem 6. Let f,g:[0,oo)rarrR,ff, g:[0, \infty) \rightarrow \mathbb{R}, f radiant and gg coradiant, and f(x) <= g(x)f(x) \leq g(x) for all x in[0,oo)x \in[0, \infty). Then there exists at least a function h:[0,oo)rarrRh:[0, \infty) \rightarrow \mathbb{R}, of the form h(x)=alpha x,alpha inRh(x)=\alpha x, \alpha \in \mathbb{R} fixed such that f(x) <= h(x) <= g(x),x in[0,oo)f(x) \leq h(x) \leq g(x), x \in[0, \infty).
If there exists x_(0) > 0x_{0}>0 such that f(x_(0))=g(x_(0))f\left(x_{0}\right)=g\left(x_{0}\right) then the unique function h:[0,oo)rarrRh:[0, \infty) \rightarrow \mathbb{R} satisfying f(x) <= h(x) <= g(x)f(x) \leq h(x) \leq g(x) is h(x)=(f(x_(0)))/(x_(0))xh(x)=\frac{f\left(x_{0}\right)}{x_{0}} x.
f(x) <= alpha x <= g(x),x in[0,oo)f(x) \leq \alpha x \leq g(x), x \in[0, \infty)
Then h(x)=alpha x,x in[0,oo)h(x)=\alpha x, x \in[0, \infty), satisfies the first conclusion of the theorem.
Let x_(0) > 0x_{0}>0 such that f(x_(0))=g(x_(0))f\left(x_{0}\right)=g\left(x_{0}\right). For every x > x_(0)x>x_{0} we have
It follows f(x)=g(x)=(f(x_(0)))/(x_(0))xf(x)=g(x)=\frac{f\left(x_{0}\right)}{x_{0}} x, for all x > x_(0)x>x_{0} and consequently, for x in[0,oo)x \in[0, \infty) the function h(x)=(f(x_(0)))/(x_(0))xh(x)=\frac{f\left(x_{0}\right)}{x_{0}} x verifies
f(x) <= h(x) <= g(x).f(x) \leq h(x) \leq g(x) .
Now we consider I=[a,0]I=[a, 0].
Theorem 7. Let f,g:[a,0]rarrR(a < 0),ff, g:[a, 0] \rightarrow \mathbb{R}(a<0), f radiant and gg coradiant, and f(x) <= g(x)f(x) \leq g(x), for all x in[a,0]x \in[a, 0]. Then there exists at least a function h:[a,0]rarrRh:[a, 0] \rightarrow \mathbb{R} having the form h(x)=alpha x,alpha inRh(x)=\alpha x, \alpha \in \mathbb{R} fixed, and verifying the inequalities:
If there exists x_(0)in[a,0)x_{0} \in[a, 0) such that f(x_(0))=g(x_(0))f\left(x_{0}\right)=g\left(x_{0}\right) then the only function h:[a,0]rarrRh:[a, 0] \rightarrow \mathbb{R} such that f(x) <= h(x) <= g(x)f(x) \leq h(x) \leq g(x) is h(x)=(f(a))/(a)xh(x)=\frac{f(a)}{a} x.
Proof. Observe that for all x in[a,0]x \in[a, 0] one obtains f(x) <= (f(a))/(a)xf(x) \leq \frac{f(a)}{a} x. Indeed, if x in[a,0)x \in[a, 0), there exists lambda in[0,1]\lambda \in[0,1] such that x=lambda ax=\lambda a. Then
f(x)=f(lambda a) <= lambda f(a)=(f(a))/(a)(lambda a)=(f(a))/(a)xf(x)=f(\lambda a) \leq \lambda f(a)=\frac{f(a)}{a}(\lambda a)=\frac{f(a)}{a} x
Analogously,
g(x)=g(lambda a) >= lambda g(a)=(g(a))/(a)(lambda a)=(g(a))/(a)xg(x)=g(\lambda a) \geq \lambda g(a)=\frac{g(a)}{a}(\lambda a)=\frac{g(a)}{a} x
But g(a) >= f(a)g(a) \geq f(a) implies (g(a))/(a) <= (f(a))/(a)\frac{g(a)}{a} \leq \frac{f(a)}{a}, and for alpha in[(g(a))/(a),(f(a))/(a)]\alpha \in\left[\frac{g(a)}{a}, \frac{f(a)}{a}\right] it follows
(g(a))/(a)x >= alpha x >= (f(a))/(a)x,x in[a,0]\frac{g(a)}{a} x \geq \alpha x \geq \frac{f(a)}{a} x, x \in[a, 0]
and, consequently f(x) <= alpha x <= g(x),x in[0,a]f(x) \leq \alpha x \leq g(x), x \in[0, a].
Now, let x_(0)in[a,0)x_{0} \in[a, 0) such that f(x_(0))=g(x_(0))f\left(x_{0}\right)=g\left(x_{0}\right).
Then
We are lead to f(a) >= g(a)f(a) \geq g(a), and because of hypothesis f(a) <= g(a)f(a) \leq g(a) it follows (f(a))/(a)=(g(a))/(a)\frac{f(a)}{a}=\frac{g(a)}{a}.
Then h(x)=(f(a))/(a)xh(x)=\frac{f(a)}{a} x is the only function having the graph between the graphs of ff and gg on [a,0][a, 0].
Suppose now that ff is radiant, gg is coradiant and f(x) >= g(x)f(x) \geq g(x) for all x in[0,a]x \in[0, a]. Because f(0) <= 0f(0) \leq 0 and g(0) >= 0g(0) \geq 0 it follows f(0)=g(0)=0f(0)=g(0)=0.
The following theorem holds.
Theorem 8. Let f,g:[0,a]rarrR,ff, g:[0, a] \rightarrow \mathbb{R}, f radiant, gg coradiant and f(x) >= g(x)f(x) \geq g(x), x in[0,a]x \in[0, a]. Then there exists at least a function h:[0,a]rarrRh:[0, a] \rightarrow \mathbb{R} having the form h(x)=alpha x,alpha inRh(x)=\alpha x, \alpha \in \mathbb{R} fixed, and such that
If there exists x_(0)in(0,a]x_{0} \in(0, a] such that f(x_(0))=g(x_(0))f\left(x_{0}\right)=g\left(x_{0}\right) then f(x)=g(x)f(x)=g(x) for every x in[0,x_(0)]x \in\left[0, x_{0}\right] and the only function h:[0,a}rarrRh:[0, a\} \rightarrow \mathbb{R} verifying 22 is h(x)=(f(x_(0)))/(x_(0))xh(x)=\frac{f\left(x_{0}\right)}{x_{0}} x.
Proof. Let x_(0)in(0,a]x_{0} \in(0, a] be such that f(x_(0))=g(x_(0))f\left(x_{0}\right)=g\left(x_{0}\right). If x in(0,x_(0)]x \in\left(0, x_{0}\right], by Lemma 3 one obtains
It follows g(x) >= f(x),x in[0,x_(0)]g(x) \geq f(x), x \in\left[0, x_{0}\right] and because f(x) >= g(x)f(x) \geq g(x) (by hypothesis) one obtains f(x)=g(x)f(x)=g(x) for every x in[0,x_(0)]x \in\left[0, x_{0}\right].
The line h(x)=(f(x_(0)))/(x_(0))x(=(g(x_(0)))/(x_(0))x)h(x)=\frac{f\left(x_{0}\right)}{x_{0}} x\left(=\frac{g\left(x_{0}\right)}{x_{0}} x\right) is between the graphs of ff and gg over the interval [0,x_(0)]\left[0, x_{0}\right].
For x > x_(0),x <= ax>x_{0}, x \leq a
(f(x_(0)))/(x_(0)) <= (f(x))/(x)" and "(f(x_(0)))/(x_(0)) <= (g(x))/(x)\frac{f\left(x_{0}\right)}{x_{0}} \leq \frac{f(x)}{x} \text { and } \frac{f\left(x_{0}\right)}{x_{0}} \leq \frac{g(x)}{x}
and consequently the function h(x)=(f(x_(0)))/(x_(0))x,x in[0,a]h(x)=\frac{f\left(x_{0}\right)}{x_{0}} x, x \in[0, a] verifies
it follows that i n f{(f(x))/(x):x in(0,a]}\inf \left\{\frac{f(x)}{x}: x \in(0, a]\right\} and s u p{(g(x))/(x):x in(0,a]}\sup \left\{\frac{g(x)}{x}: x \in(0, a]\right\} are finite, and
s u p_(0 < x <= a)(g(x))/(x) <= i n f_(0 < x <= a)(f(x))/(x).\sup _{0<x \leq a} \frac{g(x)}{x} \leq \inf _{0<x \leq a} \frac{f(x)}{x} .
By considering alpha in[s u p_(0 < x <= a)(g(x))/(x),i n f_(0 < x <= a)(f(x))/(x)]\alpha \in\left[\sup _{0<x \leq a} \frac{g(x)}{x}, \inf _{0<x \leq a} \frac{f(x)}{x}\right], the line h(x)=alpha x,x in[0,a]h(x)=\alpha x, x \in[0, a] lies between the graphs of ff and gg, i.e.
Remark 9. The result in Theorem 8 is valid also if f,g:(-oo,0]rarrRf, g:(-\infty, 0] \rightarrow \mathbb{R}. In this case there exists alpha in[lim_(a rarr-oo)s u p_(a < x <= 0)(g(x))/(x),lim_(a rarr-oo)i n f_(a < x <= 0)(f(x))/(x)]\alpha \in\left[\lim _{a \rightarrow-\infty} \sup _{a<x \leq 0} \frac{g(x)}{x}, \lim _{a \rightarrow-\infty} \inf _{a<x \leq 0} \frac{f(x)}{x}\right] such that the function h(x)=alpha xh(x)=\alpha x satisfies the inequalities
Now consider I=[a,b]I=[a, b] where a < 0 < ba<0<b, or I=RI=\mathbb{R}.
By the above results it follows:
Corollary 10. a) Let f,g:[a,b]rarrR,a < 0 < bf, g:[a, b] \rightarrow \mathbb{R}, a<0<b, ff radiant, gg coradiant and such that f(x) <= g(x),x in[a,b]f(x) \leq g(x), x \in[a, b]. Then there exists h:[a,b]rarrRh:[a, b] \rightarrow \mathbb{R},
such that f(x) <= h(x) <= g(x),x in[a,b]f(x) \leq h(x) \leq g(x), x \in[a, b].
b) Let f,g:RrarrRf, g: \mathbb{R} \rightarrow \mathbb{R}, ff radiant, gg coradiant, f(x) <= g(x)f(x) \leq g(x), for all x inRx \in \mathbb{R}.
Then there exists h:RrarrRh: \mathbb{R} \rightarrow \mathbb{R} of the form
h(x)={[alpha x","x in(-oo","0]","],[beta x","x in(0","oo)","]:}h(x)=\left\{\begin{array}{c}
\alpha x, x \in(-\infty, 0], \\
\beta x, x \in(0, \infty),
\end{array}\right.
A similar result is valid if f(x) >= g(x),ff(x) \geq g(x), f radiant and gg coradiant on [a,b][a, b], a < 0 < ba<0<b, respectively on R\mathbb{R}.
In Corollary 11 a), if [(g(a))/(a),(f(a))/(a)]nn[(f(b))/(b),(g(b))/(b)]!=O/\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\mu is a number from this set, then h(x)=mu x,x in[a,b]h(x)=\mu x, x \in[a, b] verify
f(x) <= mu x <= g(x),x in[a,b].f(x) \leq \mu x \leq g(x), x \in[a, b] .
A similar result follows in the case b).
If f:I rarrRquad(0in I)f: I \rightarrow \mathbb{R} \quad(0 \in I), and ff is a convex (concave) function on I,f(0)=0I, f(0)=0, then ff is radiant (coradiant). The above results may be enounced for convex and concave function defined on II, vanishing at zero.
Examples. 1^(0)1^{0} Let f_(m),g:[0,2]rarrRf_{m}, g:[0,2] \rightarrow \mathbb{R} be the functions defined by
{:[f_(m)(x)={[mx^(3)",",x in[0","1]],[mx",",x in(1","2]],m inR,:}],[g(x)={[-x^(2)+2",",x in[0","1]],[x",",x in(1","2]]:}]:}\begin{aligned}
& f_{m}(x)=\left\{\begin{array}{ll}
m x^{3}, & x \in[0,1] \\
m x, & x \in(1,2]
\end{array}, m \in \mathbb{R},\right. \\
& g(x)= \begin{cases}-x^{2}+2, & x \in[0,1] \\
x, & x \in(1,2]\end{cases}
\end{aligned}
Then, for m > 0m>0 the function f_(m)f_{m} is radiant, gg is coradiant and f_(m)(x) <= g(x)f_{m}(x) \leq g(x), x in[0,2]x \in[0,2]. Every function h:[0,2]rarrR,h(x)=alpha xh:[0,2] \rightarrow \mathbb{R}, h(x)=\alpha x, where alpha in[m,1]\alpha \in[m, 1] verifies
Also, for m=1,f_(1)(x) <= g(x),x in[0,2]m=1, f_{1}(x) \leq g(x), x \in[0,2] and because f_(1)(1)=g(1)=1f_{1}(1)=g(1)=1 one obtains that h(x)=xh(x)=x is the unique function verifying f_(1)(x) <= h(x) <= g(x)f_{1}(x) \leq h(x) \leq g(x), x in[0,2]x \in[0,2]. Consequently Theorem 6 is fulfilled. 2^(0)2^{0} Let a > 1,m in(0,1]a>1, m \in(0,1] and let f_(m),g_(a):[0,oo)rarrRf_{m}, g_{a}:[0, \infty) \rightarrow \mathbb{R}, be the functions:
cf_(m)(x)={[(mx^(3))/(a^(2))",",x in[0","a]","],[mx",",x in(a+oo)","]g_(a)(x)={[-x^(2)+(1+a)x",",x in[0","a]],[x",",x in(a","+oo)]:}c f_{m}(x)=\left\{\begin{array}{ll}
\frac{m x^{3}}{a^{2}}, & x \in[0, a], \\
m x, & x \in(a+\infty),
\end{array} g_{a}(x)= \begin{cases}-x^{2}+(1+a) x, & x \in[0, a] \\
x, & x \in(a,+\infty)\end{cases}\right.
Then f_(m)(x) <= g_(a)(x)f_{m}(x) \leq g_{a}(x), for x in[0,oo),f_(m)x \in[0, \infty), f_{m} is radiant and g_(a)g_{a} is coradiant.
For every alpha in[m,1]\alpha \in[m, 1] one obtains
f_(m)(x) <= h(x)=alpha x <= g(x),x in[0,oo)f_{m}(x) \leq h(x)=\alpha x \leq g(x), x \in[0, \infty)
For m=1,f_(1)(a)=g_(a)(a)=am=1, f_{1}(a)=g_{a}(a)=a and consequently h(x)=xh(x)=x is the only function verifying f_(1)(x) <= h(x)=x <= g(x),x in[0,oo)f_{1}(x) \leq h(x)=x \leq g(x), x \in[0, \infty). Theorem 7 is fulfilled. 3^(0)3^{0} Let f,g:[-2,0]rarrR,f(x)=x^(2)+xf, g:[-2,0] \rightarrow \mathbb{R}, f(x)=x^{2}+x and g(x)=-x^(2)-4xg(x)=-x^{2}-4 x. Then ff is radiant, gg is coradiant and f(x) <= g(x),x in[-2,0]f(x) \leq g(x), x \in[-2,0]. Every line h(x)=alpha xh(x)=\alpha x, where alpha in[-2,-1]\alpha \in[-2,-1] has the graph between the graphs of ff and gg.
Now let f,g:[-3,0]rarrRf, g:[-3,0] \rightarrow \mathbb{R} be the functions
{:[f(x)={[(-x)/(2)",",x in[-3","-1]],[x^(2)+(x)/(2)",",x in(-1","0]]:}],[g(x)={[(-x)/(2)",",x in[-3","-1]],[-4x^(2)(9)/(2)x",",x in(-1","0]]:}]:}\begin{aligned}
& f(x)= \begin{cases}\frac{-x}{2}, & x \in[-3,-1] \\
x^{2}+\frac{x}{2}, & x \in(-1,0]\end{cases} \\
& g(x)= \begin{cases}\frac{-x}{2}, & x \in[-3,-1] \\
-4 x^{2} \frac{9}{2} x, & x \in(-1,0]\end{cases}
\end{aligned}
Then ff is radiant, gg is coradiant and f(x) <= g(x),x in[-3,0]f(x) \leq g(x), x \in[-3,0].
The only function hh such that f(x) <= h(x) <= g(x),x in[-3,0]f(x) \leq h(x) \leq g(x), x \in[-3,0] is h(x)=-(1)/(2)xh(x)= -\frac{1}{2} x. Thus Theorem 8 is fulfilled. 4^(0)4^{0} Let f_(m),g:[0,a]rarrR,(a > 0,m >= 0)f_{m}, g:[0, a] \rightarrow \mathbb{R},(a>0, m \geq 0) be the functions
{:[f_(m)(x)=x^(2)+mx],[g(x)=-x^(3)]:}\begin{aligned}
f_{m}(x) & =x^{2}+m x \\
g(x) & =-x^{3}
\end{aligned}
Then f_(m)f_{m} is radiant, gg is coradiant and f(x) >= g(x),x in[0,a]f(x) \geq g(x), x \in[0, a]. The function h(x)=alpha xh(x)=\alpha x, where alpha in[0,m]\alpha \in[0, m] is such that
5^(0)5^{0} Let f,g:[0,2]rarrRf, g:[0,2] \rightarrow \mathbb{R} be given by
{:[f(x)={[x",",x in[0","1]],[2(x-1)",",x in(1","2]]:}],[g(x)={[x",",x in[0","1]],[-3(x-1)",",x in(1","2]]:}]:}\begin{aligned}
& f(x)= \begin{cases}x, & x \in[0,1] \\
2(x-1), & x \in(1,2]\end{cases} \\
& g(x)= \begin{cases}x, & x \in[0,1] \\
-3(x-1), & x \in(1,2]\end{cases}
\end{aligned}
Then ff is radiant, gg is coradiant, f(1)=g(1)f(1)=g(1) and f(x) >= g(x),x in[0,2]f(x) \geq g(x), x \in[0,2]. The function h(x)=x,x in[0,2]h(x)=x, x \in[0,2] is the only function having the graph between graphs of ff and gg. Thus Theorem 8 is fulfilled.
4. APPLICATIONS
Let I=[0,a]I=[0, a] and let f:I rarrRf: I \rightarrow \mathbb{R}. The function ff is called epsi\varepsilon-positively homogeneous if |f(lambda x)-lambda f(x)| < epsi|f(\lambda x)-\lambda f(x)|<\varepsilon, for all x in[0,a]x \in[0, a] and lambda in[0,1]\lambda \in[0,1].
The function f:[0,a]rarrRf:[0, a] \rightarrow \mathbb{R} is both radiant and coradiant iff ff is positively homogeneous, i.e., f(lambda x)=lambda f(x)f(\lambda x)=\lambda f(x), for all x in[0,a]x \in[0, a] and lambda in[0,1]\lambda \in[0,1].
The above results gives sufficient stability conditions of Hyers-Ulam type for positively homogeneous functions.
Corollary 11. Let f:[0,a]rarrRf:[0, a] \rightarrow \mathbb{R} such that ff is radiant or coradiant. Let epsi > 0\varepsilon>0 be a real number. In order to obtain
for all x in[0,a]x \in[0, a] and lambda in[0,1]\lambda \in[0,1] it is sufficient that
|f(x)-(f(a))/(a)x| < epsi,x in[0,a].\left|f(x)-\frac{f(a)}{a} x\right|<\varepsilon, x \in[0, a] .
Proof. Let f:[0,a]rarrRf:[0, a] \rightarrow \mathbb{R} be radiant. By Lemma 3 and Theorem 6 it follows f(x) <= (f(a))/(a)xf(x) \leq \frac{f(a)}{a} x, for all x in[0,a]x \in[0, a].
By Theorem 6, it follows that if f:[0,a]rarrRf:[0, a] \rightarrow \mathbb{R} is positively homogeneous, x_(0)in(0,a]x_{0} \in(0, a] and f(x_(0))f\left(x_{0}\right) is given, then the function is exactly f(x)=(f(x_(0)))/(x_(0))x,x in[0,a]f(x)=\frac{f\left(x_{0}\right)}{x_{0}} x, x \in [0, a].
