On a problem of extremum

Costica Mustata
(original), Approximation Theory, paper

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Costica Mustata
Institutul de Calcul

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C. Mustăţa, On a problem of extremum, ”Babeş-Bolyai” Univ., Research Seminars, Seminar on Mathematical Analysis, Preprint nr.7 (1991), 107-114 (MR # 94a: 26007).

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Preprint 7

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Babes-Bolyai University, Romania

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MR # 94a: 26007

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[1] Aronsson, G., Extension of functions satisfying Lipschitz conditions, Arkiv for Matematik 6 (1967), nr.28, 551-561.
[2] Mc Shane, E.J., Extension of range of functions, Bull. amer. Math. Soc. 40 (1934), 837-842.
[3] Mocanu, P., Variatiuni pe o temă de concurs, Seminarul ”Didactica Matematicii”, 1985-1986, 123-128.
[4] Mustata, C., On the extension problem with prescribed norm, Seminar of Functional analysis and Numerical Methods, Preprint nr.4 (1981), 93-99.
[5] Trenoguine, V., Analyse fonctionnelle, Edition Mir. Moscow, 1985.

 

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1991-Mustata-UBB-Seminar-On-a-problem-of-extremum

ON A PROBLEM OF EXTREMUM
Costica Mustata

Let [ a , b ] [ a , b ] [a,b][a, b][a,b] be an interval of the real axis and let D : a == x 0 < x 1 < x 2 < … < x n = b D : a == x 0 < x 1 < x 2 < … < x n = b D:a==x_(0) < x_(1) < x_(2) < dots < x_(n)=bD: a= =x_{0}<x_{1}<x_{2}<\ldots<x_{n}=bD:a==x0<x1<x2<…<xn=b be a division of this real interval. Let
V = { y k : k = 0 , 1 , … , n } ⊂ R V = y k : k = 0 , 1 , … , n ⊂ R V={y_(k):k=0,1,dots,n}subRV=\left\{y_{k}: k=0,1, \ldots, n\right\} \subset \mathbb{R}V={yk:k=0,1,…,n}⊂R and let M > 0 M > 0 M > 0M>0M>0 be, such that M > | y k + 1 − y k | / ( x k + 1 − x k ) , k = 0 , 1 , … , n − 1 M > y k + 1 − y k / x k + 1 − x k , k = 0 , 1 , … , n − 1 M > |y_(k+1)-y_(k)|//(x_(k+1)-x_(k)),k=0,1,dots,n-1M>\left|y_{k+1}-y_{k}\right| /\left(x_{k+1}-x_{k}\right), k=0,1, \ldots, n-1M>|yk+1−yk|/(xk+1−xk),k=0,1,…,n−1.
A A AAA function f : [ a , b ] → R f : [ a , b ] → R f:[a,b]rarrRf:[a, b] \rightarrow \mathbb{R}f:[a,b]→R is called Lipschitz on [ a , b ] [ a , b ] [a,b][a, b][a,b] if there exists a number K ≥ 0 K ≥ 0 K >= 0K \geq 0K≥0 such that:
(1) | f ( x ) − f ( y ) | ≤ K | x − y | , (1) | f ( x ) − f ( y ) | ≤ K | x − y | , {:(1)|f(x)-f(y)| <= K|x-y|",":}\begin{equation*} |f(x)-f(y)| \leq K|x-y|, \tag{1} \end{equation*}(1)|f(x)−f(y)|≤K|x−y|,
for all x , y ∈ [ a , b ] x , y ∈ [ a , b ] x,y in[a,b]x, y \in[a, b]x,y∈[a,b]. We shall denote by K f K f K_(f)K_{f}Kf the smallest of the numbers K K KKK for which the relation (1) holds and we shall call it the Lipschitz norm of the function f f fff. Obviously that K f K f K_(f)K_{f}Kf is given by:
K f = sup { | f ( x ) − f ( y ) | / | x − y | ; x , y ∈ [ a , b ] , x ≠ y } . K f = sup { | f ( x ) − f ( y ) | / | x − y | ; x , y ∈ [ a , b ] , x ≠ y } . K_(f)=s u p{|f(x)-f(y)|//|x-y|;x,y in[a,b],x!=y}.K_{f}=\sup \{|f(x)-f(y)| /|x-y| ; x, y \in[a, b], x \neq y\} .Kf=sup{|f(x)−f(y)|/|x−y|;x,y∈[a,b],x≠y}.
Denote by Lip[a,b] the set of all real-valued Lipschitz functions defined on [ a , b a , b a,b\mathrm{a}, \mathrm{b}a,b ] and let
(2) T ( D k , V , M ) = { f ∈ Lip [ a , b ] : f ( x k ) = y k , k = 0 , n ― , K f ≤ M } T D k , V , M = f ∈ Lip [ a , b ] : f x k = y k , k = 0 , n ¯ , K f ≤ M T(D^(k),V,M)={f in Lip[a,b]:f(x_(k))=y_(k),k= bar(0,n),K_(f) <= M}\mathscr{T}\left(D^{k}, V, M\right)=\left\{f \in \operatorname{Lip}[a, b]: f\left(x_{k}\right)=y_{k}, k=\overline{0, n}, K_{f} \leq M\right\}T(Dk,V,M)={f∈Lip[a,b]:f(xk)=yk,k=0,n―,Kf≤M}. The function whose graph is the polygonal line joining the points ( x k , y k ) , k = 0 , 1 , … , n x k , y k , k = 0 , 1 , … , n (x_(k),y_(k)),k=0,1,dots,n\left(x_{k}, y_{k}\right), k=0,1, \ldots, n(xk,yk),k=0,1,…,n belongs to M ( D , V , M ) M ( D , V , M ) M(D,V,M)\mathscr{M}(D, V, M)M(D,V,M) so that
ϕ ≠ ρ ( D , V , M ) ⊂ Lip [ a , b ] ⊂ c [ a , b ] , ϕ ≠ ρ ( D , V , M ) ⊂ Lip [ a , b ] ⊂ c [ a , b ] , phi!=rho(D,V,M)sub Lip[a,b]sub c[a,b],\phi \neq \rho(D, V, M) \subset \operatorname{Lip}[a, b] \subset c[a, b],ϕ≠ρ(D,V,M)⊂Lip[a,b]⊂c[a,b],
where, as usually, C [ a , b ] C [ a , b ] C[a,b]C[a, b]C[a,b] denotes the Banach space of all continuous real-valued- functions defined on [ a , b a , b a,b\mathrm{a}, \mathrm{b}a,b ], equiped with the uniform norm:
(3)
‖ f ‖ = sup { | f ( x ) | : x ∈ [ a , b ] } , f ∈ C [ a , b ] . ‖ f ‖ = sup { | f ( x ) | : x ∈ [ a , b ] } , f ∈ C [ a , b ] . ||f||=s u p{|f(x)|:x in[a,b]},quad f in C[a,b].\|f\|=\sup \{|f(x)|: x \in[a, b]\}, \quad f \in C[a, b] .‖f‖=sup{|f(x)|:x∈[a,b]},f∈C[a,b].
As a subset of the Banach space C [ a , b ] C [ a , b ] C[a,b]C[a, b]C[a,b] the set S ( D , V , M ) S ( D , V , M ) S(D,V,M)\mathscr{S}(D, V, M)S(D,V,M) has the following properties:
THEOREM 1. a) The set P ( D , V , M ) P ( D , V , M ) P(D,V,M)\mathscr{P}(\mathrm{D}, \mathrm{V}, \mathrm{M})P(D,V,M) is a convex subset of C
b) The functions F i F i F_(i)F_{i}Fi and F s F s F_(s)F_{s}Fs given by
(4) F i ( x ) = max { f ( x k ) − M | x − x k | , k = 0 , 1 , … , n } , F s ( x ) = min { f ( x k ) + M | x − x k | , k = 0 , 1 , … , n } , (4) F i ( x ) = max f x k − M x − x k , k = 0 , 1 , … , n , F s ( x ) = min f x k + M x − x k , k = 0 , 1 , … , n , {:[(4)F_(i)(x)=max{f(x_(k))-M|x-x_(k)|,k=0,1,dots,n}","],[F_(s)(x)=min{f(x_(k))+M|x-x_(k)|,k=0,1,dots,n}","]:}\begin{align*} & F_{i}(x)=\max \left\{f\left(x_{k}\right)-M\left|x-x_{k}\right|, k=0,1, \ldots, n\right\}, \tag{4}\\ & F_{s}(x)=\min \left\{f\left(x_{k}\right)+M\left|x-x_{k}\right|, k=0,1, \ldots, n\right\}, \end{align*}(4)Fi(x)=max{f(xk)−M|x−xk|,k=0,1,…,n},Fs(x)=min{f(xk)+M|x−xk|,k=0,1,…,n},
for x ∈ [ a , b ] x ∈ [ a , b ] x in[a,b]x \in[a, b]x∈[a,b], are extremal points of S ( D , V , M ) S ( D , V , M ) S(D,V,M)\mathscr{S}(D, V, M)S(D,V,M);
c) The set S ( D , V , M ) S ( D , V , M ) S(D,V,M)S(\mathrm{D}, \mathrm{V}, \mathrm{M})S(D,V,M) is compact with respect to the uniform topol ngy of the space c [ a , b ] c [ a , b ] c[a,b]c[\mathrm{a}, \mathrm{b}]c[a,b].
Proof. a) Let f 1 , f 2 ∈ S ( D , V , M ) , λ ∈ [ 0 , 1 ] f 1 , f 2 ∈ S ( D , V , M ) , λ ∈ [ 0 , 1 ] f_(1),f_(2)in S(D,V,M),lambda in[0,1]f_{1}, f_{2} \in S(D, V, M), \lambda \in[0,1]f1,f2∈S(D,V,M),λ∈[0,1] and f = λ f 1 + + ( 1 − λ ) f 2 f = λ f 1 + + ( 1 − λ ) f 2 f=lambdaf_(1)++(1-lambda)f_(2)f=\lambda f_{1}+ +(1-\lambda) f_{2}f=λf1++(1−λ)f2. Then, obviously, f ( x k ) = y k , k = 0 , 1 , … , n f x k = y k , k = 0 , 1 , … , n f(x_(k))=y_(k),k=0,1,dots,nf\left(x_{k}\right)=y_{k}, k=0,1, \ldots, nf(xk)=yk,k=0,1,…,n and
| f ( x ) − f ( y ) | ≤ λ | f 1 ( x ) − f 1 ( y ) | + ( 1 − λ ) ⋅ | f 2 ( x ) − f 2 ( y ) | ≤ ≤ ( λ K f 1 + ( 1 − λ ) K f 2 ) ⋅ | x − y | ≤ ≤ ( λ M + ( 1 − λ ) M ) | x − y | = M ⋅ | x − y | | f ( x ) − f ( y ) | ≤ λ f 1 ( x ) − f 1 ( y ) + ( 1 − λ ) ⋅ f 2 ( x ) − f 2 ( y ) ≤ ≤ λ K f 1 + ( 1 − λ ) K f 2 ⋅ | x − y | ≤ ≤ ( λ M + ( 1 − λ ) M ) | x − y | = M ⋅ | x − y | {:[|f(x)-f(y)| <= lambda|f_(1)(x)-f_(1)(y)|+(1-lambda)*|f_(2)(x)-f_(2)(y)| <= ],[ <= (lambdaK_(f_(1))+(1-lambda)K_(f_(2)))*|x-y| <= ],[ <= (lambda M+(1-lambda)M)|x-y|=M*|x-y|]:}\begin{aligned} |f(x)-f(y)| & \leq \lambda\left|f_{1}(x)-f_{1}(y)\right|+(1-\lambda) \cdot\left|f_{2}(x)-f_{2}(y)\right| \leq \\ & \leq\left(\lambda K_{f_{1}}+(1-\lambda) K_{f_{2}}\right) \cdot|x-y| \leq \\ & \leq(\lambda M+(1-\lambda) M)|x-y|=M \cdot|x-y| \end{aligned}|f(x)−f(y)|≤λ|f1(x)−f1(y)|+(1−λ)⋅|f2(x)−f2(y)|≤≤(λKf1+(1−λ)Kf2)⋅|x−y|≤≤(λM+(1−λ)M)|x−y|=M⋅|x−y|
for all x , y ∈ [ a , b ] x , y ∈ [ a , b ] x,y in[a,b]x, y \in[a, b]x,y∈[a,b], impling K f ≤ M K f ≤ M K_(f) <= MK_{f} \leq MKf≤M. It follows that f ∈ S ( D , V , M ) f ∈ S ( D , V , M ) f inS(D,V,M)f \in \mathscr{S}(\mathrm{D}, \mathrm{V}, \mathrm{M})f∈S(D,V,M).
b) By a theorem of McShane [2], the functions F i F i F_(i)F_{i}Fi and F s F s F_(s)F_{s}Fs defined by (4) are in S ( D , V , M ) S ( D , V , M ) S(D,V,M)\mathscr{S}(D, V, M)S(D,V,M) and furthermore
(5)
F i ( x ) ≤ f ( x ) ≤ F s ( x ) , x ∈ [ a , b ] F i ( x ) ≤ f ( x ) ≤ F s ( x ) , x ∈ [ a , b ] F_(i)(x) <= f(x) <= F_(s)(x)quad,x in[a,b]F_{i}(x) \leq f(x) \leq F_{s}(x) \quad, x \in[a, b]Fi(x)≤f(x)≤Fs(x),x∈[a,b]
for all f ∈ F ( D , V , M ) f ∈ F ( D , V , M ) f inF(D,V,M)f \in \mathscr{F}(D, V, M)f∈F(D,V,M). To prove the second inequality in (5), suppose, on the contrary, that there exists a function f ∈ P ( D , V , M ) f ∈ P ( D , V , M ) f inP(D,V,M)f \in \mathscr{P}(D, V, M)f∈P(D,V,M) on a point c ∈ [ a , b ] c ∈ [ a , b ] c in[a,b]c \in[a, b]c∈[a,b] such that f ( c ) > F s ( c ) f ( c ) > F s ( c ) f(c) > F_(s)(c)f(c)>F_{s}(c)f(c)>Fs(c). As F i ( x k ) = f ( x k ) = F s ( x k ) = y k , k = 0 , 1 , … , n F i x k = f x k = F s x k = y k , k = 0 , 1 , … , n F_(i)(x_(k))=f(x_(k))=F_(s)(x_(k))=y_(k),k=0,1,dots,nF_{i}\left(x_{k}\right)=f\left(x_{k}\right)=F_{s}\left(x_{k}\right)=y_{k}, k=0,1, \ldots, nFi(xk)=f(xk)=Fs(xk)=yk,k=0,1,…,n, it follows that
there exists k 0 ∈ { 0 , 1 , … , n } k 0 ∈ { 0 , 1 , … , n } k_(0)in{0,1,dots,n}k_{0} \in\{0,1, \ldots, n\}k0∈{0,1,…,n} such that c ∈ ( x k 0 , x k 0 + 1 ) c ∈ x k 0 , x k 0 + 1 c in(x_(k_(0)),x_(k_(0)+1))c \in\left(x_{k_{0}}, x_{k_{0}+1}\right)c∈(xk0,xk0+1). But then
f ( c ) − f ( x k 0 ) c − x k 0 > F s ( c ) − F s ( x k 0 ) c − x k 0 = M or f ( x k 0 + 1 ) − f ( c ) x k 0 + 1 − c < F s ( x k 0 + 1 ) − F s ( c ) x k 0 + 1 − c = − M , f ( c ) − f x k 0 c − x k 0 > F s ( c ) − F s x k 0 c − x k 0 = M  or  f x k 0 + 1 − f ( c ) x k 0 + 1 − c < F s x k 0 + 1 − F s ( c ) x k 0 + 1 − c = − M , {:[(f(c)-f(x_(k_(0))))/(c-x_(k_(0))) > (F_(s)(c)-F_(s)(x_(k_(0))))/(c-x_(k_(0)))=M],[" or "quad(f(x_(k_(0)+1))-f(c))/(x_(k_(0)+1)-c) < (F_(s)(x_(k_(0)+1))-F_(s)(c))/(x_(k_(0)+1)-c)=-M","]:}\begin{aligned} & \frac{f(c)-f\left(x_{k_{0}}\right)}{c-x_{k_{0}}}>\frac{F_{s}(c)-F_{s}\left(x_{k_{0}}\right)}{c-x_{k_{0}}}=M \\ \text { or } \quad & \frac{f\left(x_{k_{0}+1}\right)-f(c)}{x_{k_{0}+1}-c}<\frac{F_{s}\left(x_{k_{0}+1}\right)-F_{s}(c)}{x_{k_{0}+1}-c}=-M, \end{aligned}f(c)−f(xk0)c−xk0>Fs(c)−Fs(xk0)c−xk0=M or f(xk0+1)−f(c)xk0+1−c<Fs(xk0+1)−Fs(c)xk0+1−c=−M,
according as c c ccc belongs to the interval
( x k , x k o + 1 + x k o 2 + y k o + 1 − y k o 2 M ) x k , x k o + 1 + x k o 2 + y k o + 1 − y k o 2 M (x_(k),(x_(ko+1)+x_(ko))/(2)+(y_(ko+1)-y_(ko))/(2M))quad\left(x_{k}, \frac{x_{k o+1}+x_{k o}}{2}+\frac{y_{k o+1}-y_{k o}}{2 M}\right) \quad(xk,xko+1+xko2+yko+1−yko2M) or
( x k 0 + 1 + x k 0 2 + y k 0 + 1 − y k 0 2 M , x k 0 + 1 ) x k 0 + 1 + x k 0 2 + y k 0 + 1 − y k 0 2 M , x k 0 + 1 ((x_(k_(0)+1)+x_(k_(0)))/(2)+(y_(k_(0)+1)-y_(k_(0)))/(2M),x_(k_(0)+1))quad\left(\frac{x_{k_{0}+1}+x_{k_{0}}}{2}+\frac{y_{k_{0}+1}-y_{k_{0}}}{2 M}, x_{k_{0}+1}\right) \quad(xk0+1+xk02+yk0+1−yk02M,xk0+1),
respectively. In both of the cases it follows K f > M K f > M K_(f) > MK_{f}>MKf>M, contradicting the hypothesis f ∈ P ( D , V , M ) f ∈ P ( D , V , M ) f inP(D,V,M)f \in \mathscr{P}(D, V, M)f∈P(D,V,M). The first, inequality in (5), F i ( x ) ≤ f ( x ) F i ( x ) ≤ f ( x ) F_(i)(x) <= f(x)F_{i}(x) \leq f(x)Fi(x)≤f(x), for all x ∈ [ a , b ] x ∈ [ a , b ] x in[a,b]x \in[a, b]x∈[a,b], can be proved similary.
To prove that F g F g F_(g)F_{\mathrm{g}}Fg is an extreme point of the convex set I ( D , V , H ) I ( D , V , H ) I(D,V,H)\mathscr{I}(D, V, H)I(D,V,H) suppose that F s = λ f 1 + ( 1 − λ ) f 2 F s = λ f 1 + ( 1 − λ ) f 2 F_(s)=lambdaf_(1)+(1-lambda)f_(2)F_{s}=\lambda f_{1}+(1-\lambda) f_{2}Fs=λf1+(1−λ)f2 for two functions f 1 f 1 f_(1)f_{1}f1, f 2 ∈ S ( D , V , M ) f 2 ∈ S ( D , V , M ) f_(2)inS(D,V,M)f_{2} \in \mathcal{S}(D, V, M)f2∈S(D,V,M) and a number λ ∈ ( 0 , 1 ) λ ∈ ( 0 , 1 ) lambda in(0,1)\lambda \in(0,1)λ∈(0,1). We have to show that f 1 = f 2 = F s f 1 = f 2 = F s f_(1)=f_(2)=F_(s)f_{1}= f_{2}=F_{s}f1=f2=Fs, but this follows immediately from the inequalities (5).
c) By the Arzela - Ascoli theorem (see e.g. [5]) it is sufficient to show that ℜ ( D , V , H ) ℜ ( D , V , H ) ℜ(D,V,H)\Re(D, V, H)ℜ(D,V,H) is a closed, uniformly bounded and equicontinuous subset of c [ a , b ] c [ a , b ] c[a,b]c[a, b]c[a,b]. By the definition of S ( D , V , M ) S ( D , V , M ) S(D,V,M)\mathscr{S}(D, V, M)S(D,V,M) it is obvious that if ( f n ) f n (f_(n))\left(f_{n}\right)(fn) is a sequence in S ( D , V , M ) S ( D , V , M ) S(D,V,M)S(D, V, M)S(D,V,M) converging to f ∈ C [ a , b ] f ∈ C [ a , b ] f in C[a,b]f \in C[a, b]f∈C[a,b] then f f fff is in M ( D , V , M ) M ( D , V , M ) M(D,V,M)\mathscr{M}(D, V, M)M(D,V,M) too, and by (5)
‖ f ‖ ≤ max { ‖ F i ‖ , ‖ F s ‖ } ‖ f ‖ ≤ max F i , F s ||f|| <= max{||F_(i)||,||F_(s)||}\|f\| \leq \max \left\{\left\|F_{i}\right\|,\left\|F_{s}\right\|\right\}‖f‖≤max{‖Fi‖,‖Fs‖}
showing that T ( D , V , M ) T ( D , V , M ) T(D,V,M)\mathscr{T}(D, V, M)T(D,V,M) is a closed and uniformly bounded subset of c [ a , b ] c [ a , b ] c[a,b]c[a, b]c[a,b].
Now, for ε > 0 let δ = ε / ( M + 1 ) . Then | f ( x ) − f ( y ) | ≤ M | x − y | < M ε M + 1 < ε  Now, for  ε > 0  let  δ = ε / ( M + 1 ) . Then  | f ( x ) − f ( y ) | ≤ M | x − y | < M ε M + 1 < ε {:[" Now, for "epsi > 0" let "delta=epsi//(M+1)". Then "],[|f(x)-f(y)| <= M|x-y| < M(epsi)/(M+1) < epsi]:}\begin{aligned} & \text { Now, for } \varepsilon>0 \text { let } \delta=\varepsilon /(M+1) \text {. Then } \\ & |f(x)-f(y)| \leq M|x-y|<M \frac{\varepsilon}{M+1}<\varepsilon \end{aligned} Now, for ε>0 let δ=ε/(M+1). Then |f(x)−f(y)|≤M|x−y|<MεM+1<ε
for all x , y ∈ [ a , b ] x , y ∈ [ a , b ] x,y in[a,b]x, y \in[a, b]x,y∈[a,b] with | x − y | < δ | x − y | < δ |x-y| < delta|x-y|<\delta|x−y|<δ and all f f fff in S ( D , V , M ) S ( D , V , M ) S(D,V,M)\mathscr{S}(D, V, M)S(D,V,M) proving the equicontinuity of the set S ( D , V , M ) S ( D , V , M ) S(D,V,M)\mathscr{S}(D, V, M)S(D,V,M) and, by the above quated result of Arzelà - Ascoli, also its compactness.
Exemple. In the paper [3] there are given several solutions to the following problem: let f : [ 0 , 2 ] → R f : [ 0 , 2 ] → R f:[0,2]rarrRf:[0,2] \rightarrow \mathbf{R}f:[0,2]→R be a continuous function derivable on ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) and such that | f ′ ( x ) | ≤ 1 f ′ ( x ) ≤ 1 |f^(')(x)| <= 1\left|f^{\prime}(x)\right| \leq 1|f′(x)|≤1, for all x ∈ ( 0 , 2 ) x ∈ ( 0 , 2 ) x in(0,2)x \in(0,2)x∈(0,2) and f ( 0 ) = f ( 2 ) = 1 f ( 0 ) = f ( 2 ) = 1 f(0)=f(2)=1f(0)=f(2)=1f(0)=f(2)=1. Show that 1 < ∫ 0 2 f ( x ) d x < 3 1 < ∫ 0 2   f ( x ) d x < 3 1 < int_(0)^(2)f(x)dx < 31<\int_{0}^{2} f(x) d x<31<∫02f(x)dx<3.
The hypotesis of the problem show that f f fff belongs to a class of the type S ( D , V , M ) S ( D , V , M ) S(D,V,M)\mathscr{S}(D, V, M)S(D,V,M), namely for D = { 0 , 2 } , V = { 1 , 1 } D = { 0 , 2 } , V = { 1 , 1 } D={0,2},V={1,1}D=\{0,2\}, V=\{1,1\}D={0,2},V={1,1} and M = 1 M = 1 M=1M=1M=1.
In this case the exremal functions F i F i F_(i)F_{i}Fi and F B F B F_(B)F_{B}FB are not derivable in the point x = 1 x = 1 x=1x=1x=1, explaining why the inequalities in the conclusion of the problem are strict.
Consider now for p ∈ N p ∈ N p in Np \in Np∈N the functional I p : P ( D , V , M ) → R I p : P ( D , V , M ) → R I_(p):P(D,V,M)rarrRI_{p}: \mathscr{P}(D, V, M) \rightarrow \mathbb{R}Ip:P(D,V,M)→R, defined by:
(6) I p ( f ) = ∫ a b | f ( x ) | p d x (6) I p ( f ) = ∫ a b   | f ( x ) | p d x {:(6)I_(p)(f)=int_(a)^(b)|f(x)|^(p)dx:}\begin{equation*} I_{p}(f)=\int_{a}^{b}|f(x)|^{p} d x \tag{6} \end{equation*}(6)Ip(f)=∫ab|f(x)|pdx
One asks to find the minimal and the maximal values of this functional. The solution of this problem is given by:
THEOREM 2. a) If the numbers α k = y k + 1 + y k 2 − M 2 ( x k + 1 − x k ) α k = y k + 1 + y k 2 − M 2 x k + 1 − x k alpha_(k)=(y_(k+1)+y_(k))/(2)-(M)/(2)(x_(k+1)-x_(k))\alpha_{k}=\frac{y_{k+1}+y_{k}}{2}-\frac{M}{2}\left(x_{k+1}-x_{k}\right)αk=yk+1+yk2−M2(xk+1−xk) are non-negative for all k = 0 , 1 , … , n − 1 k = 0 , 1 , … , n − 1 k=0,1,dots,n-1k=0,1, \ldots, n-1k=0,1,…,n−1 then
(7) max I p ( f ) = ∫ a b ( F s ( x ) ) p d x , and min I p ( f ) = ∫ a b ( F i ( x ) ) p d x ; (7) max I p ( f ) = ∫ a b   F s ( x ) p d x ,  and  min I p ( f ) = ∫ a b   F i ( x ) p d x ; {:[(7) maxI_(p)(f)=int_(a)^(b)(F_(s)(x))^(p)dx","quad" and "],[ minI_(p)(f)=int_(a)^(b)(F_(i)(x))^(p)dx;]:}\begin{align*} & \max I_{p}(f)=\int_{a}^{b}\left(F_{s}(x)\right)^{p} d x, \quad \text { and } \tag{7}\\ & \min I_{p}(f)=\int_{a}^{b}\left(F_{i}(x)\right)^{p} d x ; \end{align*}(7)maxIp(f)=∫ab(Fs(x))pdx, and minIp(f)=∫ab(Fi(x))pdx;
b) If the numbers B k = Y k + 1 + Y k 2 + M 2 ( x k + 1 − x k ) B k = Y k + 1 + Y k 2 + M 2 x k + 1 − x k B_(k)=(Y_(k+1)+Y_(k))/(2)+(M)/(2)(x_(k+1)-x_(k))B_{k}=\frac{Y_{k+1}+Y_{k}}{2}+\frac{M}{2}\left(x_{k+1}-x_{k}\right)Bk=Yk+1+Yk2+M2(xk+1−xk) are non - positive for all k = 0 , 1 , … , n − 1 k = 0 , 1 , … , n − 1 k=0,1,dots,n-1k=0,1, \ldots, n-1k=0,1,…,n−1, then
(8)
max I p ( f ) = ∫ a b | F i ( x ) | p d x , max I p ( f ) = ∫ a b   F i ( x ) p d x , maxI_(p)(f)=int_(a)^(b)|F_(i)(x)|^(p)dx,quad\max I_{p}(f)=\int_{a}^{b}\left|F_{i}(x)\right|^{p} d x, \quadmaxIp(f)=∫ab|Fi(x)|pdx, and min I p ( f ) = ∫ a b | F s ( x ) | p d x ; min I p ( f ) = ∫ a b   F s ( x ) p d x ; minI_(p)(f)=int_(a)^(b)|F_(s)(x)|^(p)dx;\min I_{p}(f)=\int_{a}^{b}\left|F_{s}(x)\right|^{p} d x ;minIp(f)=∫ab|Fs(x)|pdx;
c) If the numbers y k y k y_(k)y_{k}yk are non - negative for all k = k = k=\mathrm{k}=k=
(9)
max x p ( ε ) = ∫ a b ( F s ( x ) ) p d x max x p ( ε ) = ∫ a b   F s ( x ) p d x maxx_(p)(epsi)=int_(a)^(b)(F_(s)(x))^(p)dx\max x_{p}(\varepsilon)=\int_{a}^{b}\left(F_{s}(x)\right)^{p} d xmaxxp(ε)=∫ab(Fs(x))pdx,
min I p ( f ) = ∫ a b ( max { F 1 ( x ) , 0 } ) p d x ; min I p ( f ) = ∫ a b   max F 1 ( x ) , 0 p d x ; minI_(p)(f)=int_(a)^(b)(max{F_(1)(x),0})^(p)dx;\min I_{p}(f)=\int_{a}^{b}\left(\max \left\{F_{1}(x), 0\right\}\right)^{p} d x ;minIp(f)=∫ab(max{F1(x),0})pdx;
d) If the numbers y k y k y_(k)y_{k}yk are non - positive for all k = k = k=k=k=
= 0 , 1 , … , n = 0 , 1 , … , n =0,1,dots,n=0,1, \ldots, n=0,1,…,n then
(10)
max I p ( f ) = ∫ a b | F i ( x ) | p d x , and min I p ( f ) = ∫ a b ( max { | F s ( x ) | , 0 } ) p d x . max I p ( f ) = ∫ a b   F i ( x ) p d x ,  and  min I p ( f ) = ∫ a b   max F s ( x ) , 0 p d x . {:[maxI_(p)(f)=int_(a)^(b)|F_(i)(x)|^(p)dx","" and "],[minI_(p)(f)=int_(a)^(b)(max{|F_(s)(x)|,0})^(p)dx.]:}\begin{aligned} \max I_{p}(f) & =\int_{a}^{b}\left|F_{i}(x)\right|^{p} d x, \text { and } \\ \min I_{p}(f) & =\int_{a}^{b}\left(\max \left\{\left|F_{s}(x)\right|, 0\right\}\right)^{p} d x . \end{aligned}maxIp(f)=∫ab|Fi(x)|pdx, and minIp(f)=∫ab(max{|Fs(x)|,0})pdx.
Proof. a) The numbers α k , k = 0 , 1 , … , n − 1 α k , k = 0 , 1 , … , n − 1 alpha_(k),k=0,1,dots,n-1\alpha_{k}, k=0,1, \ldots, n-1αk,k=0,1,…,n−1 are the relative minima of the function F i F i F_(i)F_{i}Fi on the interval [ a , b ] [ a , b ] [a,b][a, b][a,b]. If α k ≥ 0 α k ≥ 0 alpha_(k) >= 0\alpha_{k} \geq 0αk≥0 for k = 0 , 1 , … , n − 1 k = 0 , 1 , … , n − 1 k=0,1,dots,n-1k=0,1, \ldots, n-1k=0,1,…,n−1, then
0 ≤ F i ( x ) ≤ f ( x ) ≤ F B ( x ) , x ∈ [ a , b ] , 0 ≤ F i ( x ) ≤ f ( x ) ≤ F B ( x ) , x ∈ [ a , b ] , 0 <= F_(i)(x) <= f(x) <= F_(B)(x),x in[a,b],0 \leq F_{i}(x) \leq f(x) \leq F_{B}(x), x \in[a, b],0≤Fi(x)≤f(x)≤FB(x),x∈[a,b],
implying the inequalities
0 ≤ ( F i ( x ) ) P ≤ ( f ( x ) ) P ≤ ( F B ( x ) ) P , x ∈ [ a , b ] , 0 ≤ F i ( x ) P ≤ ( f ( x ) ) P ≤ F B ( x ) P , x ∈ [ a , b ] , 0 <= (F_(i)(x))^(P) <= (f(x))^(P) <= (F_(B)(x))^(P),x in[a,b],0 \leq\left(F_{i}(x)\right)^{P} \leq(f(x))^{P} \leq\left(F_{B}(x)\right)^{P}, x \in[a, b],0≤(Fi(x))P≤(f(x))P≤(FB(x))P,x∈[a,b],
which by integration over [ a , b ] [ a , b ] [a,b][a, b][a,b] yield á).
b) The numbers β k β k beta_(k)\beta_{k}βk are the relative maxima of the function F s F s F_(s)F_{s}Fs on [ a , b ] [ a , b ] [a,b][a, b][a,b]. If β k ≤ 0 β k ≤ 0 beta_(k) <= 0\beta_{k} \leq 0βk≤0 for all k = 0 , 1 , … , n − 1 k = 0 , 1 , … , n − 1 k=0,1,dots,n-1k=0,1, \ldots, n-1k=0,1,…,n−1, then F i ( x ) ≤≤ f ( x ) ≤ F s ( x ) ≤ 0 , x ∈ [ a , b ] F i ( x ) ≤≤ f ( x ) ≤ F s ( x ) ≤ 0 , x ∈ [ a , b ] F_(i)(x)≤≤f(x) <= F_(s)(x) <= 0,x in[a,b]F_{i}(x) \leq \leq f(x) \leq F_{s}(x) \leq 0, x \in[a, b]Fi(x)≤≤f(x)≤Fs(x)≤0,x∈[a,b], implying:
− F i ( x ) ≥ − f ( x ) ≥ − F g ( x ) ≥ 0 , x ∈ [ a , b ] . − F i ( x ) ≥ − f ( x ) ≥ − F g ( x ) ≥ 0 , x ∈ [ a , b ] . -F_(i)(x) >= -f(x) >= -F_(g)(x) >= 0,x in[a,b].-F_{i}(x) \geq-f(x) \geq-F_{g}(x) \geq 0, x \in[a, b] .−Fi(x)≥−f(x)≥−Fg(x)≥0,x∈[a,b].
Rising to the power p p ppp and integrating over [ a , b ] [ a , b ] [a,b][a, b][a,b] one obtains b).
c) If the numbers y k , k = 0 , 1 , … , n y k , k = 0 , 1 , … , n y_(k),k=0,1,dots,ny_{k}, k=0,1, \ldots, nyk,k=0,1,…,n are all non - negative
then the inequalities
max { F i ( x ) , 0 } ≤ | f ( x ) | ≤ F s ( x ) , x ∈ [ a , b ] , max F i ( x ) , 0 ≤ | f ( x ) | ≤ F s ( x ) , x ∈ [ a , b ] , max{F_(i)(x),0} <= |f(x)| <= F_(s)(x),x in[a,b],\max \left\{F_{i}(x), 0\right\} \leq|f(x)| \leq F_{s}(x), x \in[a, b],max{Fi(x),0}≤|f(x)|≤Fs(x),x∈[a,b],
hold, which rised to the power p p ppp and integrated over [ a , b ] [ a , b ] [a,b][a, b][a,b] give (9).
d) The proof is similar to that in the case c).
Remark 1. The set P ( D , V , M ) P ( D , V , M ) P(D,V,M)\mathscr{P}(\mathrm{D}, \mathrm{V}, \mathrm{M})P(D,V,M) being compact every continuous functional defined on S ( D , V , M ) S ( D , V , M ) S(D,V,M)\mathscr{S}(D, V, M)S(D,V,M) attains its extrema.
Remark 2. All the integrals appearing in the calculation of the extrema of the functional I p I p I_(p)I_{p}Ip (the formulae (7) - (10) ) can be easily calculated, taking into account the fact that the functions F i F i F_(i)F_{i}Fi and F s F s F_(s)F_{s}Fs are segmentary linear functions and have very simple expression. For instance, in the case a) :
max I p ( f ) = ∑ k = 0 n − 1 ∫ X k X k + 1 ( F s ( x ) ) p d x = ∑ k = 0 n − 1 ∬ X k X M k ( F s ( x ) ) p d x + max I p ( f ) = ∑ k = 0 n − 1   ∫ X k X k + 1   F s ( x ) p d x = ∑ k = 0 n − 1   ∬ X k X M k   F s ( x ) p d x + maxI_(p)(f)=sum_(k=0)^(n-1)int_(X_(k))^(X_(k+1))(F_(s)(x))^(p)dx=sum_(k=0)^(n-1)∬_(X_(k))^(X_(M_(k)))(F_(s)(x))^(p)dx+\max I_{p}(f)=\sum_{k=0}^{n-1} \int_{X_{k}}^{X_{k+1}}\left(F_{s}(x)\right)^{p} d x=\sum_{k=0}^{n-1} \iint_{X_{k}}^{X_{M_{k}}}\left(F_{s}(x)\right)^{p} d x+maxIp(f)=∑k=0n−1∫XkXk+1(Fs(x))pdx=∑k=0n−1∬XkXMk(Fs(x))pdx+
+ ∫ X M k X k + 1 ( F 5 ( x ) ) P d x ∣= ∑ k = 0 n − 1 ∫ X k X M k [ M ( x − x k ) + y k ] P ⋅ d x + + ∫ X M k X k + 1   F 5 ( x ) P d x ∣= ∑ k = 0 n − 1   ∫ X k X M k   M x − x k + y k P ⋅ d x + +int_(X_(M_(k)))^(X_(k+1))(F_(5)(x))^(P)dx∣=sum_(k=0)^(n-1)int_(X_(k))^(X_(M_(k)))[M(x-x_(k))+y_(k)]^(P)*dx++\int_{X_{M_{k}}}^{X_{k+1}}\left(F_{5}(x)\right)^{P} d x \mid=\sum_{k=0}^{n-1} \int_{X_{k}}^{X_{M_{k}}}\left[M\left(x-x_{k}\right)+y_{k}\right]^{P} \cdot d x++∫XMkXk+1(F5(x))Pdx∣=∑k=0n−1∫XkXMk[M(x−xk)+yk]P⋅dx+
+ ∑ k = 0 n − 1 ∫ x M k x k + 1 [ − M ( x − x k + 1 ) + y k + 1 ] p d x + ∑ k = 0 n − 1   ∫ x M k x k + 1   − M x − x k + 1 + y k + 1 p d x +sum_(k=0)^(n-1)int_(x_(M_(k)))^(x_(k+1))[-M(x-x_(k+1))+y_(k+1)]^(p)dx+\sum_{k=0}^{n-1} \int_{x_{M_{k}}}^{x_{k+1}}\left[-M\left(x-x_{k+1}\right)+y_{k+1}\right]^{p} d x+∑k=0n−1∫xMkxk+1[−M(x−xk+1)+yk+1]pdx,
where x M k = x k + x k + 1 2 + y k + 1 − y k 2 M x M k = x k + x k + 1 2 + y k + 1 − y k 2 M x_(Mk)=(x_(k)+x_(k+1))/(2)+(y_(k+1)-y_(k))/(2M)x_{M k}=\frac{x_{k}+x_{k+1}}{2}+\frac{y_{k+1}-y_{k}}{2 M}xMk=xk+xk+12+yk+1−yk2M is the point of relative
maximum of the function F k F k F_(k)F_{k}Fk on the interval [ x k , x k + 1 ] , k == 0 , 1 , … , n − 1 x k , x k + 1 , k == 0 , 1 , … , n − 1 [x_(k),x_(k+1)],k==0,1,dots,n-1\left[x_{k}, x_{k+1}\right], k= =0,1, \ldots, n-1[xk,xk+1],k==0,1,…,n−1.
In the case D = { 0 , 2 } , V = { 1 , 1 } D = { 0 , 2 } , V = { 1 , 1 } D={0,2},V={1,1}D=\{0,2\}, V=\{1,1\}D={0,2},V={1,1} and M = 1 , p ∈ N M = 1 , p ∈ N M=1,p in NM=1, p \in NM=1,p∈N one obtains the following result:
The inequalities
m p ≤ I p ( f ) ≤ M p , m p ≤ I p ( f ) ≤ M p , m_(p) <= I_(p)(f) <= M_(p),m_{p} \leq I_{p}(f) \leq M_{p},mp≤Ip(f)≤Mp,
hold for every f ∈ S ( D , V , M ) f ∈ S ( D , V , M ) f inS(D,V,M)f \in \mathcal{S}(D, V, M)f∈S(D,V,M), where
m p = ∫ 0 2 | F i ( x ) | p d x = ∫ 0 1 ( − x + 1 ) p d x + ∫ 1 2 ( x − 1 ) p d x = = 2 / ( p + 1 ) M p = ∫ 0 2 | F s ( x ) | p d x = ∫ 0 1 ( x + 1 ) p d x + ∫ 1 2 ( − x + 3 ) p d x = = 2 ( 2 p + 1 − 1 ) / ( p + 1 ) m p = ∫ 0 2   F i ( x ) p d x = ∫ 0 1   ( − x + 1 ) p d x + ∫ 1 2   ( x − 1 ) p d x = = 2 / ( p + 1 ) M p = ∫ 0 2   F s ( x ) p d x = ∫ 0 1   ( x + 1 ) p d x + ∫ 1 2   ( − x + 3 ) p d x = = 2 2 p + 1 − 1 / ( p + 1 ) {:[m_(p)=int_(0)^(2)|F_(i)(x)|^(p)dx=int_(0)^(1)(-x+1)^(p)dx+int_(1)^(2)(x-1)^(p)dx=],[=2//(p+1)],[M_(p)=int_(0)^(2)|F_(s)(x)|^(p)dx=int_(0)^(1)(x+1)^(p)dx+int_(1)^(2)(-x+3)^(p)dx=],[=2(2^(p+1)-1)//(p+1)]:}\begin{aligned} m_{p} & =\int_{0}^{2}\left|F_{i}(x)\right|^{p} d x=\int_{0}^{1}(-x+1)^{p} d x+\int_{1}^{2}(x-1)^{p} d x= \\ & =2 /(p+1) \\ M_{p} & =\int_{0}^{2}\left|F_{s}(x)\right|^{p} d x=\int_{0}^{1}(x+1)^{p} d x+\int_{1}^{2}(-x+3)^{p} d x= \\ & =2\left(2^{p+1}-1\right) /(p+1) \end{aligned}mp=∫02|Fi(x)|pdx=∫01(−x+1)pdx+∫12(x−1)pdx==2/(p+1)Mp=∫02|Fs(x)|pdx=∫01(x+1)pdx+∫12(−x+3)pdx==2(2p+1−1)/(p+1)
For p = 1 p = 1 p=1p=1p=1 we find
1 ≤ ∫ 0 2 f ( x ) d x ≤ 3 1 ≤ ∫ 0 2   f ( x ) d x ≤ 3 1 <= int_(0)^(2)f(x)dx <= 31 \leq \int_{0}^{2} f(x) d x \leq 31≤∫02f(x)dx≤3
for all f ∈ G ( D , V , M ) f ∈ G ( D , V , M ) f inG(D,V,M)f \in \mathscr{G}(D, V, M)f∈G(D,V,M), i.e a non - sharp version of the inequality proved in [3].
Considering the L p L p L_(p)L_{p}Lp - norm of a function f ∈ I ( D , V , M ) f ∈ I ( D , V , M ) f inI(D,V,M)f \in \mathscr{I}(D, V, M)f∈I(D,V,M) one gets
2 p + 1 p ≤ ‖ f ‖ ≤ 2 2 p + 1 − 1 p + 1 p 2 p + 1 p ≤ ‖ f ‖ ≤ 2 2 p + 1 − 1 p + 1 p root(p)((2)/(p+1)) <= ||f|| <= root(p)(2(2^(p+1)-1)/(p+1))\sqrt[p]{\frac{2}{p+1}} \leq\|f\| \leq \sqrt[p]{2 \frac{2^{p+1}-1}{p+1}}2p+1p≤‖f‖≤22p+1−1p+1p
which for p → ∞ p → ∞ p rarr oop \rightarrow \inftyp→∞ yields the uniform bounds of the set P ( D , V , M ) P ( D , V , M ) P(D,V,M)\mathscr{P}(D, V, M)P(D,V,M) : 1 ≤ ‖ f ‖ ≤ 2 1 ≤ ‖ f ‖ ≤ 2 1 <= ||f|| <= 21 \leq\|f\| \leq 21≤‖f‖≤2 for every f ∈ S ( D , V , M ) f ∈ S ( D , V , M ) f inS(D,V,M)f \in \mathscr{S}(D, V, M)f∈S(D,V,M).

REFEREMCES

  1. ARONSSON, G., Extension of functions satisfying Lipschitz conditions, Arkiv för Mathematik 6 (1967) Nr.28, 551 - 561.
  2. MC SHANE, E.J., Extension of range of functions, Bull. Amer. Math. Soc. 40 (1934), 837-842.
  3. MOCraNU, P., Variaţiuni pe o temă de concurs, Seminarul "Didactica Matematicii", 1985-1986, 123-128.
  4. MUSTĂTA, C., On the extension problem with prescribed norm, Seminar of Functional Analysis and Numerical Methods, Preprint Nr. 4 (1981), 93-99..
  5. TRENOGUINE, V., Analyse fonctionnelle, Edition Mir, Moscow 1985.
Institutul de Calcul Oficiul Poştal 1 C.P. 68 3400 Cluj - Napoca Romania  Institutul de Calcul   Oficiul Poştal  1  C.P.  68 3400  Cluj - Napoca   Romania  {:[" Institutul de Calcul "],[" Oficiul Poştal "1],[" C.P. "68],[3400" Cluj - Napoca "],[" Romania "]:}\begin{aligned} & \text { Institutul de Calcul } \\ & \text { Oficiul Poştal } 1 \\ & \text { C.P. } 68 \\ & 3400 \text { Cluj - Napoca } \\ & \text { Romania } \end{aligned} Institutul de Calcul  Oficiul Poştal 1 C.P. 683400 Cluj - Napoca  Romania 
1991

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